Content for Capacitors and Inductors- Circuits

Circuits

Knowledge of the variation of current and potential difference with time in a CR circuit during charging and discharging.

Definition of the time constant for a CR circuit.

Numerical and graphical determination of the time constant for a CR circuit.

Definition of capacitive reactance.

Use of appropriate relationships to solve problems involving capacitive reactance, voltage, current, frequency and capacitance.

Electromagnetic Induction

Inductors in d.c. circuits

Self-inductance (inductance) of a coil

Lenz’s law

Definition of inductance and of back e.m.f.

Energy stored by an inductor

Inductors in a.c. circuits

Inductive reactance.

Use of an appropriate relationship to solve problems involving back e.m.f., inductance (self-inductance) and rate of change of current.

Use of appropriate relationships to solve problems relating to inductive reactance, voltage, current, frequency, energy and inductance (self-inductance).

Investigate the current and potential difference in CR circuits during charging and discharging — possible use of datalogging to determine the time constant for a CR circuit.

Investigate applications of capacitors in d.c. circuits.

Experiments to investigate the relationship between current, frequency and capacitive reactance.

Investigate the factors affecting the size of the induced e.m.f. in a coil.

Demonstration — neon bulb lit from 1·5 V cell.

Investigate the growth and decay of current in a d.c. circuit containing an inductor.

Determine the inductance (self inductance) of a coil by use of datalogging or waveform capture

Experiments to investigate the relationship between current, frequency and inductive reactance.

Applications: induction cookers, electromagnetic braking, LC filters, tuned circuits, etc.

t = RC and an understanding of the definition of time constant as the time to increase the charge stored by 63% of the difference between initial charge and full charge, or the time taken to discharge the capacitor to 37% of initial charge).

Capacitors- Higher Review

Background to Capacitors

What is a capacitor?

A capacitor is a useful device for storing charge, and hence energy. They have numerous uses and so obviously lots of potential for exam questions; many of which are quite tricky, so it is quite important that you have a sound understanding of how they work. They can also be used for an integration question where capacitors are linked with another part of the course.

Capacitor circuit symbol

Capacitance is measured in FARADS.

1 Farad = 1 Coulomb per volt. A capacitor of 10 pF can store 10 pF of charge at a voltage of 1 Volt.

N.B. 1 F is a very large unit. C would normally be expressed as F or nF.

Capacitance

Units of capacitance are therefore C V-1 or Farads (F)

Here is some information on capacitors that is for interest and help in understanding.

Capacitors are analogous with a beaker:

The size of beaker corresponds to the maximum capacitance.

Amount of liquid in the beaker corresponds to the charge on the capacitor.

Show and describe

metal plate insulator another metal plate in front of this to compete a “sandwich” Most capacitors can be connected either way around, but ELECTROLYTIC capacitors must be connected correctly according to the + and – labels.

If the rated p.d. across a capacitor is exceeded it will break down (puncturing the insulation).

As an aside:

Capacitors in parallel:	Capacitors in series:

4)	Factors affecting capacitance:

area of plates;

distance between plates;

material between plates.

So how do Capacitors work?

The capacitor stores electric charge - note that this is entirely different to current & voltage. Once charged, as in the circuit below - the capacitor will store the charge. This charge will not go anywhere (actually, it will slowly leak away - but at an insignificant rate) & can be used to power a circuit for a very short while - lighting a small torch light bulb for 5 seconds is an impressive feat for a capacitor. They are usually used in a delay circuit. Most capacitors are polarity sensitive - ie. They won't work properly if you insert them the wrong way round.

Note that these little dudes will quite happily ‘blow up’ /break if you put too much current through it or voltage across it, or if you connect it the wrong way round (although that only applies to the electrolytic capacitor). They are quite difficult to break/ ‘blow up’ without trying, as long as you take care when designing your circuits.

No one bothered to explain to me when I was at school how a capacitor worked. It is easy to confuse them with the work that was completed on electric fields and moving charge between two plates. THIS IS NOT THE SAME. With a capacitor charge is NOT moved from one plate to the other plate.

- - - - - - + + + + + + Electrons are transferred from the cell to this plate Electrons are repelled from this plate by the electrons on the other plate. The first thing to realise is that the electrons do not pass between the capacitor plates. Electrons travel from the power supply to one of the capacitor plates. Because the plates are in such close proximity the electrons on that plate repel electrons from the other plate which pass around the circuit and back to the power supply. The first electron is very easy to put on the plate and the capacitor offers no resistance. As the charge builds up on one plate it becomes more difficult to add charge and so the reactance (resistance) of the capacitor increases. Eventually the voltage on the capacitor equals the supply voltage. There is then no potential difference between the plates and the source so no more charge can be added. As the reactance (resistance) of the capacitor increases the current in the circuit decreases. In this circuit we have also placed a resistor as this is used to control the maximum current passing through the capacitor. Too much current can destroy a capacitor.

Charging the capacitor requires a potential difference to be placed across it. Work is done transferring charge (Q) onto the plates. The rate at which the charge is transferred is controlled by the capacity of the capacitor and the value of the resistor.

Capacitance

VS VC RR I

constants

RR = RR

When the switch is open and

the capacitor holds no charge:

I = 0 A

Vc = 0 V

Qc = 0 C

VR = 0 V

Rc = 0

The instant the switch is closed

Vc = 0 V

Qc = 0 C

VR = Vs

Rc = 0

thereafter

I decreases as charge builds up on the capacitor

Vc increases

Qc increases

VR decreases

one plate of the capacitor becomes positive

one plate of the capacitor becomes negative

When the capacitor is fully charged

Vc = Vs

VR = 0 V

I = 0 A

During the charging process the voltage across the capacitor gradually increases until it reaches a maximum which is equal to the supply voltage Vs. The rate of change in voltage across the capacitor is greatest at the beginning. Vs remains constant (ignoring lost volts so there would be a difficult question!) therefore as Vc increases, VR must decrease.

The total of these two voltages (Vc and VR) must remain equal to the supply voltage so if you know the supply voltage at any time and one of the voltages Vc or VR then the other can be found.

To find VR use the current given for that time multiplied by RR.

Charging and DischargingIf we circuit so that at point B the capacitor is DISCHARGED through the same resistor we would observe the following voltage and current graphs.charge the capacitor by closing the switch at time A then arrange the

The area under the current-time graph (= I x t) is the total charge, Q, stored on the capacitor. The area under the graph in the charging phase must be equal to the area under the graph in the discharging phase (assuming the capacitor is COMPLETELY discharged). In the above example the areas are obviously the same since the capacitor is charged and discharged through the same resistor. However, if the capacitor is discharged through a smaller value resistor it will discharge more quickly and have a larger initial current. BUT THE AREA UNDER THE DISCHARGE CURVE WILL BE EXACTLY THE SAME AS BEFORE.

When discharging a capacitor, if the load resistor is halved in value then the initial current will be doubled and therefore the time for discharging must be halved, as the total charge is the same.

If values can be put on the graph then make sure you do put them on!

Measuring the capacitance of a capacitor.

mA V constant current supply + - C R 0:00:00

Discharge C.

Connect up. Start clock and take regular values of V and I at set time intervals.

Find the charge that has accumulated on the capacitor at each time interval according to Q = I x t.

Plot a graph of Q against V. Since the gradient of the graph will be the capacitance

Therefore the capacitor has a value of 3000 μF (that is quite a large capacitor!)

Q and V are directly proportional.

1 Farad = 1 Coulomb per volt.

N.B. 1 F is a very large unit. C would normally be expressed as F or nF.

Using your knowledge work out what the area under this graph would represent. We will come back to it later.

V Q 0

Charging a capacitor on a d.c. supply.

Discharging a capacitor.

The current and voltage fall to zero, but be aware that the current flows in the opposite

direction to the original charging current.

Energy stored in a Capacitor

When a charge is moved between 2 charged, parallel plates, work is done. If we move a charge from one plate to the other against the uniform field then work done is given by:

work done = QV

When a capacitor is being charged, work is done moving charges against opposing forces. Electrons are pushed onto the already negative plate. This repels them. They are also removed from the positive plate which tends to attract them.

Energy Stored

The voltage across the capacitor is, however, constantly changing so:

E QV

We average the voltage over the charging period so that

V Q 0 i.e. it is the area under a V-Q graph for a capacitor.

In summary, and using Q = CV,

Capacitors and a.c.

NB.	Remember that “REACTANCE” is used to describe the “RESISTANCE” of a capacitor.

Charging on a.c.

With a.c. a capacitor will begin to charge up, then current starts flowing in the opposite direction. The capacitor therefore discharges and then begins to charge up again.

Hopefully you can imagine that the storage of charge will be greatly affected by how quickly this change of direction is occurring.

Low frequency a.c.

When the frequency of the supply is low the capacitor is fully charged up (voltage is high across the capacitor and the current zero) before the current flows in the opposite direction.

High frequency a.c.

The capacitor does not get time to fully charge up or discharge before the current direction is changed.

This shows us that capacitors block d.c. but allows a.c. through.

Resistance and Frequency

A Signal Generator V R Part 1 Resistance and Frequency

Set up the above circuit and use it to see how the voltages across and the current in R change as you vary the frequency of the supply.

Examine your results to see if there is any connection between an increase in frequency and the resisting effect of R.

Write up your experiment briefly.

Part 2 Resistance and frequency with an oscilloscope

Replace the voltmeter with an oscilloscope and explain how to measure the frequency.

Write up this experiment INCLUDING ALL YOUR WORKING.

NB.	THE RESISTANCE OF A RESISTOR IS UNAFFECTED BY FREQUENCY.

This can therefore be a method of determining whether a component is a capacitor or a resistor.

Reactance of capacitors in a.c. circuits

Resistance (properly called reactance when used with capacitors) of a capacitor can be found out in the usual way.

With a LOW FREQUENCY supply the average VOLTAGE is LARGE and the average CURRENT is SMALL.

Reactance is large.

With a HIGH FREQUENCY supply the average VOLTAGE is SMALL and the average CURRENT is LARGE.

Reactance is small.

We tend therefore to say that capacitors allow current to flow in an a.c. circuit but not in a d.c. circuit.

Read up on BLOCKING and SMOOTHING.

Relationship Between Current and Frequency in an A.C capacitor Circuit.

Blocking and Smoothing

The change in reactance of a capacitor with frequency can be exploited as follows.

Blocking

An electrical signal consists of a steady d.c. voltage with an a.c. voltage superimposed on it.

If this signal is now fed into the following circuit:

then the d.c. component of the signal is removed (or blocked).

Smoothing

A capacitor can also be used in to smooth an a.c. voltage as follows.

A normal a.c. signal is fed into the following circuit.

In the absence of the capacitor the four diodes would rectify this signal thus:

But with the capacitor present this signal is smoothed by the repeated charging and discharging of the capacitor. This output is known as the ripple voltage.

Tutorial 1: capacitance

1.	What is the capacitance of a capacitor storing 0.005C when the p.d. between its plates is 50V?

2.	A capacitor has a p.d. of 20V between its plates. If its capacitance is 25µF, what charge is on it?

3.	What is the p.d. across a 50µF capacitor storing 2.5mC?

Tutorial 2: capacitance

1.	How much work is done in charging up a 30µF capacitor which stores 0.001C at 20V?

2.	What is the capacitance of a capacitor storing 0.016J at a p.d. of 40V?

3.	What is the charge on a 30µF capacitor storing 1.35J?

4.	Calculate the p.d. across a 10µF capacitor if 0.0125J is required to charge it.

Tutorial 3: capacitance

1.	What is the reactance of a capacitor when the p.d. across it is 20V and the current in the circuit 5mA?

2.

What is the reactance of the capacitor?

3.	A 2000Ω resistor is connected in series with a capacitor whose reactance measures 1500Ω. If the a.c. supply is quoted as 14V, what is the peak voltage across the capacitor?

Tutorial 4: CAPACITANCE

1.

How do you show that the component in the box is a resistor?

2.	A 24V a.c. power supply sends a 100Hz current through a 480Ω resistor:- What is the current when the frequency is doubled to 200Hz?

3.

The 2µF capacitor holds 20mC of charge. What is the initial current when S is closed?

4.	If the switch S is moved to A and held for a time, then moved quickly to B, what do the graphs of current against time and p.d. against time look like for the resistor R while the switch is held at B?

5.

How does the brightness of each lamp change when the supply voltage is increased to 25V?

6.	In the diagram below, the bulbs are identical and the three capacitors have the values shown. How do the bulbs compare in brightness?

7.

8.

What is the p.d. of the supply at the instant C is storing 1.125x10-4J in the circuit shown here? : -

Tutorial 5: Exam Questions

In an experiment to measure the capacitance of a capacitor a student sets up the following circuit:

C coulombmeter Y X V VS

When the switch is in position X the capacitor charges up to the supply voltage, VS. When the switch is in position Y the coulombmeter indicates the charge stored by the capacitor.
The student records the following measurements and uncertainties.
Reading on voltmeter	= (2·56 ± 0.01) V
Reading on coulombmeter	= (32 ± 1) μC
Calculate the value of the capacitance and the percentage uncertainty in this value. You must give your answer in the form	value ± percentage uncertainty. 3

V + - 12 V S relay 2200 F 3·3 k The student designs the circuit shown below to switch off a lamp after a certain time.
c)	The 12 V battery has negligible internal resistance.
The relay contacts are normally open. When there is a current in the relay coil the contacts close and complete the lamp circuit. Switch S is initially closed and the lamp is on.

What is the maximum energy stored in the capacitor?

Switch S is now opened. Explain why the lamp stays lit for a few seconds.

The 2200 F capacitor is replaced with a 1000 F capacitor. Describe and explain the effect of this change on the operation of the circuit. 6 (9)

6·0 V 470 F 1·5 k V A S The following diagram shows a circuit that is used to investigate the charging of a capacitor.
The capacitor is initially uncharged. It has a capacitance of 470 F and the resistor has a resistance of 1.5 k. The battery has an e.m.f of 6 V and negligible internal resistance.

Switch S is now closed. What is the initial current in the circuit?

How much energy is stored in the capacitor when it is fully charged?

What change could be made to this circuit to ensure that the same capacitor stores more energy?	5

A capacitor is used to provide the energy for an electronic flash in a camera.
When the flash is fired, 6·35 x 10-3 J of the stored energy is emitted as light.
The mean value of the frequency of photons of light from the flash is 5·80 x 1014 Hz.
Calculate the number of photons emitted in each flash of light. 3 (8)

R S V A 2000 F + - The circuit below is used to investigate the charging of a 2000 F capacitor. The d.c. supply has negligible internal resistance.
The graphs below show how the potential difference, VR across the resistor and the current, I, in the circuit vary with time from the instant switch S is closed.
R S V A C + - power supply I)	What is the potential difference across the capacitor when it is fully charged?

II)	Calculate the energy stored in the capacitor when it is fully charged.

III)	Calculate the resistance of R in the circuit above. 5

The circuit below is used to investigate the charging and discharging of a capacitor.
The graph below shows how the power supply voltage varies with time after switch S is closed.
The capacitor is initially uncharged.
The capacitor charges fully in 0·3 s and discharges fully in 0·3 s.
Sketch a graph of the reading on the voltmeter for the first 2·5 s after switch S is closed.
The axes on your graph must have the same numerical values as those in the above graph 2

A technician carries out an experiment to measure the capacitance of a capacitor C.
The capacitor, initially uncharged, is charged up to 2 V using the circuit below.
The charging current is kept constant at 0·20 mA during the charging process by adjusting the resistance of R. The capacitor is fully charged in 10 seconds.

Explain whether the resistance of R is increased or decreased during the charging period.	1

What is the charge supplied to the capacitor?

Calculate the capacitance of the capacitor.	2

The capacitor is then used in a circuit where it is connected across a 10 V supply.

Calculate:

the charge stored on the capacitor

the energy stored on the capacitor.	(9)

An audio engineer obtains the results shown on the graph below. The graph shows how the current in a circuit containing an 8 F capacitor varies with frequency. The output of the electrical supply is 2 V r.m.s.

Describe an experiment to obtain such a graph. Your answer should include the following:

a circuit diagram of the apparatus required

a statement of the variables measured and controlled

a description of how the measurements were taken

conclusions which can be drawn from the graph.

An engineer has two loudspeakers, LS1 and LS2, to connect to an audio amplifier. One of the speakers is designed so that it is able to produce low frequency sounds. The engineer connects the loudspeakers to the amplifier using the circuit shown below.

Which of the loudspeakers, LS1 or LS2, is intended to emit low frequency sounds. You must explain your answer. (7)

Tutorial Answers Capacitance

Tutorial 1

1.	The capacitance is 1x10-4F (100µF)

2.	The capacitor is storing 0.0005C

3.	The p.d. across the capacitor is 50V

Tutorial 2

1.	The work done during charging is 0.01J

2.	The capacitance is 2x10-5F (20µF)

3.	The stored charge is 0.009C

.	The p.d. across the capacitor is 50V

Tutorial 3

1.	The reactance of the capacitor is 4000Ω

2.	Since the resistance of the whole circuit is 2500Ω, the reactance of the capacitor must be 1000Ω

3.	The peak voltage must = 8V

Tutorial 4

1.	Note the voltage and current at some frequency. Use them to calculate the resistance of the component. Repeat the measurements for different supply frequencies. If the resistance remains constant, then the component is a resistor.

2.	Since the resistance is independent of the frequency, the current remains at 0.005A when the frequency doubles to 200Hz.

3.	The initial current is 1A

4.

5.	Each is 2.5 times brighter since neither the resistance nor the capacitive reactance depend on the supply voltage.

6.	The bulbs increase in brightness from (1) to(3) since the capacitive reactance becomes less with increasing capacitance.

7.	Bulb (1) remains at constant brightness while bulb (2) increases in brightness as the capacitive reactance decreases with the increase in supply frequency.

8.	The supply voltage is 8V.

Tutorial Work Answers

Capacitance/ tutorial answers

Answers for tutorial 1

1.

The capacitance is 1x10-4F (100µF)

2.

The capacitor is storing 0.0005C

3.	The p.d. across the capacitor is 50V

ANSWERS FOR TUTORIAL 2

1. The work done during charging is 0.01J

2. The capacitance is 2x10-5F (20µF)

3.

The stored charge is 0.009C

4.

The p.d. across the capacitor is 50V

ANSWERS FOR TUTORIAL 3

1.

The reactance of the capacitor is 4000Ω

2.	Apply V=RI to the whole circuit to get:-

Since the resistance of the whole circuit is 2500Ω, the reactance of the capacitor must be 1000Ω.

3.	Since the circuit is behaving as a voltage divider, the p.d. across the capacitor is This is the r.m.s. voltage across the capacitor so the peak voltage must by 6x√2 = 8V

ANSWERS FOR TUTORIAL 4

1.	Note the voltage and current at some frequency. Use them to calculate the resistance of the component. Repeat the measurements for different supply frequencies. If the resistance remains constant, then the component is a resistor.

2. Since the resistance is independent of the frequency, the current remains at 0.005A when the frequency doubles to 200Hz.

3.	The initial current depends only on the p.d. and the resistance.

Since the p.d. is 10000V,

The initial current is 1A

4.

5.	Each is 2.5 times brighter since neither the resistance nor the capacitive reactance depends on the supply voltage.

6.	The bulbs increase in brightness from (1) to (3) since the capacitive reactance becomes less with increasing capacitance.

7.	Bulb (1) remains at constant brightness while bulb (2) increases in brightness as the capacitive reactance decreases with the increase in supply frequency.

8.

The p.d. across the capacitor is 3V.

V=RI applied to the resistor gives V = 2000 x 2.5x10-3 = 5 volts.

Since the supply voltage equals the sum of the individual p.d.'s round the circuit, its value is 8V.

Tutorial Answers: Exam question soutions

2000

24.a.	To calculate the capacitance uses the mean values of charge and voltage.

C = Q/V

C = (32/2.56)F

C=12.5F

The percentage error in capacitance value can be taken to be equal

to that of largest individual percentage error.

% error in voltage =	(0.01/2.56)x100 = 0.39%

% error in charge =	(1/32)x100	= 3.125%

C=12.5F+-3.125%

OR

3.125% of 12.5F = 0.39F =>	C = (12.5+-0.4)F

NB/	Only quote the error to the same number of decimal places as the capacitance value.

b.i.The maximum energy is stored in the capacitor when the voltage across the capacitor is equal to the supply voltage.

Vc= 12V

C = 2200x10-6

E = ?

E = 1/2(QVc)

Q = CVc

=>E = 1/2(CVc2)

E = 0.5(2200x10-6x122)

E = 0.1584J

b.ii.(A)	When the switch is opened the capacitor discharges through the resistor and relay coil. The discharge current magnetises the coil closing the switch in the lamp circuit, causing the lamp	to light. As the discharge current gradually falls the coil loses its magnetism and the switch in the lamp circuit opens. When this happens the lamp goes off.

(B)	Increasing the value of the capacitor increases the discharge time. The energy stored in the capacitor is also greater. This means that the lamp will stay lit for longer.

2001

25.a.i.	The initial charging current(Imax) occurs when all of the supply voltage(Vsupply) is across the 1.5k resistor(R).

Imax = Vsupply/R

Imax = 6/1500

Imax = 4x10-3A

a.ii.	When fully charged the voltage across the supply voltage is equal to the voltage across the capacitor.

Vsupply = Vc = 6V

Ecapacitor = QVc/2

Q = CVc

=>Ecapacitor = CVc2/2

Ecapacitor = 470x10-6x62/2

Ecapacitor = 8.46x10-3J

a.iii.	Increasing the supply voltage would increase the energy storing capacity of the capacitor. This is because the final voltage, across the fully charged capacitor, would be higher.

25.b.	Etotal = 6.35x10-3J

fphoton = 5.80x1014Hz

h = 6.63x10-34Js

Ephoton = ?

Ephoton = hfphoton

Ephoton = 6.63x10-34 x 5.80x1014

Ephoton = 3.84x10-19J

Etotal = NEphoton

N = Etotal/Ephoton

N = 6.35x10-3/3.84x10-19

N = 1.65x1016

2002

25.a.i.	Initially all the supply voltage is across the resistor.

VR = Vsupply = 6V

When the capacitor is fully charged: Vsupply = Vcapacitor

Vcapacitor = 6V

a.ii. E = CV2/2

E = (2000x10-6x62)/2

E = 0.072/2

E = 0.036J

a.iii.	Imax = Vsupply/R

R = Vsupply/Imax

R = 6/7.5x10-3

R = 800

Circuits

Knowledge of the variation of current and potential difference with time in a CR circuit during charging and discharging.

The capacitor now starts to charge up as shown, with the rise in the RC charging curve steeper at the beginning because the charging rate is fastest at the start and then tapers off as the capacitor takes on additional charge at a slower rate.

As the capacitor charges up, the potential difference across its plates slowly increases with the actual time taken for the charge on the capacitor to reach 63% of its maximum possible voltage, in our curve 0.63Vs being known as one Time Constant, ( T ).

This 0.63Vs voltage point is given the abbreviation of 1T, (one time constant).

The capacitor continues charging up and the voltage difference between Vs and Vc reduces, so to does the circuit current, i. Then at its final condition greater than five time constants ( 5T ) when the capacitor is said to be fully charged, t = ∞, i = 0, q = Q = CV. Then at infinity the current diminishes to zero, the capacitor acts like an open circuit condition therefore, the voltage drop is entirely across the capacitor.

So mathematically we can say that the time required for a capacitor to charge up to one time constant, ( 1T ) is given as:

RC Time Constant, Tau

This RC time constant only specifies a rate of charge where, R is in Ω‘s and C in Farads.

Since voltage V is related to charge on a capacitor given by the equation, Vc = Q/C, the voltage across the value of the voltage across the capacitor ( Vc ) at any instant in time during the charging period is given as:

Where:

Vc is the voltage across the capacitor

Vs is the supply voltage

t is the elapsed time since the application of the supply voltage

RC is the time constant of the RC charging circuit

After a period equivalent to 4 time constants, ( 4T ) the capacitor in this RC charging circuit is virtually fully charged and the voltage across the capacitor is now approx 98% of its maximum value, 0.98Vs. The time period taken for the capacitor to reach this 4T point is known as the Transient Period.

After a time of 5T the capacitor is now fully charged and the voltage across the capacitor, ( Vc ) is equal to the supply voltage, ( Vs ). As the capacitor is fully charged no more current flows in the circuit. The time period after this 5T point is known as the Steady State Period.

Then we can show in the following table the percentage voltage and current values for the capacitor in a RC charging circuit for a given time constant.

Time	RC Value	Percentage of Maximum
Constant	Voltage	Current
0.5 time constant	0.5T = 0.5RC	39.30%	60.70%
0.7 time constant	0.7T = 0.7RC	50.30%	49.70%
1.0 time constant	1T = 1RC	63.20%	36.80%
2.0 time constants	2T = 2RC	86.50%	13.50%
3.0 time constants	3T = 3RC	95.00%	5.00%
4.0 time constants	4T = 4RC	98.20%	1.80%
5.0 time constants	5T = 5RC	99.30%	0.70%

Note that as the charging curve for a RC charging circuit is exponential, the capacitor in reality never becomes 100% fully charged due to the energy stored in the capacitor. So for all practical purposes, after five time constants a capacitor is considered to be fully charged.

As the voltage across the capacitor Vc changes with time, and is a different value at each time constant up to 5T, we can calculate this value of capacitor voltage, Vc at any given point, for example.

Definition of the time constant for a CR circuit.

t = RC

(units =second)

definition of time constant as the time to increase the charge stored by 63% of the difference between initial charge and full charge, or the time taken to discharge the capacitor to 37% of initial charge.

After one time constant the current has fallen to 37% of the initial charging current.

After one time constant the voltage across the capacitor will be 63% of the Supply voltage.

Numerical and graphical determination of the time constant for a CR circuit.

E.g RC Charging Circuit Example

Calculate the RC time constant, τ of the following circuit.

The time constant, τ is found using the formula t = R x C in seconds.
Therefore the time constant τ is given as:

t = R x C = 47000 0.001F = 47 s

What will the current be after one time constant

What value will be the voltage across the capacitor at 0.7 time constants?

Or from the table

At 0.7 time constants ( 0.7T ) Vc = 0.5Vs. Therefore, Vc = 0.5 x 5V = 2.5V

What value will be the voltage across the capacitor at 1 time constant?

At 1 time constant ( 1T ) Vc = 0.63Vs. Therefore, Vc = 0.63 x 5V = 3.2V

How long will it take to “fully charge” the capacitor?

The capacitor will be fully charged at 5 time constants.
1 time constant ( 1T ) = 47 seconds, (from above). Therefore, 5T = 5 x 47 = 235 s

The voltage across the Capacitor after 100 seconds?

The voltage formula is given as

which equals:

RC = 47 seconds from above, Therefore, Vc = 4.4 V

We have seen that the charge on a capacitor is given by the expression: Q = CV and that when a voltage is firstly applied to the plates of the capacitor it charges up at a rate determined by its time constant, τ. In the next tutorial we will examine the current-voltage relationship of a discharging capacitor and look at the curves associated with it when the capacitors plates are shorted together.

RC Discharging Circuit

In the previous RC Charging Circuit tutorial, we saw how a Capacitor, C charges up through the resistor until it reaches an amount of time equal to 5 time constants or 5T and then remains fully charged. If this fully charged capacitor was now disconnected from its DC battery supply voltage it would store its energy built up during the charging process indefinitely (assuming an ideal capacitor and ignoring any internal losses), keeping the voltage across its terminals constant.

If the battery was now removed and replaced by a short circuit, when the switch was closed again the capacitor would discharge itself back through the resistor, R as we now have a RC discharging circuit. As the capacitor discharges its current through the series resistor the stored energy inside the capacitor is extracted with the voltage Vc across the capacitor decaying to zero as shown below.

RC Discharging Circuit

As with the previous RC charging circuit, in a RC Discharging Circuit, the time constant ( τ ) is still equal to the value of 63%. Then for a RC discharging circuit that is initially fully charged, the voltage across the capacitor after one time constant, 1T, has dropped to 63% of its initial value which is 1 – 0.63 = 0.37 or 37% of its final value.

So now this is given as the time taken for the capacitor to discharge down to within 37% of its fully discharged value which will be zero volts (fully discharged), and in our curve this is given as 0.37Vc.

As the capacitor discharges, it loses its charge at a declining rate. At the start of discharge the initial conditions of the circuit, are t = 0, i = 0 and q = Q. The voltage across the capacitors plates is equal to the supply voltage and Vc = Vs. As the voltage across the plates is at its highest value maximum discharge current flows around the circuit.

RC Discharging Circuit Curves

With the switch closed, the capacitor now starts to discharge as shown, with the decay in the RC discharging curve steeper at the beginning because the discharging rate is fastest at the start and then tapers off as the capacitor looses charge at a slower rate. As the discharge continues, Vc goes down and there is less discharge current.

As with the previous charging circuit the voltage across the capacitor, C is equal to 0.5Vc at 0.7T with the steady state fully discharged value being finally reached at 5T.

For a RC discharging circuit, the voltage across the capacitor ( Vc ) as a function of time during the discharge period is defined as:

Where:

Vc is the voltage across the capacitor

Vs is the supply voltage

t is the elapsed time since the removal of the supply voltage

RC is the time constant of the RC discharging circuit

Just like the previous RC Charging circuit, we can say that in a RC Discharging Circuit the time required for a capacitor to discharge itself down to one time constant is given as:

Where, R is in Ω’s and C in Farads.

Then we can show in the following table the percentage voltage and current values for the capacitor in a RC discharging circuit for a given time constant.

RC Discharging Table

Time Constant	RC Value	Percentage of Maximum
Voltage	Current
0.5 time constant	0.5T = 0.5RC	60.7%	39.3%
0.7 time constant	0.7T = 0.7RC	49.7%	50.3%
1.0 time constant	1T = 1RC	36.6%	63.4%
2.0 time constants	2T = 2RC	13.5%	86.5%
3.0 time constants	3T = 3RC	5.0%	95.0%
4.0 time constants	4T = 4RC	1.8%	98.2%
5.0 time constants	5T = 5RC	0.7%	99.3%

Note that as the discharging curve for a RC discharging circuit is exponential, for all practical purposes, after five time constants a capacitor is considered to be fully discharged.

So a RC circuit’s time constant is a measure of how quickly it either charges or discharges.

RC Discharging Circuit Example No1

Calculate the RC time constant, τ of the following RC discharging circuit.

The time constant, τ is found using the formula T = R x C in seconds.
Therefore the time constant τ is given as:
T = R x C = 100k x 22uF = 2.2 Seconds

a) What value will be the voltage across the capacitor at 0.7 time constants?

At 0.7 time constants ( 0.7T ) Vc = 0.5Vc. Therefore, Vc = 0.5 x 10V = 5V

b) What value will be the voltage across the capacitor after 1 time constant?

At 1 time constant ( 1T ) Vc = 0.37Vc. Therefore, Vc = 0.37 x 10V = 3.7V

c) How long will it take for the capacitor to “fully discharge” itself, (equals 5 time constants)

1 time constant ( 1T ) = 2.2 seconds. Therefore, 5T = 5 x 2.2 = 11 Seconds

Factors affecting the rate of charge and discharge

The time taken for a capacitor to charge is controlled by the resistance of the resistor R (because it controls the size of the current, i.e. the charge flow rate) and the capacitance of the capacitor (since a larger capacitor will take longer to fill and empty). As an analogy, consider charging a capacitor as being like filling a jug with water. The size of the jug is like the capacitance and the resistor is like the tap you use to control the rate of flow.

The values of R and C can be multiplied together to form what is known as the time constant. Can you prove that R × C has units of time, seconds? The time taken for the capacitor to charge or discharge is related to the time constant.

Large capacitance and large resistance both increase the charge or discharge time.

The I/t graphs for capacitors of different value during charging are shown below:

current 0 time small resistor large resistor current 0 time large capacitor small capacitor

The effect of capacitance on charging current	The effect of resistance on charging current

Note that since the area under the I/t graph is equal to charge, for a given capacitor the area under the graphs must be equal.

Tutorial Capacitors

Capacitors in d.c. circuits

1.	A 50 µF capacitor is charged until the p.d. across it is 100 V.

(a)	Calculate the charge on the capacitor when the p.d. across it is 100 V.

(b)	(i)	The capacitor is now ‘fully’ discharged in a time of 4·0 ms.

Calculate the average current during this time.

(ii)	Why is this average current?

2.	A capacitor stores a charge of 3·0 × 10–4 C when the p.d. across its terminals is 600 V. What is the capacitance of the capacitor?

3.	A 15 µF capacitor is charged using a 1·5 V battery. Calculate the charge stored on the capacitor when it is fully charged.

4.	(a)	A capacitor stores a charge of 1·2 × 10–5 C when there is a p.d. of 12 V across it. Calculate the capacitance of the capacitor.

(b)	A 0·10 µF capacitor is connected to an 8·0 V d.c. supply. Calculate the charge stored on the capacitor when it is fully charged.

5.	In the circuit below the capacitor C is initially uncharged.

9 V C A V S + –

Switch S is now closed. By carefully adjusting the variable resistor R a constant charging current of 1·0 mA is maintained.

The reading on the voltmeter is recorded every 10 s. The results are shown in the table.

Time / s	0	10	20	30	40
V / V	0	1·9	4·0	6·2	8·1

(a)	Plot a graph of the charge on the capacitor against the p.d. across the capacitor.

(b)	Use the graph to calculate the capacitance of the capacitor.

Tutorial Answers Capacitors

Capacitors in d.c. circuits

	(a)	C=Q/V	Q=CV=	Q=50 × 10–6 100	Q=50 × 10–3 C….

(b)	(i)	Q=It	Q/t=I	Q=50 × 10–3 / 40 × 10–3	I=125 A

(ii)	current decreases exponentially

2.	C=Q/V	C=3·0 × 10–4 /	C=0.5 F

3.	C=Q/V	Q=CV	Q=15·0 × 10–6 1·5	Q=225 × 10–5 C

4.	(a)	C=Q/V	C=1·2 × 10–5/12	C=10 F

(b)	C=Q/V	Q=CV	Q=0·10 × 10–6 8·0	C=08 C

5.

(a)

Time / s	Charge (C )	Voltage (V)
0	0	0.0
10	0.01	1.9
20	0.02	4.0
30	0.03	6.2
40	0.04	8.1

(b)	49 mF

6.	(e)	(i)	40 A

(ii)	40 × 102 C or 40 × 10–4 C

7.	(e)	12 V

10.	8 ms

12.	(b)	(i)	8 s

(ii)	73 V

Definition of capacitive reactance.

Consider the following circuit:

The signal generator is set to a low frequency and the potential difference across the resistor to a known voltage. The current in the circuit is then measured using the ammeter.

The frequency of the signal generator is altered but the potential difference kept constant and the current measured.

0

The current in the resistive remains constant with frequency with I = V/R. Resistors are unaffected by the frequency of the supply and behave in the same way in both d.c. and a.c. circuits.

The resistor is replaced with a capacitor and the experiment repeated.

0

The current in the circuit increases in direct proportion to the frequency.

The opposition to current in a capacitive circuit is the capacitive reactance, XC.

Use of appropriate relationships to solve problems involving capacitive reactance, voltage, current, frequency and capacitance.

Current in a capacitive circuit is determined by I = V/XC and therefore XC = V/I. XC is measured in ohms, the same as resistance, but it is not appropriate to refer to the opposition to current in a capacitive circuit as a resistance.

I f and XC 1/I therefore XC 1/f.

XC = 1 / 2fC

Capacitors in a.c. circuits

1.	A capacitor is connected to a variable frequency a.c. supply as shown below. The amplitude of the output voltage from the supply is kept constant.

Variable frequency Constant amplitude supply ˜ C

(a)	The capacitor has reactance. State what is meant by the term ‘reactance’.

(b)	The frequency of the output from the a.c. supply is increased.

Sketch a graph to show how:

(i)	the reactance of the capacitor varies with the frequency of the supply

(ii)	the current in the circuit varies with the frequency of the supply.

2.	A 1·0 F capacitor is connected to 5·0 V a.c. power supply. The frequency of the a.c. supply is 50 Hz.

(a)	Calculate the capacitive reactance of the capacitor.

(b)	Calculate the current in the circuit.

3.	A capacitor is connected across a 250 V r.m.s supply having a frequency of 50 Hz. The current in the capacitor is 0·50 A r.m.s.

Calculate:

(a)	the reactance of the capacitor at this frequency

(b)	the capacitance of the capacitor.

4.	A 500  resistor and a capacitor are connected in series across an a.c. supply. The frequency of the a.c. is 50 Hz. The p.d. across the resistor is 120 V. The p.d. across the capacitor is 160 V.

(a)	Calculate the current in the circuit.

(b)	Calculate the capacitance of the capacitor.

5.	A 300  resistor and a capacitor are connected in series with an a.c. supply of frequency 100 Hz.

The p.d. across the capacitor is 5.00 V. When the frequency of the output from the supply is 100 Hz the capacitive reactance of the capacitor is 265 

Calculate:

(a)	the capacitance of the capacitor

(b)	the current in the circuit

(c)	the p.d. across the resistor.

6.	A resistor and a capacitor are connected in series with a variable frequency constant amplitude a.c. supply. The combined effect of the resistance R and capacitive reactance XC in series is known as the impedance Z of the circuit where

Z can be calculated from Z = V/I where V is the voltage of the supply and I is the current in the circuit.

The value of Z is found for different frequencies and the data used to plot the following graph.

Use the graph and knowledge of the relationship for Z and to estimate:

(a)	the value of the resistance;

(b)	the value of the capacitance.

Electromagnetic Induction

Electromagnetic induction is the production of an induced e.m.f. in a conductor when it is present in a changing magnetic field. Electromagnetic induction involves the transformation of mechanical energy into electrical energy.

One method of generating an induced voltage is by relative moving a permanent magnet inside a coil of wire.

Note that the induced e.m.f. occurs when there is relative motion between the magnetic field and a conductor.

A Simple Experiment on Electromagnetic Induction

Apparatus: coil, magnet, centre-zero meter

Observations When the magnet is moving into coil - meter needle moves to the right (say). We say a current has been induced. When the magnet is moving out of the coil - induced current is in the opposite direction (left). Magnet stationary, either inside or outside the coil - no induced current. When the magnet is reversed, i.e. the south pole is nearest the coil, - induced current reversed. Moving the magnet faster makes the induced current bigger.

Note:	Moving the coil instead of the magnet produces the same effect. It is the relative movement which is important.

The induced currents that are observed are said to be produced by an induced electromotive force, e.m.f. This electrical energy must come from somewhere. The work done by the person pushing the magnet at the coil is the source of the energy. In fact the induced current sets up a magnetic field in the coil which opposes the movement of the magnet.

Summary

The size of the induced e.m.f. depends on:

the relative speed of movement of the magnet and coil

the strength of the magnet

the number of turns on the coil.

Inductors in d.c. circuits

Growth and Decay of Current in an Inductive Circuit.

R S2 S1 _ + L A

The circuit symbol is Vs	An inductor is a coil of wire wound on a soft iron core. An inductor is denoted by the letter L. An inductor has inductance, see later.
time large inductance small inductance current Vs R Growth of current	Switch S2 is left open. When switch S1 is closed the ammeter reading rises slowly to a final value, showing that the current takes time to reach its maximum steady value. With no current there is no magnetic field through the coil. When S1 is closed the magnetic field through the coil will increase and an e.m.f. will be generated to prevent this increase. The graph opposite shows how the circuit current varies with time for inductors of large and small inductance.
Vs R large inductance small inductance time current Decay of current	After the current has reached its steady value S2 is closed, and then S1 is opened. The ammeter reading falls slowly to zero. The current does not decay immediately because there is an e.m.f. generated which tries to maintain the current, that is the induced current opposes the change. The graph opposite shows how the circuit current varies with time for inductors of large and small inductance.

Notes

For both the growth and decay, the induced e.m.f. opposes the change in current.

For the growth of current, the current tries to increase but the induced e.m.f. acts to prevent the increase. It takes time for the current to reach its maximum value. Notice that the induced e.m.f. acts in the opposite direction to the circuit current.

For the decay of current the induced e.m.f acts in the same direction as the current in the circuit. Now the induced e.m.f. is trying to prevent the decrease in the current.

Experiment to show build up current in an inductive circuit

R S _ + B2 L B1	The switch is closed and the variable resistor adjusted until the lamps B1 and B2 have the same brightness. The supply is switched off. The supply is switched on again and the brightness of the lamps observed.

Lamp B2 lights up immediately. There is a time lag before lamp B1 reaches its maximum brightness. An e.m.f. is induced in the coil because the current in the coil is changing. This induced e.m.f. opposes the change in current, and it is called a back e.m.f. It acts against the increase in current, hence the time lag.

The experiment is repeated with an inductor of more turns. Lamp B1 takes longer to light fully. If the core is removed from the inductor, lamp B1 will light more quickly.

Induced e.m.f. when the current in a circuit is switched off

When the current in a circuit, containing an inductor, is switched off the magnetic field through the inductor will fall very rapidly to zero. There will be a large change in the magnetic field leading to a large induced e.m.f. For example a car ignition coil produces a high e.m.f. for a short time when the circuit is broken.

Lighting a neon lamp

The 1.5 V supply in the circuit below is insufficient to light the neon lamp. A neon lamp needs about 80 V across it before it will light.

1.5 V	The switch is closed and the current builds up to its maximum value. When the switch is opened, the current rapidly falls to zero. The magnetic field through the inductor collapses (changes) to zero producing a very large induced e.m.f. for a short time. The lamp will flash.

Self-inductance (inductance) of a coil

self inductance

This may seem like a rather abstract concept so it is best demonstrated with the experiment below. This experiment needs to be done with an analogue meter as a digital meter can fluctuate.

Safety

Ensure that all apparatus is away from the edge of benches.

Ensure that you do not touch of any of the terminals when the switch is being opened.

VOLTAGE AT SWITCH ON Introduction

In this experiment you will investigate how the voltage across an inductor changes at switch on in a series Resistor - Inductor (R-L) circuit.

A Voltage/Time graph will be plotted.

You can investigate the effect of changing the values of the inductor and resistor.

You have several questions to answer.

Background Information

You should be familiar with Ohm's law - the voltage across a resistor is equal to the current through the resistor multiplied by the value of the resistor. (V = IR)

Apparatus

For this experiment you will require:

ALBA, serial lead and plugtop power supply.

djb microtech's Inductor Investigation board or equivalent.

1200 - 0 - 1200 turn coil or two separate coils e.g. 1200 and 2400-turn coils.

Resistors

330R (orange, orange, brown)

430R (yellow, orange, brown)

560R (green, blue, brown)

Connecting leads.

Examine the following Circuit Diagram and Photograph to see how the apparatus is used - you will be directed how to set up the apparatus in the Procedure section.

How it Works

ALBA has four analogue input channels and each of these acts as a voltmeter.

The values on the 'voltmeter' can be read many times per second and the results sent to the computer for processing.

Each 'voltmeter' measures the voltage above ground.

If you examine the circuit diagram you will see that channel 1 samples the supply voltage and channel 2 samples the voltage across the resistor.

Subtracting the voltage across the resistor from the supply voltage gives the voltage across the inductor.

Procedure 1

Do not start the experiment until you have read all of Procedures 1 and 2.

Set up the circuit. Use the 430R resistor and the 2400-turn coil.

If you are having difficulty, follow the wires in the photograph.

Make sure that you have the diode connected across the coil. This is important.

Ensure that the switch beside the battery is set to 3 V.

Procedure 2

When you hold down switch A, you will see the graph build up on the screen. Keep the switch closed until the graph is complete.

Investigate what happens when a 1200-turn coil, i.e. a smaller inductance, is used.

Investigate the effect of changing the value of the circuit resistance.

Overlay your graphs. If you are uncertain how to do this then read the Help file.

Read the Safety Notes then select the GO icon to start the experiment.

More to Do

Predict what will happen if the experiments are repeated using a 1.5V battery.

Now repeat the experiments to test your predictions

Write down a sentence to describe how the voltage across an inductor in an R-L circuit behaves at switch on.

Describe the effect of using a smaller inductor on the rate of decay of voltage.

Why is the final voltage different for the 1200-turn coil compared with the 2400-turn coil?

If the winding of the coil had no resistance what would the final voltage across it be?

What effect does increasing the value of circuit resistance have on the rate of rise of current?

CURRENT AT SWITCH ON Introduction

In this experiment you will investigate how the current through an inductor changes at switch on in a series Resistor - Inductor (R-L) circuit.

A Current/Time graph will be plotted.

You can investigate the effect of changing the values of the inductor and resistor.

You have several questions to answer.

Apparatus

For this experiment you will require:

ALBA, serial lead and plugtop power supply.

djb microtech's Inductor Investigation board or equivalent.

1200 - 0 - 1200 turn coil or two separate coils e.g. 1200 and 2400-turn coils.

Resistors

330R (orange, orange, brown)

430R (yellow, orange, brown)

560R (green, blue, brown)

Connecting leads.

Examine the following Circuit Diagram and Photograph to see how the apparatus is used - you will be directed how to set up the apparatus in the Procedure section.

CURRENT AT SWITCH OFF Introduction

In this experiment you will investigate how the current through an inductor changes at switch off in a series Resistor - Inductor (R-L) circuit.

A Current/Time graph will be plotted.

You can investigate the effect of changing the values of the inductor and resistor.

You have several questions to answer.

Background Information

You should be familiar with Ohm's law - the voltage across a resistor is equal to the current through the resistor multiplied by the value of the resistor. (V = IR)

Apparatus

For this experiment you will require:

ALBA, serial lead and plugtop power supply.

djb microtech's Inductor Investigation board or equivalent.

1200 - 0 - 1200-turn coil or two separate coils e.g. 1200 and 2400-turn coils.

Resistors

330R (orange, orange, brown)

430R (yellow, orange, brown)

560R (green, blue, brown)

Connecting leads.

Examine the following Circuit Diagram and Photograph to see how the apparatus is used - you will be directed how to set up the apparatus in the Procedure section.

How it Works

ALBA has four analogue input channels and each of these acts as a voltmeter.

The values on the 'voltmeter' can be read many times per second and the results sent to the computer for processing.

Each 'voltmeter' measures the voltage above ground.

If you examine the circuit diagram you will see that channel 1 measures the voltage across the resistor. The program knows the value of the resistor and it uses Ohm's Law to calculate the current.

The current is made to flow through the inductor by closing switch A. This causes the magnetic field to grow around the inductor. Closing switch B has the effect of shorting the terminals of the R-L circuit. The current from the battery will no longer flow through the inductor and resistor. Note that closing switch B also has the effect of shorting the battery and consequently you must release this switch as soon as the results are displayed.

Procedure 1

Do not start the experiment until you have read all of Procedures 1 and 2.

Set up the circuit. Use the 430R resistor and the 2400-turn coil.

If you are having difficulty, follow the wires in the photograph.

Do not connect the diode across the coil. Ensure that the switch beside the battery is set to 3 V.

Procedure 2

The software will direct you to hold down switch A and while holding it down press switch B. As soon as the graph is displayed on the screen release switch A then release switch B.

Investigate what happens when a 1200-turn coil, i.e. a smaller inductance, is used.

Investigate the effect of changing the value of the circuit resistance.

Overlay your graphs. If you are uncertain how to do this then read the Help file.

Read the Safety Notes then select the GO icon to start the experiment.

Predict what will happen if the experiments are repeated using a 1.5V battery.

Now repeat the experiments to test your predictions

Write down a sentence to describe how the current in an R-L circuit behaves at switch off.

Describe the effect of using a smaller inductor on the rate of change of current.

dI/dt experiment Background Information

You should be familiar with Ohm’s law – the voltage across a resistor is equal to the current through the resistor multiplied by the value of the resistor. (V = IR)

From your previous work you should know that it takes time for the current to reach its final value in an R-L circuit and while the current is changing a voltage is induced across a coil.

Introduction

In this experiment you will investigate how the rate of rise of current through an inductor is related to the voltage induced across the inductor.

Voltage/Time and Current/Time graphs will be plotted.

You can investigate the effect of changing the values of the inductor and resistor.

You have several questions to answer.

Apparatus

For this experiment you require:

ALBA, serial lead and plugtop power supply.

djb microtech’s Inductor Investigation board or equivalent.

1200 - 0 – 1200-turn coil or two separate coils e.g. 1200 and 2400-turn coils.

Two ‘C’ cores and clip

A 0.5H air core coil - if available.

Resistors

330R (orange, orange, brown)

430R (yellow, orange, brown)

560R (green, blue, brown)

Connecting leads.

Examine the following Circuit Diagram and Photograph to see how the apparatus is used - you will be directed how to set up the apparatus in the Procedure section.

How it Works

Each analogue channel acts like a 'voltmeter' and measures the voltage above ground.

If you examine the circuit diagram you will see that channel 1 samples the supply voltage and channel 2 samples the voltage across the resistor. Knowing the voltage across the resistor and the value of the resistor, the current can be calculated using Ohm’s Law.

Subtracting the voltage across the resistor from the supply voltage gives the voltage across the inductor.

Procedure 1

Do not start the experiment until you have read all of Procedures 1 and 2.

Set up the circuit. Use the 430R resistor and the 2400-turn coil.

If you are having difficulty, follow the wires in the photograph.

Make sure that you have the diode connected across the coil. This is important.

Ensure that the switch beside the battery is set to 3 V.

Procedure 2

When you hold down switch A, you will see the graph build up on the screen. Keep the switch closed until the graph is complete.

When the graphs are complete, click on the restore button of the Table window.

The ‘Analysis of Results’ section will direct you on the next steps to take.

Read the Safety Notes then select the GO icon to start the experiment.

Safety

Ensure that all apparatus is away from the edge of benches.

Ensure that you do not touch of any of the terminals when the switch is being opened.

Explanation of Table

The Table is minimized – click the Restore button and drag the left hand side of the Table so that you can see all the columns.

An ideal inductor is one whose winding resistance is zero. Think upon a real inductor as one that can be represented by the series combination of an ideal inductor and a resistor representing the winding resistance. You will have observed from the voltage time graph that the final voltage across the inductor is not zero and hence the inductor is not an ideal one.

The program calculates the resistance of the inductor windings. Then at each instant in time the program takes the current and multiplies it by the winding resistance to get the voltage across the winding resistance. This voltage is then subtracted from the voltage across the inductor to give the voltage across the ideal inductor. These values are shown in column 4.

Column 5 is the gradient at each point on the current/time graph.

Analysis of Results

Draw a graph of the voltage across the ideal inductor (i.e. induced voltage) against the rate of change of current. Note that the induced voltage is on the y-axis.

Fit a straight line to data points.

Go to the end column in the Table and double click on the ‘use’ in the first six rows. This has the effect of removing these points from the graph.

Questions

What is the relationship between induced voltage and rate of change of current?

What is the effect of removing the first six points in the table?

Suggest a reason for the behaviour of the coil over the first 30ms.

More to Do

Repeat the experiment using a 0.5H air cored coil.

Compare the gradient of induced voltage/rate of change of current graph with the value of the inductance of the coil.

Explain any difference in shape between the graphs for the air-cored coil and the iron- cored coil.

Complete the following sentences:

The slope of a Voltage/Current graph gives ---

The slope of a Charge/Voltage graph gives ---

The slope of an Induced voltage/Rate of change of current graph gives ---

The self induced e.m.f. induced is given by

Where L is the inductance, dI is the change in current and dt is the time for this change in current or dI/dt is the rate of change of current.

You probably met transformers and generators when studying lower down the school but did not appreciate that these were examples of induced e.m.f. The generator was shown to produce an e.m.f when there is relative motion between the wire and the magnetic field.

The mutual inductance of the two coils in a transformer can be a good starting point to discuss the self inductance of a coil. Whenever a magnetic field changes through an area of space there will be an e.m.f. developed around that area. This will be the case whether or not there is a conductor or free charges present.

It might be conceptually easier for you to think of pictures of magnetic field lines and conducting circuits with free electrons in them. If the wire is fixed and in the form of a coil, a changing field may be visualised as the field lines growing or shrinking through the turns of coil (this is not accurate but might help you picture what is going on!) This, in a sense , can be thought of as a relative motion between the field lines and the free charges in the wires. In the case of a transformer, the current in the primary coil produces the changing field that threads the secondary coil. In the case of the single isolated coil, it is the field produced by its own current that changes through its own turns, admittedly hard to grasp.

As far as individual electrons are concerned electrons do not “know” what the source of the changing field is. The e.m.f. generated by the changing field acts on them regardless of its origin. It should be obvious to realise that this e.m.f cannot be in a direction that encourages growth of current as an e.m.f. in this direction would represent energy being generated without being converted from another form. This e.m.f. forms a “back e.m.f.” and that Lenz’s law applies is evidence that energy cannot be spontaneously created. THE ESTABLISHMENT OF THE MAGNETIC FIELD REQUIRES ENERGY TO BE STORED IN THE FIELD. You must realise that the energy stored in the field is not “lost” but can be recovered when the current ceases, and the field collapses. So if you receive a shock from an appliance from its back e.m.f it will be short lived! However, the size of the shock is often so large as to be fatal!

When doing the experiment to measure the self inductance of a heavily inductive coil, the time constant of many inductors is so short that often a data-capture device (such as ALBA) is required.

It should be noted that:

A pure inductor (one that has no resistance) does not dissipate energy

when the steady state is reaches, charges pass through the inductor but not through a potential difference- a pure inductor has no potential difference across it in the steady state, and so there is no further energy transfer involved

when the circuit is eventually broken, this stored energy will be converted to some other form.

If the current through an inductor is switched off then the magnetic field will collapse, As it does so, it will induce an e.m.f. across the coil. The direction of this will be such as to maintain the current. If the current has been switched off by opening the circuit, then the field will collapse very rapidly to begin with, and the e.m.f. generated –L dI/dt, may create an electric field of sufficient intensity to break down the insulation of the air across the switch terminals.

A spark will jump across the gap, and the resulting flow of charge will act to convert the stored energy to heat and possibly other forms. Under normal circumstances this is of little consequence, however, where repeated sparking like this occurs material may be eroded from one switch contact and possibly deposited on the other, The points in older car ignition systems required regular readjustment as a result. If equipment might be damaged by exposure to high voltages generated then precautions must be taken to provide a safe route for this energy. A diode can be connected across the relay coil so that when the relay switches off, the back e.m.f drive a current through the diode.

This is why you should always close down you pen drives rather than pulling them out of the drive.

Experiment to show build up current in an inductive circuit

R S _ + B2 L B1	The switch is closed and the variable resistor adjusted until the lamps B1 and B2 have the same brightness. The supply is switched off.

The supply is switched on again and the brightness of the lamps observed.

Lamp B2 lights up immediately. There is a time lag before lamp B1 reaches its maximum brightness. An e.m.f. is induced in the coil because the current in the coil is changing. This induced e.m.f. opposes the change in current, and it is called a back e.m.f. It acts against the increase in current, hence the time lag.

The experiment is repeated with an inductor of more turns. Lamp B1 takes longer to light fully. If the core is removed from the inductor, lamp B1 will light more quickly.

Induced e.m.f. when the current in a circuit is switched off

When the current in a circuit, containing an inductor, is switched off the magnetic field through the inductor will fall very rapidly to zero. There will be a large change in the magnetic field leading to a large induced e.m.f. For example a car ignition coil produces a high e.m.f. for a short time when the circuit is broken.

Lighting a neon lamp

The 1.5 V supply in the circuit below is insufficient to light the neon lamp. A neon lamp needs about 80 V across it before it will light.

1.5 V The switch is closed and the current builds up to its maximum value. When the switch is opened, the current rapidly falls to zero. The magnetic field through the inductor collapses (changes) to zero producing a very large induced e.m.f. for a short time. The lamp will flash.

DO QUESTION FROM 2007 paper!

Self inductance

A current in a coil sets up a magnetic field through and round the coil. When the current in the coil changes the magnetic field changes. A changing magnetic field induces an e.m.f across the coil. This is called a self induced e.m.f. because the coil is inducing an e.m.f. in itself due to its own changing current.

The coil, or inductor as it is called, is said to have the property of inductance, L.

The inductance of an inductor depends on its design. Inductance is a property of the device itself, like resistance of a resistor or capacitance of a capacitor. An inductor will tend to have a large inductance if it has many turns of wire, a large area and is wound on an iron core.

Electromagnetic induction was investigated independently by the English physicist Michael Faraday and the German physicist Heinrich Lenz in the mid-19th century. The laws which bear their names tell us the magnitude and direction of the induced e.m.f. produced by electromagnetic induction.

Faraday's law of electromagnetic induction states that the magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux through the coil or circuit.

Lenz's law states that the induced current is always in such a direction as to oppose the change that is causing it.

These two laws are summed up in the relationship

where is the induced e.m.f. Lenz's law is essentially a statement of conservation of energy: to induce a current, we have to put work into a system.

Looking back at, Faraday's law tells us that the faster we move the magnet up and down, the larger the induced e.m.f. will be. A current around a coil produces its own magnetic field and Lenz's law tells us that this field will cause a force that will oppose the motion of the bar magnet towards the coil. Similarly, when the bar magnet is being removed from the coil, the induced current causes an attractive force on the bar magnet, again opposing its motion.

Lenz’s law

https://www.youtube.com/watch?v=nqMnDfNWlLM&ebc=ANyPxKrdhDJj_H3K0Um74TBtrD3q9PaSKnXxPNac6sUoEOKwgK6A44dv1azJpfqavepKepTyNhej2PO_7t35BpfufZVJDWVAsA

Lenz’s Law states that the induced e.m.f opposes the change in current causing it.

Lenz’s laws summarises this. The induced e.m.f. always acts in such a direction as to oppose the change which produced it. Anything which causes the magnetic field in a coil to change will be opposed.

This emf opposing the change in current is often called back emf.

Conservation of Energy and Direction of Induced e.m.f.

In terms of energy the direction of the induced e.m.f. must oppose the change in current. If it acted in the same direction as the increasing current we would be able to produce more current for no energy! This would violate the conservation of energy.

The source has to do work to drive the current through the coil. It is this work done which appears as energy in the magnetic field of the coil and can be obtained when the magnetic field collapses, e.g. the large e.m.f. generated for a short time across the neon lamp.

Uses

An inductor is sometimes called a ‘choke’ because of its opposing effects.

However it must be remembered that when the current decreases the effect of an inductor is to try and maintain the current. Now the induced e.m.f. acts in the same direction as the current.

An airport metal detector uses eddy currents to generate a magnetic field, and it is this field that is actually detected.

Figure: Schematic of an airport metal detector

Each passenger passes between two coils . The steady current in the transmitter coil creates a magnetic field B. If a passenger walking between the coils is carrying a metal object, then eddy currents are induced in the object, and these currents in turn produce their own (moving) magnetic field. This new magnetic field induces a current in the receiver coil, triggering the alarm.

Magnetic induction/ flux - an aside to clarify terminology

Magnetic induction (flux) may be thought of as the number of lines of magnetic field which pass through a coil.

Large magnetic flux,
many lines through the coil Small magnetic flux, only a few lines through the coil

Faradays laws refer to the magnetic flux  rather than the magnetic induction B. His two laws are given below.

1.	When the magnetic flux through a circuit is changing an e.m.f. is induced.

2.	The magnitude of the induced e.m.f. is proportional to the rate of change of the magnetic flux.

Definition of inductance and of back e.m.f.

Energy stored by an inductor

Energy stored by an Inductor

In situations where the current in an inductor is suddenly switched off large e.m.f.s are produced and can cause sparks. At the moment of switch off the change in current is very large. The inductor tries to maintain the current as the magnetic field collapses and the energy stored by the magnetic field is given up. A magnetic field can be a source of energy. To set up the magnetic field work must have been done.

Equation for the energy stored in an inductor

For an inductor with a current I the energy stored is given by the equation below:

Energy = L I2

where L is the inductance of the inductor and I the steady current.

Example

An inductor is connected to a 6.0 V direct supply which has a negligible internal resistance. The inductor has a resistance of 0.8 . When the circuit is switched on it is observed that the current increases gradually. The rate of growth of the current is 200 A s-1 when the current in the circuit is 4.0 A.

L resistance of inductor 0.8  _ + 6 V

(a)	Calculate the induced e.m.f. across the coil when the current is 4.0 A.

(b)	Hence calculate the inductance of the coil.

(c)	Calculate the energy stored in the inductor when the current is 4.0 A.

(d)	(i)	When is the energy stored by the inductor a maximum?

(ii)	What value does the current have at this time?

Solution

(a)	Potential difference across the resistive element of the circuit V	= I R

= 4 x 0.8	= 3.2 V

Thus p.d. across the inductor = 6.0 - 3.2 = 2.8 V

(b)	Using	E = - L gives L = = 0.014 H = 14 mH

(c)	Using	E = L I2 = 0.5 x 0.014 x 4 x 4 = 0.11 J

(d)	(i)	The energy will be a maximum when the current reaches a steady value.

(ii)	Imax = = = 7.5 A

The energy stored in the magnetic field of an inductor can itself be a source of e.m.f. When the current is switched off, there is a rapid change in current so a self-induced e.m.f. (or back emf) will appear across the inductor opposing the change in current. The energy used to create this e.m.f. comes from the energy that has been stored in the magnetic field.

Inductors in a.c. circuits

In an a.c. circuit the current is continually changing. This means that the magnetic field through the inductor is continually changing. Hence an e.m.f. is continually induced in the coil.

Consider the applied alternating voltage at the point in the cycle when the voltage is zero. As the current tries to increase the induced e.m.f. will oppose this increase. Later in the cycle as the voltage decreases the current will try to fall but the induced e.m.f. will oppose the fall. The induced e.m.f. produced by the inductor will continually oppose the current.

If the frequency of the applied voltage is increased then the rate of change of current increases. The magnitude of the induced e.m.f. will also increase. Hence there should be a greater opposition to the current at a higher frequency.

Frequency response of inductor

A	An inductor is connected in series with an alternating supply of variable frequency and constant amplitude. Readings of current and frequency are taken.

As the frequency is increased the current is observed to decrease. The opposition to the current is greater at the higher frequencies. Graphs of current against frequency and current against 1/frequency are shown below.

0 current frequency	0 current 1/frequency

Note: these graphs show the inductive effects only. Considering the construction of an inductor, it is likely that the inductor has some resistance. A 2400 turns coil has a resistance of about 80 . The opposition to the current at ‘zero’ frequency will be the resistance of the inductor. In practice if readings were taken at low frequencies, the current measured would be a mixture of the inductive and resistive effects.

An inductor can be used to block a.c. signals while transmitting d.c. signals, because the inductor produces large induced e.m.f.s at high frequencies.

For a capacitor in an a.c. circuit the current increases when the frequency increases. The inductor has the opposite effect to a capacitor.

Reactance

The reactance of a capacitor or an inductor is its opposition to alternating current.

The resistance of a resistor is given by R = .

The reactance of an inductor or capacitor is given by X = (XL or XC respectively).

The opposition of a capacitor, the capacitive reactance, decreases as the frequency increases. At high frequencies there is a greater transfer of charge per second on and off the capacitor plates. For an inductor, the inductive reactance increases with frequency. At higher frequencies there is a larger induced e.m.f. opposing the current.

Note: In inductive and capacitive circuits reactances and resistance cannot be added.

Uses of inductors and capacitors in a.c. circuits

The circuit below shows a capacitor and inductor in series with an alternating supply.

L C VL VC At low frequencies the opposition to the current by the inductor is low, so the p.d across L will be low. At low frequencies the opposition to the current by the capacitor is high so the p.d. across C will be high. At low frequencies XL<XC.

At high frequencies, the reverse is the case. The p.d. across the inductor VL will be the higher than the p.d. across the capacitor. At high frequencies XL>XC.

This circuit could be used to filter high and low frequency signals.

The transformer - this discussion is for interest only

The principle of operation of the transformer can be given in terms of induced e.m.f.

S a.c. ammeter primary coil secondary coil a.c. supply soft iron core S centre zero meter primary coil secondary coil d.c. supply soft iron core
When S is closed, the meter needle ‘kicks’ momentarily then returns to zero. When the current is steady the meter reads zero. When S is opened, the meter needle kicks briefly in the opposite direction and returns to zero. A changing magnetic field is produced in the coil when the current changes. This changing magnetic field will produce an induced e.m.f. during this short time. However when the current is steady, there is no changing field hence no induced e.m.f.	With an a.c. supply the current is continually changing. This sets up a continually changing magnetic field in the soft iron core. Hence an induced e.m.f. is produced in the secondary coil. From the conservation of energy the direction of the induced e.m.f. will oppose the change which sets it up. Hence the direction of a current in the secondary, at any time, will always be in the opposite direction to the current in the primary.

We can now understand why a transformer only operates with an alternating supply.

The transformer is an example of mutual inductance.

ELECTROMAGNETISM- HOMEWORK

1. Define the henry, the unit of self induction.	(1)

2. A circuit contains a switch, a d.c. power supply and an inductor.

Draw graphs to show the growth and decay of current when the switch is

a) closed	b) opened	(2)

3. Describe how you could measure the self inductance of a coil.	(3)

4. Find the e.m.f. induced in a 0.1 H coil when the current is changing at a rate of 80 A/s.	(3)

5. A particular coil has a 10 V e.m.f. induced in it when the current is changing at 50 A/s.

What is the self inductance of the coil?	(3)

6. A resistor and a 0.2 H inductor are connected in series with a 6 V dc power supply.

a) What is the initial back e.m.f. at the instant the supply is connected?	(1)

b) At a time later the pd across the resistor is 4 V. What is the rate of change of current at this time?	(3)

7. An inductor of self inductance 0.2 H and negligible resistance is connected in series with a 9 Ω resistor, a cell and a switch. The cell has an e.m.f. of 1.5 V and an internal resistance of 1 Ω.

a) Calculate the initial growth of current.	(2)

b) Calculate the final current.	(3)

c) Calculate the rate of growth of current when the current is 0.1 A.	(2)

8. A 0.2 H inductor is connected to a 12 V dc supply. The inductor has a resistance of 2 Ω.

a) When is the energy stored in the inductor a maximum?	(1)

b) What is the current when this happens?	(2)

c) What is the maximum energy stored in the inductor?	(2)

http://www.pa.msu.edu/courses/2000spring/PHY232/lectures/induction/self.html

Inductors in a.c. circuits

In an a.c. circuit the current is continually changing. This means that the magnetic field through the inductor is continually changing. Hence an e.m.f. is continually induced in the coil.

Consider the applied alternating voltage at the point in the cycle when the voltage is zero. As the current tries to increase the induced e.m.f. will oppose this increase. Later in the cycle as the voltage decreases the current will try to fall but the induced e.m.f. will oppose the fall. The induced e.m.f. produced by the inductor will continually oppose the current.

If the frequency of the applied voltage is increased then the rate of change of current increases. The magnitude of the induced e.m.f. will also increase. Hence there should be a greater opposition to the current at a higher frequency.

Frequency response of inductor

A	An inductor is connected in series with an alternating supply of variable frequency and constant amplitude. Readings of current and frequency are taken.

As the frequency is increased the current is observed to decrease. The opposition to the current is greater at the higher frequencies. Graphs of current against frequency and current against 1/frequency are shown below.

0 current 1/frequency 0 current frequency

Note: these graphs show the inductive effects only. Considering the construction of an inductor, it is likely that the inductor has some resistance. A 2400 turns coil has a resistance of about 80 . The opposition to the current at ‘zero’ frequency will be the resistance of the inductor. In practice if readings were taken at low frequencies, the current measured would be a mixture of the inductive and resistive effects.

An inductor can be used to block a.c. signals while transmitting d.c. signals, because the inductor produces large induced e.m.f.s at high frequencies.

For a capacitor in an a.c. circuit the current increases when the frequency increases. The inductor has the opposite effect to a capacitor.

Inductive reactance.

The reactance of a capacitor or an inductor is its opposition to alternating current.

The resistance of a resistor is given by R = .

The reactance of an inductor or capacitor is given by X = (XL or XC respectively).

The opposition of a capacitor, the capacitive reactance, decreases as the frequency increases. At high frequencies there is a greater transfer of charge per second on and off the capacitor plates. For an inductor, the inductive reactance increases with frequency. At higher frequencies there is a larger induced e.m.f. opposing the current.

Current in an inductive circuit is determined by I = V/XL and therefore XL = V/I. XL is measured in ohms, the same as resistance, but it is not appropriate to refer to the opposition to current in a capacitive circuit as a resistance.

I 1/f and XL 1/I therefore XL f.

XL = 2fC

In inductive and capacitive circuits reactances and resistance cannot be added arithmetically. There are phase differences between inductive and capacitive reactances and resistances and therefore vector addition must be used. The total impedance Z, measured in ohms, in an a.c. circuit is given by:

Use of an appropriate relationship to solve problems involving back e.m.f., inductance (self-inductance) and rate of change of current.

XL = 2fC

Energy = L I2

Use of appropriate relationships to solve problems relating to inductive reactance, voltage, current, frequency, energy and inductance (self-inductance).

#Uses of inductors and capacitors in a.c. circuits

The circuit below shows a capacitor and inductor in series with an alternating supply.

L C VL VC

At low frequencies the opposition to the current by the inductor is low, so the p.d across L will be low. At low frequencies the opposition to the current by the capacitor is high so the p.d. across C will be high. At low frequencies XL<XC.

At high frequencies, the reverse is the case. The p.d. across the inductor VL will be the higher than the p.d. across the capacitor. At high frequencies XL>XC.

This circuit could be used to filter high and low frequency signals.

Capacitor C1 allows high frequency signals to pass to loudspeaker LS1. High frequency signals can also pass more easily through capacitor C2 than loudspeaker LS2. Low frequency signals are ‘blocked’ by C1 and C2 but pass easily through inductor L to loudspeaker LS2. Cross-over Networks in Loudspeakers

C1 LS1 LS2 L C2

Capacitors in Radio Circuits

L C1 C2 aerial diode (decoder) to amplifier

Capacitor C1 is a variable capacitor, which when used in conjunction with inductor L, allows the radio to be tuned to one particular radio frequency. Capacitor C2 allows the high frequency radio carrier signal to flow to earth but ‘blocks’ the low frequency audio signal which must then pass on to the amplifier and loudspeaker of the radio.

Amplifier Bias Network

+Vs Vin 0 V Vout When an a.c. signal is to be amplified by a simple transistor amplifier the a.c. signal should be input to the transistor via a capacitor. The capacitor will allow the a.c. signal to pass but ‘block’ any unwanted d.c. signal.

The transformer

The principle of operation of the transformer can be given in terms of induced e.m.f.

S centre zero meter primary coil secondary coil d.c. supply soft iron core S a.c. ammeter primary coil secondary coil a.c. supply soft iron core
When S is closed, the meter needle ‘kicks’ momentarily then returns to zero. When the current is steady the meter reads zero. When S is opened, the meter needle kicks briefly in the opposite direction and returns to zero. A changing magnetic field is produced in the coil when the current changes. This changing magnetic field will produce an induced e.m.f. during this short time. However when the current is steady, there is no changing field hence no induced e.m.f.	With an a.c. supply the current is continually changing. This sets up a continually changing magnetic field in the soft iron core. Hence an induced e.m.f. is produced in the secondary coil. From the conservation of energy the direction of the induced e.m.f. will oppose the change which sets it up. Hence the direction of a current in the secondary, at any time, will always be in the opposite direction to the current in the primary.

We can now understand why a transformer only operates with an alternating supply.

The transformer is an example of mutual inductance.

Summary

You should now be able to:

sketch graphs showing the growth and decay of current in a simple d.c. circuit containing an inductor;

describe the principles of a method to illustrate the growth of current in a d.c. circuit;

state that an e.m.f. is induced across a coil when the current through the coil is varying;

explain the production of the induced e.m.f across a coil;

explain the direction of the induced e.m.f in terms of energy;

state that the inductance of an inductor is one henry if an e.m.f. of one volt is induced when the current is changing at a rate of one ampere per second;

use the equation and explain why a minus sign appears in this equation;

state that the work done in building up the current in an inductor is stored in the magnetic field of the inductor, and that this energy is given by the equation ;

calculate the maximum values of current and induced e.m.f. in a d.c. LR circuit;

use the equations for inductive reactance and ;

describe the response of an a.c. inductive circuit to low and high frequency signals.

results obtained are used to produce the following graphs.

Simple a.c. inductive circuit

Self Inductance (additional material)

Whenever you change the current in a coil of wire, the magnetic field it produces will change. That will change the magnetic flux through the coil and hence produce a voltage across the coil. This phenomenon is called self-inductance, and the coil is referred to as an inductor.

The quantitative measure of this effect is the inductance L defined by

An inductor can store energy. The power (= energy / time) being stored in an inductor is

This implies (by a little calculus) that the energy stored in an inductor is

As an example, we can derive the inductance of a solenoid. The magnetic field inside a solenoid of length l is

as found previously. The flux through each turn is equal to B times the area A of the turn. The voltage across the solenoid is given by the rate of change of this flux times the number of loops:

Therefore, the inductance L can be given in terms of the geometry of the solenoid.

Longer solenoids and solenoids with more cross-sectional area have more inductance. The strongest dependence is on the number of turns per unit length. One can also increase the inductance by using iron instead of a vacuum; in that case, the parameter 0 would be replaced by a larger value  .