North Berwick High School

Department of Physics

Higher Physics

Unit 2 Electricity

Section 1 Voltage, Current and Resistance

Section 1 Voltage, Current and Resistance

Note Making

Make a dictionary with the meanings of any new words.

Alternating current1. State the difference between a.c. and d.c. and the main way in which

a.c. is produced.2. Copy both graphs.

Measuring frequency and peak voltage1. Copy the worked example.

Alternating current – peak and rms1. Define the term rms and copy both formulae.

Graphical method

1. Copy the 3 important notes.

Current, voltage, power and resistance1. Define the terms current and voltage.2. State the meanings of emf and pd (not just what the letters

stand for).

Sources of emf1. State some sources of emf.

Ohm’s law1. You do not need to define Ohm's law but you must be able to

use the equation.

Electric power

1. State the 3 equations for power.2. State the significance of I2R.3. Copy the example if you think it will help.4. You should already know the circuit rules but copy the table if

you don't.5. Copy the addition of power sentence.

Voltage and current by proportion1. Copy the examples. Make sure that you understand them.

The potential divider

1. State the function of a potential divider circuit.2. Copy the example and calculations of the potential divider

circuit. Note that you can also use the formulae on the next page.

3. Copy the diagram with the variable resistor and explain how it works.

4. You should also copy the worked example (a former exam question).

Electrical sources and internal resistance1. Describe what is meant by internal resistance, lost volts and

tpd.2. Explain how the emf of a cell may be measured.3. Copy the under load example. (It will help you visualise some

terms.)4. Show how the internal resistance equation is produced.5. Show how the short circuit equation is produced.

Measuring E and r by a graphical method1. Copy both graphs and questions 14 and 16 to explain both

methods. (pages 53 and 55)

Capacitors

1. What is meant by capacitance.2. Copy the graph, equation and the formal definition of

capacitance.

Energy stored in a capacitor

1. Copy the diagrams to show a capacitor charging.2. Explain why work has to be done when charging a capacitor.3. Copy the graph of Q v V and the energy equations. (Read the

paragraph in the middle of this page.

Charging and discharging a capacitor

1. Copy the charging and discharging graphs. You must be able to clearly explain why they are this shape.

Factors affecting the rate of charge and discharge1. Copy both graphs and again, explain their shapes.

Section 1 Voltage, Current and Resistance

Contents

Content Statements....................................................................................1

Alternating current......................................................................................2

Measuring frequency and peak voltage...................................................2

Alternating current – peak and rms..........................................................4

Graphical method to derive the relationship between peak and rms values of alternating current.....................................................................6

Current, voltage, power and resistance...................................................9

Sources of emf...........................................................................................10

Ohm’s law...................................................................................................11

Electric power............................................................................................12

Circuit rules................................................................................................14

Addition of power.....................................................................................14

Voltage and current by proportion.........................................................15

The potential divider................................................................................19

Electrical sources and internal resistance.............................................23

Under load.................................................................................................25

Measuring E and r by a graphical method.............................................27

Second graphical method........................................................................28

Capacitors..................................................................................................31

Relationship between charge and pd.....................................................31

Energy stored in a capacitor....................................................................33

Charging and discharging a capacitor....................................................35

Factors affecting the rate of charge and discharge..............................37

Problems.....................................................................................................39

Solutions.....................................................................................................68

Content Statements

Content Notes Context

a) Monitoring and measuring a.c.

b) Current, potential difference (p.d.), power and resistance

c) Electrical sources and internal resistance.

d) Capacitors

a.c. as a current which changes direction and instantaneous value with time. Monitoring a.c. signals with an oscilloscope, including measuring frequency, and peak and r.m.s. values.

Current, potential difference and power in series and parallel circuits. Calculations involving p.d., current and resistance may involve several steps.Potential dividers as voltage controllers.

Electromotive force, internal resistance and terminal potential difference. Ideal supplies, short circuits and open circuits. Determining internal resistance and electromotive force using graphical analysis.

Capacitors and the relationship between capacitance, charge and potential difference.The total energy stored in a charged capacitor is the area under the charge against potential difference graph. Use the relationships between energy, charge, capacitance and potential difference.Variation of current and potential difference against time for both charging and discharging. The effect of resistance and capacitance on charging and discharging curves.

Using a multimeter as an ammeter, voltmeter and ohmmeter.Oscilloscope as a voltmeter and waveform monitor.

Investigating simple a.c. or d.c. circuits with switches and resistive components. Potential dividers in measurement circuits and used with variable resistors to set and control voltages in electronic circuits.

Investigating the reduction in voltage as additional devices are connected.Investigating internal resistance of low voltage power supplies.Load matching.

Energy storage. Flash photography.Smoothing and suppressing. Capacitance based touch screens.

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Section 1 Voltage, Current and Resistance

Alternating currentElectric current is the rate of electron flow past a given point in an electric circuit, measured in Coulombs/second which is named Amperes.

An alternating current (a.c.) periodically changes direction. Contrast this to direct current (d.c.), which only flows in one direction. Although it is possible that current could vary in any way, we will consider a sinusoidal function, ie the voltage constantly changes in the form of a sine wave. This is the most common form of a.c. since electricity generators work by spinning in circles, which means that the current is pushed one way then the other, with the instantaneous value of voltage constantly varying. Examples of each type of current when displayed on an oscilloscope are shown below.

a.c. waveform d.c. waveform

Measuring frequency and peak voltageAn oscilloscope plots a graph of changing voltage on the y-axis versus time on the x-axis. Just like a graph we need to read the scale, except the numbers aren't along the axis, they are on a dial underneath the

2

Timebase = 5 ms cm–1

Volts/div = 2 V cm–1

screen. The voltage axis scale (y) is usually called the volts/div or volts/cm and the time axis scale (x) is called the timebase.

To calculate the frequency of an a.c. signal, we first of all have to find its period. This is the time for one complete cycle of to and fro current, so we measure the horizontal distance on the screen between crests.

An alternating voltage varies between the same maximum negative and positive value as the voltage pushes first one way (positive) then the other (negative). This maximum is defined as the peak voltage. It can be measured from an oscilloscope by measuring the vertical distance on the screen from the bottom of the signal to the top. This then has to be halved to get the peak voltage.

Worked example

In the picture below each box on the CRO screen has a side of 1 cm.

The distance between crests is 4 cm. The distance from bottom to top is 8 cm.

The time base was set at 5 ms cm–1, The volts/div was set at 2 V cm–1,

3

then the period of the wave is: then the peak voltage is:

T = 4 × 5 ms = 4 × 0.005 = 0.02 s Vpeak = ½ × 8 × 2

Vpeak = 8 V

f = 50 Hz

If the time base is switched off, the a.c. signal will not be spread along the x-axis. However, the voltage variation will continue, meaning a straight vertical line will be displayed on the screen. The peak voltage can still be calculated in exactly the same way as above, ie halved then converted into voltage using the volts/div.

Alternating current – peak and rmsElectricity is a method of transferring energy, so a.c. is capable of transferring energy in the same way as d.c. The instantaneous amount of power being supplied can be calculated using the value of voltage and/or current at a particular point. However, in a.c. these values change constantly, completing 50 cycles every second in the UK. Our eye

4

A A

couldn’t possibly follow such quick changes. We need an average, but the average value of the voltage during any complete cycle is zero!

We know that homes, schools and industry use a.c. supplies, so there must be an average that is not zero. The ‘average’ value of an a.c. voltage is called the root mean square (rms) voltage (Vrms).

The definition of the rms voltage is that it is that value of direct voltage that produces the same power (eg heating or lighting) as the alternating voltage. The rms value is what is quoted on a power supply so that a fair comparison between a.c. and d.c. can be made, eg a 6 V battery will produce the same brightness of light bulb as a 6 V rms a.c. supply. The rms current has a similar definition.

Consider the following two circuits, which contain identical lamps.

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The variable resistors are altered until the lamps are of equal brightness. As a result the d.c. has the same value as the effective a.c. (ie the lamps have the same power output). Both voltages are measured using an oscilloscope, giving the voltage equation below.

Also, since V = IR applies to the rms valves and to the peak values, a similar equation for currents can be deduced.

Note: A multimeter switched to a.c. mode will display rms values.

Graphical method to derive the relationship between peak and rms values of alternating currentThe power produced by a current I in a resistor of resistance R is given by I2R. A graph of I2 against t for an alternating current is shown below. A similar method can be used for voltage.

The average value of I2

is and therefore the average power

supplied is .

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An identical heating effect (power output) for a d.c. supply is I rms2R (since

Irms is defined as the value of d.c. current that will supply equivalent power).

Setting both of these equal to each other gives:

Important notes:

1. Readings on meters that measure a.c. are rms values, not peak values.

2. For power calculations involving a.c. always use rms values:

3. The mains supply is usually quoted as 230 V a.c. This is of course 230 V rms. The peak voltage rises to approximately 325 V. Insulation must be provided to withstand this peak voltage.

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Example

A transformer is labelled with a primary of 230 V rms and secondary of 12 Vrms. What is the peak voltage which would occur in the secondary?

Vpeak = √2 × Vrms

Vpeak = √2 × 12

Vpeak = 17.0 V

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Current, voltage, power and resistanceBasic definitions

In the circuit below, when S is closed the free electrons in the conductor experience a force that causes them to move. Electrons will tend to drift away from the negatively charged end towards the positively charged end.

Wire magnified(free electrons)

S

Current is a flow of electrons and is given by the flow of charge (coulombs) per second:

I=Qt

The energy required to drive the electron current around this circuit is provided by a chemical reaction in the battery. The electrical energy that is supplied by the source is transformed into other forms of energy in the components that make up the circuit.

Voltage is defined as the energy transferred (work done) per unit charge.

V = W/Q

In this section W will be used for the work done, ie energy transferred, therefore 1 volt = 1 joule per coulomb (J C–1).

When energy is being transferred from an external source to the circuit, the voltage is referred to as an electromotive force (emf). When energy is transformed into another form of energy by a component in the circuit, the voltage is referred to as a potential difference (pd).

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Energy is supplied to the circuit:chemical electricalAn emf (voltage) can be measured

Energy is provided by the circuit:electrical l ight + heatA pd (voltage) can be measured

Sources of emfEmfs can be generated in a great variety of ways. See the table below for examples.

Chemical cell Chemical energy drives the current(eg battery)

Thermocouple Heat energy drives the current(eg temperature sensor in an oven)

Piezo-electric generator Mechanical vibrations drive the current(eg acoustic guitar pickup)

Solar cell Light energy drives the current(eg solar panels on a house)

Electromagnetic generatorChanges in magnetic field drive the current(eg power stations)

Ohm’s law

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Current

Voltage

In any circuit, providing the resistance of a component remains constant, if the potential difference V across the component increases, the current I through the component will increase in direct proportion.

At a fixed temperature, for a given conductor:

V ¿ I, ie V/I is constant

The constant of proportion is defined to be the resistance,

ie R=V

I or V=IR

This is Ohm’s law.

A component that has a constant resistance when the current through it is increased is said to be ohmic. Some components do not have a constant resistance; their resistance changes as the pd across the component is altered, for example a light bulb, a transistor or a diode. A graph of V versus I for such components will not be a straight line.

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Electric powerFor a given component P = IV since:

power = work done = QV = Q =IV time t t

SinceV = IR

P = I × IR (substituting for V)

= I 2 R

or

P = V x V R

P = V 2 R

The expression I2R gives the energy transferred in one second due to resistive heating. Apart from obvious uses in electric fires, cookers, toasters etc, consideration has to be given to heating effects in resistors, transistors and integrated circuits, and care taken not to exceed the maximum ratings for such components. The expression V 2/R is particularly useful when the voltage of a power supply is fixed and you are considering changes in resistance, eg different power of heating elements.

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x V

15 Ω

230 V

30 Ω

Example

An electric heater has two heating elements allowing three settings: low, medium and high. Show by calculation which element(s) would be switched on to provide each power setting.

for 30 Ω: = = 230 2 1800 W → low power 30

for 15 Ω: = = 230 2 3500 W → medium power 15

for both:

1 = 1 + 1 = 1 Rp 15 30 10

Rp = 10 Ω

power

= 230 2 5300 W → high power 10

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P = V 2 R

P = V 2 R

P = V 2 R

R

R

≡ ½R

Circuit rules

Series Parallel

Current

Voltage Vs = VB1 = VB2 = VB3

Resistance

Addition of powerBy conservation of energy, the total power used in both series and parallel circuits is the sum of the power used in each component.

Useful to remember

Identical resistors in parallel

Two resistors of the same value (R) wired in parallel:

RP = half the value of one of them (½R).

eg 2 × 1000 Ω in parallel, RP = 500 Ω

2 × 250 Ω in parallel, RP = 125 Ω

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2 Ω

4 Ω

3A

8 Ω 4 Ω 12 Ω

12 V

Voltage and current by proportionThe voltage across an individual resistor in a series circuit is proportional to the resistance. The current through one of two resistors in parallel is inversely proportional to the resistance, ie proportional to the other resistance.

Examples

Current in parallel Voltage in series

3A will split in the ratio: RS = 8 + 4 + 12 = 24 Ω

24+2

: 44+2

, ie 13

: 23

V 1=8

24×12=4 V

13׿ ¿

3 A = 1 AV 2=

424

×12=2 V

23׿ ¿

3 A = 2 AV 3=

1224

×12=6 V

The smallest resistor takes the largest current, so:

I1 = 2 A, I2 = 1 A

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6 Ω 4 Ω

5 Ω

40 V

10 Ω

Worked example 1

In the circuit shown opposite calculate:

(a) the current in each resistor

(b) the pd across each resistor

(c) the power dissipated in each resistor.

Solution(a) 4 Ω and 6 Ω in series are equivalent to 10 Ω so the resistor network

can be represented as:

1Rp

= 110

+ 110

= 15

RP = 5 Ω

RT = 5 + 5 = 10 Ω

I T=V TRT

+ 4010

=4 A

I1 meets a resistance of 10 Ω and so does I2.

so I1 = I2 = 2 A

(b) For the 5 Ω resistor: V = IR = 4 × 5 = 20 V

for the 4 Ω resistor: V = IR = 2 × 4 = 8 V

for the 6 Ω resistor: V = IR = 2 × 6 = 12 V

for the 10 Ω resistor: V = IR = 2 × 10 = 20 V

16

(current through 5 Ω)

(current through 4 Ω, 6 Ω and 10 Ω)

12 V

M

(c) For 5 Ω: P = IV = 20 × 4 = 80 W

for 4 Ω: P = I2R = 22 × 4 = 16 W

for 6 Ω: P = = = 24 W

for 10 Ω: P = IV = 20 × 2 = 40 W

(note: you can use any of the electrical power formulae that are appropriate.)

Worked example 2A 10 W model car motor operates on a 12 V supply. In order to slow the car it is connected in series with a controller (rheostat).

The controller is adjusted to reduce the motor’s power to 4 W.

Assuming that the resistance of the motor does not change, calculate:

(a) the resistance of the controller at this setting

(b) the power wasted in the controller.

Solution(a) We cannot find the resistance of the controller directly. We need

to find the resistance of the motor and the total resistance of the circuit when the motor is operating at 10 W.

For the motor at 10 W, P = 10 W, V = 12 V, R = ?

P=V2

R , so R=V

2

P=122

10

R = 14 Ω

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V 2 R

12 2 6

For the motor at 4 W, P = I2R, so I = √ PR=√ 414⋅4

= 0.53 A

For the whole circuit: RT = VT = 12 = 23 Ω

I 0.527

Resistance of controller = 22.8 – 14.4 = 8.4 Ω

(b) Power in controller = I2R = 0.5272 × 8.4 = 2.3 W

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10 Ω 20 Ω

6 V

V1 V2

The potential dividerA potential divider provides a convenient way of obtaining a variable voltage from a fixed voltage supply.

ExampleConsider first two fixed resistors, R1 = 10 Ω and R2 = 20 Ω, connected in series across a 6 V supply:

The current in this circuit can be calculated using Ohm’s law (total R = 10 + 20 = 30 Ω).

so,I=V SR

= 630

= 0.2 A

The voltage across R1 V1 = IR1 = 0.2 × 10

= 2 V

The voltage across R2 V2 = IR2 = 0.2 × 20

= 4 V

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If you look closely you can see that the voltages across the resistors are in the ratio of their resistances (½ in this case),

ie,

V 1

V 2=R1

R2

The voltage across R1 is given by:

V 1=R1

R1+R2V S

similarly, the voltage across R2 is

V 1=R2

R1+R2V S

This is known as the potential divider rule.

A variable resistor connected as shown on the next page provides another way of changing the ratio R1/R2. The resistance between A and W represents R1 and that between W and B represents R2.

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A

W W

A

6 V

B

A variable voltage (eg from 0 to 6 V) is available between A and W as the wiper (or slider) is rotated. For example, it will be 3 V when the wiper is halfway round the track.

Note that the above only holds true as long as the load connected across AW has a very high resistance. If it has a resistance comparable with that of AW the situation becomes more complicated. The effective resistance at the output must then be calculated before the potential divider rule is applied.

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A10 kΩ

Y

X

12 V

20 kΩ

Worked example

(a) Calculate the voltage across XY.

R1 = 10 kΩ R2 = 20 kΩ VS = 12 V V1 = ?

V 1=R1

R1+R2V S

V 1=1010+20

12

V1 = 4 V

(b) The circuit attached to XY has a resistance of 10 kΩ. Calculate the effective resistance between XY and thus the voltage across XY.

The resistance between XY comes from the two identical 10 kΩ resistors in parallel. The combined resistance is therefore 5 kΩ. Use this resistance as R1 in the potential divider formula.

R1 = 5 kΩ R2 = 20 kΩ VS = 12 V V1 = ?

V 1=R1

R1+R2V S

V 1=5

5+2012

V1 = 2.4 V

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Electrical sources and internal resistanceUp to this stage in your study of physics we have assumed that power supplies are ideal. This means that their voltage remains constant and they can supply any current we wish if we connect the correct resistance. In most cases these are good assumptions, but if you take a small battery and connect bulbs one after another in parallel you will notice that eventually their brightness dims. A similar effect can be noticed if you start a car with the headlights on. Both these situations have one thing in common: they involve significant current being supplied by the battery. If you are able to touch the battery you may notice it getting warm when in use. This is a wasted transfer of energy and can be modelled by considering that the battery, like all conductors, has some resistance of its own. This model explains why some of the chemical energy converted is dissipated as heat and is not available to the circuit – resistors convert electrical energy to heat energy. We say that the power supply has an internal resistance, r. In most cases this is so small (a few ohms) that it can be considered negligible. However, when the current in the circuit is large, meaning the external resistance is small, its effects can be significant. In the past we have assumed that any cell could deliver an unlimited current, but clearly this is not the case. The greater the current the more energy will be dissipated in the power supply until eventually all the available energy (the emf) is wasted and none is available outside the power supply.

Energy will be wasted in getting the charge through the supply (this energy appears as heat) and so the energy per unit charge available at the output (the terminal potential difference or tpd) will fall. There will be ‘lost volts’. The lost volts = Ir.

Note: A common confusion is that Ir stands for internal resistance. r is the symbol for internal resistance and Ir is the value of the lost volts.

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E

r

Real cel l

Internal resistanceIdeal cell

E

r I = 0

The fact that a power supply has an internal resistance is not a very difficult problem to deal with because we can represent a real supply as:

real cell = ideal cell + a resistance r

Special case – open circuit (I = 0)

We now need to consider how we can measure the key quantities. Firstly, it would be very useful to know the voltage of the ideal cell. This is called the emf of the cell, which was defined previously as the energy (eg chemical energy) supplied per unit of charge. It is found by finding the voltage across the cell on ‘open circuit’, ie when it is delivering no current. This can be done in practice by using a voltmeter or oscilloscope.

Why should this be so? Look at this diagram:

Since I = 0

V across r = I × r = 0 × r

= 0

So no voltage is dropped across the internal resistance and a voltmeter across the real cell would register the voltage of the ideal cell, the emf.

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I = 0.5 A

E

r

11.4 Ω

V

0.6 Ω

Under loadWhen a real cell is delivering current in a circuit we say that it is under load and the external resistance is referred to as the load.

Consider the case of a cell with an internal resistance of 0.6 Ω delivering current to an external resistance of 11.4 Ω:

Cell emf. = 6 V

current = 0.5 A

The voltage measured across the terminals of the cell will be the voltage across the 11.4 Ω resistor, ie,

V = IR = 0.5 × 11.4

= 5.7 V

and the voltage across the internal resistance:

V = Ir = 0.5 × 0.6

= 0.3 V

so the voltage across the terminals is only 5.7 V (this is the terminal potential difference, or tpd) and this happens because of the voltage dropped across the internal resistance (0.3 V in this case). This is called the lost volts.

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E

r

I

R

If Ohm’s law is applied to a circuit containing a battery of emf E and internal resistance r with an external load resistance R:

and if I is the current in the circuit and V is the pd across R (the tpd.), then using conservation of energy:

emf = tpd + lost volts

E = V across R + V across r

and: E = V + Ir

This is the internal resistance equation, which can clearly be rearranged into different forms, for example:

V = E – Ir

or, since V = IR,

E = IR + Ir

ie E = I(R + r)

Special case – short circuit (R = 0)

The maximum current is the short-circuit current. This is the current that will flow when the terminals of the supply are joined with a short piece of thick wire (ie there is no external resistance).

By substituting R = 0 in the above equation we get:

E = I(0 + r)

ie E = Ir

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Current

Voltage

Gradient (m) = –r ( internal resistance)

Intercept (c) = E (emf)

Measuring E and r by a graphical methodWhen we increase the current in a circuit the tpd will decrease. We can use a graph to measure the emf and internal resistance. If we plot V on the y-axis and I on the x-axis, we get a straight line of negative gradient. From above:

V = E – Ir

V = (–r) × I + E

Comparing with the equation of a straight line

y = mx + c

we can see that the gradient of the line (m) is equivalent to –r and the y-intercept (c) is E.

This graph can also be modified to show that the external resistance (R) and internal resistance (r) act as a potential divider. As R decreases then lost volts must increase.

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Current

Voltage

r (internal resistance)

R (external resistance)

We can then use the potential divider relationships:

emf = tpd + lost volts

lost voltstpd

= rR

lost volts= rR+r

×emf

tpd= RR+r

×emf

Second graphical method

When we change the external resistance (R) in a circuit, there will be a corresponding change in the current flowing. If we plot R on the y-axis and 1/I on the x-axis, we get a straight line of positive gradient and negative intercept. From above:

E = IR + Ir

Divide both sides by I:

E/I = R + r

R = (E × 1/I) – r

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R (external resistance)

Ideal cell

1 I

Resistance

Gradient (m) = E (emf)

Intercept (c) = –r (internal resistance)

E

r

28 Ω

V

Comparing with the equation of a straight line:

y = mx + c

we can see that the gradient of the line (m) is equivalent to E and the y-intercept (c) is –r.

Worked example

A cell of emf 1.5 V is connected in series with a 28 Ω resistor. A voltmeter measures the voltage across the cell as 1.4 V.

Calculate:

(a) the internal resistance of the cell

(b) the current if the cell terminals are short circuited

(c) the lost volts if the external resistance R is increased to 58 Ω.

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or use

V1 = R1V2 R2

1.5 V

r

I2 Ω

Solution(a) E = V + Ir

In this case we do not know the current, I, but we do know that the voltage across the 28 Ω resistor is 1.4 V, and I = V/R , so I = 0.05 A.

1.5 = 1.4 + 0.05r

r= 0.10 .05

= 2 Ω

(b) Short circuit:

Total resistance = 2 Ω

I= Er=1 .5

2=

0.75 A

(c) E = I(R + r)

1.5= I(58 + 2)

= 1.5/60 = 0.025 A

lost volts = Ir

= 0.025 × 2 = 0.05 V

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V C

A B

+-

CapacitorsCapacitance is the ability (or capacity) to store charge. A device that stores charge is called a capacitor.

Practical capacitors are conductors separated by an insulator. The simplest type consists of two metal plates with an air gap between them. The symbol for a capacitor is based on this:

Relationship between charge and pd

The capacitor is charged to a chosen voltage by setting the switch to A. The charge stored can be measured directly by discharging through the coulometer with the switch set to B. In this way pairs of readings of voltage and charge are obtained.

It is found that the charge stored on a capacitor and the pd (voltage) across it are directly proportional:

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so Q = a constantV

This constant is defined as the capacitance, C:

C=QV or Q = CV

The formal definition of capacitance is therefore the charge stored per unit voltage. Notice that this means that capacitance is the gradient of the above graph. The unit of capacitance is the farad (F).

From the above formula: 1 F = 1 coulomb per volt.

The farad is too large a unit for practical purposes and the following submultiples are used:

1 mF (microfarad) = 1 × 10–6 F

1 nF (nanofarad) = 1 × 10–9 F

Note: When a capacitor is charging, the current is not constant (more on this later). This means the formula Q = It cannot be used to work out the charge stored.

Worked exampleA capacitor stores 4 × 10–4 C of charge when the potential difference across it is 100 V. Calculate the capacitance.

C=QV

=4×10−4

100=4×10−6

F (4 μF)

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Vs Vs

Electroncurrent

Vs

Electroncurrent

+-+-

Vc Vc

= Vs+- +-+-+-

Energy stored in a capacitorWhy work must be done to charge a capacitor

Consider this:

When current is switched on electrons flow onto one plate of the capacitor and away from the other plate.

This results in one plate becoming negatively charged and the other plate positively charged.

Eventually the current ceases to flow. This happens when the pd across the plates of the capacitor is equal to the supply voltage.

The negatively charged plate will tend to repel the electrons approaching it. In order to overcome this repulsion work has to be done and energy supplied. This energy is supplied by the battery. Note that current does not flow through the capacitor, electrons flow onto one plate and away from the other plate.

For a given capacitor the pd across the plates is directly proportional to the charge stored. Consider a capacitor being charged to a pd of V and holding a charge Q.

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Q

V p.d.

charge

The energy stored in the capacitor is given by the area under the graph.

This work is stored as electrical energy, so:

E = ½QV

(Contrast this with the work done moving a charge in an electric field where W = QV. In a capacitor the amount of charge and voltage are constantly changing rather than fixed and therefore the ½ is needed as an averaging factor.)

Since Q = CV there are alternative forms of this relationship:

energy = ½CV2

and energy = ½

Q2

C

Worked exampleA 40 mF capacitor is fully charged using a 50 V supply. Calculate the energy stored in the capacitor.

energy = ½CV2

= ½ × 40 × 10–6 × 502

= 0.05 J

34

A

As soon as the switch is closed there is no chargecharge on the capacitor. The current is limited only by the resistance in the circuit and can be found using Ohm’s

0

A

As soon as the switch is closed, there is no charge on the capacitor. The current is limited only by the resistance in the circuit and can be found using Ohm's law.As the capacitor charges, a pd develops across the plates which opposes the pd of the cell. As a result the supply current decreases .The capacitor becomes fully charged and the pd across the plates is equal and opposite to that across the cell and the charging current becomes zero.

Charging and discharging a capacitorCharging

Consider the following circuit:

When the switch is closed the current flowing in the circuit and the voltage across the capacitor behave as shown in the graphs below.

Consider the circuit at three different times.

35

If the cell is taken out of the circuit and the switch is set to A, the capacitor will discharge

Consider this circuit when the capacitor is fully charged, switch to position B

A A

- - - -

+ + + + B A A B

Discharging

Consider the circuit opposite in whichthe capacitor is fully charged:

If the cell is shorted out of the circuitthe capacitor will discharge.

While the capacitor is discharging, the current in the circuit and the voltage across the capacitor behave as shown in the graphs below:

Although the current/time graph has the same shape as that during charging, the currents in each case are flowing in opposite directions. The discharging current decreases because the pd across the plates decreases as charge leaves them.

A capacitor stores charge, but unlike a cell it has no capability to supply more energy. When it discharges, the energy stored will be used in the circuit, eg in the above circuit it would be dissipated as heat in the resistor.

36

Factors affecting the rate of charge and dischargeThe time taken for a capacitor to charge is controlled by the resistance of the resistor R (because it controls the size of the current, ie 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:

The effect of capacitance on The effect of resistance on charge current 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.

37

1 MΩ

VS = 10 V

2 μF

Worked exampleThe switch in the following circuit is closed at time t = 0.

(a) Immediately after closing the switch what is

(i) the charge on C?

(ii) the pd across C?

(iii) the pd across R?

(iv) the current through R?

(b) When the capacitor is fully charged what is

(i) the pd across the capacitor?

(ii) the charge stored?

Solution(a) (i) The initial charge on the capacitor is zero.

(ii) The initial pd across the capacitor is zero since there is no charge.

(iii) pd across the resistor is 10 V

(VR = VS – VC = 10 – 0 = 10 V)

(iv)I=VR

=101 x106

=1 × 10–5 A

(b) (i) Final pd across the capacitor equals the supply voltage, 10 V.

(ii) Q = VC = 2 × 10–6 × 10 = 2 × 10–5 C

38

Problems

Monitoring and measuring a.c.

1. (a) What is the peak voltage of the 230 V mains supply?

(b) The frequency of the mains supply is 50 Hz. How many times does the voltage fall to zero in 1 second?

2. The circuit below is used to compare a.c. and d.c. supplies.

The variable resistor is used to adjust the brightness of the lamp until the lamp has the same brightness when connected to either supply.

(a) Explain why the brightness of the lamp changes when the setting on the variable resistor is altered.

(b) What additional apparatus would you use to ensure the brightness of the lamp is the same when connected to either supply?

(c) The time-base of the oscilloscope is switched off. Diagram 1 shows the oscilloscope trace obtained when the switch is in position B. Diagram 2 shows the oscilloscope trace obtained when the switch is in position A.

39

3.6 cm

Diagram 1

10.2 cm

Diagram 2

Y gain set to 1 V cm–1

Using information from the oscilloscope traces, find the relationship between the root mean square (r.m.s.) voltage and the peak voltage of a voltage supply.

(d) The time-base of the oscilloscope is now switched on. Redraw diagrams 1 and 2 to show what happens to the traces.

3. The root mean square voltage produced by a low voltage power supply is 10 V.

(a) Calculate the peak voltage of the supply.

(b) An oscilloscope, with its time-base switched off, is connected across the supply. The Y-gain of the oscilloscope is set to 5 V cm–1. Describe the trace seen on the oscilloscope screen.

4. (a) A transformer has a peak output voltage of 12 V.

Calculate the r.m.s. value of this voltage.

(b) An oscilloscope, with the time base switched off, is connected across another a.c. supply. The Y gain of the oscilloscope is set to 20 V cm–1. A vertical line 6 cm high appears on the oscilloscope screen.

Calculate:

(i) the peak voltage of the input(ii) the r.m.s. voltage of the input.

40

5. An oscilloscope is connected across a signal generator. The time-base switch is set at 2·5 ms cm–1.

The diagram shows the trace on the oscilloscope screen.

(a) (i) What is the frequency of the output from the signal generator?

(ii) What is the uncertainty in the frequency to the nearest Hz?

(b) The time base switch is now changed to:

(i) 5 ms cm–1

(ii) 1·25 ms cm–1.

Sketch the new traces seen on the screen.

6. An a.c. signal of frequency 20 Hz is connected to an oscilloscope. The time-base switch on the oscilloscope is set at 0.01 s cm–1.

Calculate the distance between the neighbouring peaks of this waveform when viewed on the screen.

41

6 ± 0.1 cm

A10 V

+ –

70

10

20 (a)

A12 V+ –

15

9

V

(b)

A+ –

3

5

V

6 V (c)

Current, voltage, power and resistance

1. There is a current of 40 mA in a lamp for 16 s. Calculate the quantity of charge that passes any point in the circuit in this time.

2. A flash of lightning lasts for 1 ms. The charge transferred between the cloud and the ground in this time is 5 C. Calculate the value of the average current in this flash of lightning.

3. The current in a circuit is 2.5 × 10–2 A. How long does it take for 500 C of charge to pass any given point in the circuit?

4. There is a current of 3 mA in a 2 kΩ resistor. Calculate the p.d. across the resistor.

5. Calculate the values of the readings on the meters in the following circuits.

42

A40 V

+ –

4

10

R = ?(b)

V

20 V

2 A

(a)

A40 V

+ –

5

10

R = ?

(a) 10

X

Y

20

10 (b) 10

X

Y

10

10

10

(d) 25

X

Y

5

20

10

(c)

20

X10

4

10

5 1

6. Calculate the unknown values R of the resistors in the following circuits.

7. Calculate the total resistance between X and Y for the following combinations of resistors.

43

8

(f) 12

X

Y

5

6 3

10

(e) 10

X

Y

10

5 25

+

5 k

A

V

3 k

–

8. In the following circuit the reading on the ammeter is 2 mA. Calculate the reading on the voltmeter.

44

A

V 6 6 6

6 V+ –

9. Calculate the power in each of the following situations.

(a) A 12 V battery is connected to a motor. There is a current of 5 A in the motor.

(b) A heater of resistance 60 Ω connected across a 140 V supply.

(c) A current of 5 A in a heater coil of resistance 20 Ω.

10. The heating element in an electric kettle has a resistance of 30 Ω.

(a) What is the current in the heating element when it is connected to a 230 V supply?

(b) Calculate the power rating of the element in the kettle.

11. A 15 V supply produces a current of 2 A in a lamp for 5 minutes. Calculate the energy supplied in this time.

12. Calculate the readings on the ammeter and the voltmeter in the circuit shown below.

13. Each of the four cells in the circuit shown is identical.

45

A

20 5 2

6 V

A

B

R

Calculate

(a) the reading on the ammeter

(b) the current in the 20 Ω resistor

(c) the voltage across the 2 Ω resistor.

14. A voltage of 12 V is applied across a resistor. The current in the resistor is 50 mA. Calculate the resistance of the resistor.

15. The LED in the circuit below is to emit light.

(a) What is the required polarity of A and B when connected to a 5 V supply so that the LED emits light?

(b) What is the purpose of the resistor R in the circuit?

(c) The LED rating is 20 mA at 1·5 V. Calculate the value of resistor R.

46

V1

V2

R1

R2

10V

R110V

V1

V2

16. Write down the rules which connect the (a) potential differences and (b) the currents in series and parallel circuits.

17. What is the name given to the circuit shown? Write down the relationship between V1, V2, R1 and R2.

18. Calculate the values of V1 and V2 of the circuit in question 17 when:

(a) R1 = 1 kΩ R2 = 49 kΩ

(b) R1 = 5 kΩ R2 = 15 kΩ

19. The light dependent resistor in the circuit is in darkness.Light is now shone on the LDR. Explain what happens to the readings on V1 and V2.

20. Calculate the p.d. across resistor R2 in each of the following circuits.

(a) (b) (c)

47

B B

+12 V

0V

3 k

3 k

V

9 k

3 k

A

+10 V

0V

3 k

2 k

V6 k

4 k

A B

+5 V

0V

5 k

2 k

V

10 k

8 k

A

(a) (b) (c)

+9 V0 V

6 k

3 k

V

9 k

6 k

A

B

21. Calculate the p.d. across AB (voltmeter reading) in each of the following circuits.

22. A circuit consisting of two potential dividers is set up as shown.

(a) Calculate the reading on the voltmeter.

(b) (i) Suggest a value of a resistor to replace the 9 kΩ resistor that would give a reading of 0 V on the voltmeter.

(ii) Suggest a value of resistor to replace the 3 kΩ resistor that would give a reading of 0 V on the voltmeter.

23. In the circuits shown the reading on the voltmeters is zero. Calculate the value of the unknown resistors X and Y in each of the circuits.

48

3

X

Y

A 1 A 2

12

4

12 V

Electrical sources and internal resistance

1. State what is meant by:

(a) the e.m.f. of a cell

(b) the p.d. between two points in a circuit.

2. A circuit is set up as shown.

(a) Calculate the total resistance of the circuit.

(b) Calculate the readings on the ammeters.

(c) What is the value of the p.d. between X and Y?

(d) Calculate the power supplied by the battery.

3. The circuit shown uses a 230 V alternating mains supply.

49

9

12 V

120

120

V

X

12 V

4 k

12 k

V

Y

15 k

12

8

24

S

230 V

Calculate the current in each resistor when:

(a) switch S is open

(b) switch S is closed.

4. An electric cooker has two settings, high and low. This involves two heating elements, R1 and R2. On the low setting the current from the supply is 1 A. On the high setting the current from the supply is 3 A.

(a) Calculate the resistance of R1 and R2.

(b) What is the power consumption at each setting?

5. A lamp is rated at 12 V, 36 W. It is connected in a circuit as shown.

50

(a) Calculate the value of the resistor R that allows the lamp to operate at its normal rating.

(b) Calculate the power dissipated in the resistor.

6. In the circuit shown, r represents the internal resistance of the cell and R represents the external resistance (or load resistance) of the circuit.

When S is open, the reading on the voltmeter is 2 0 V.

When S is closed, the reading on the voltmeter is 1 6 V and the reading on the ammeter is 0 8 A.

(a) What is the value of the emf of the cell?

(b) When S is closed what is the terminal potential difference across the cell?

(c) Calculate the values of r and R.

51

12 V36 WR

24 V

R

rE = 1.5 V lost volts = 0.2 V

V

r

A

(d) The resistance R is now halved in value. Calculate the new readings on the ammeter and voltmeter when S is closed.

7. The battery in the circuit shown has an emf of 5 0 V.

The current in the lamp is 0.20 A and the reading on the voltmeter is 3.0 V. Calculate the internal resistance of the battery.

8. A battery of emf of 4.0 V is connected to a load resistor with a resistance of 15 Ω. There is a current of 0.2 A in the load resistor.Calculate the internal resistance of the battery.

9. A signal generator has an emf of 8.0 V and an internal resistance of 4. 0 . A load resistor is connected across the terminals of the generator. The current in the load resistor is 0.50 A. Calculate the resistance of the load resistor.

10. A cell is connected in a circuit as shown below.

52

RS

V

r

A

6 V

1

(a) Calculate the terminal p.d. across the cell.

(b) The resistance of the variable resistor R is now increased.

(i) Describe and explain what happens to the current in the circuit.

(ii) Describe and explain what happens to the p.d. across the terminals of the cell.

11. A cell has an emf of 1·5 V and an internal resistance of 2·0 Ω. A 3·0 Ω resistor is connected across the terminals of the cell. Calculate the current in the circuit.

12. A student is given a voltmeter and a torch battery. When the voltmeter is connected across the terminals of the battery the reading on the voltmeter is 4·5 V.When the battery is connected across a 6·0 Ω resistor the reading on the voltmeter decreases to 3·0 V.

(a) Calculate the internal resistance of the battery.

(b) What value of resistor when connected across the battery reduces the reading on the voltmeter to 2·5 V?

13. In the circuit shown, the battery has an emf of 6·0 V and an internal resistance of 1·0 Ω.

53

When the switch is closed, the reading on the ammeter is 2·0 A.

What is the corresponding reading on the voltmeter?

14. To find the internal resistance of a cell a load resistor is connected across the terminals of the cell. A voltmeter is used to measure Vtpd, the voltage measured across the terminals of the cell. An ammeter is used to measure I, the current in the variable resistor. The table below shows the results obtained as the resistance of the variable resistor is changed.

Vtpd (V) 1·02 0·94 0·85 0·78 0·69 0·60

I (A) 0·02 0·04 0·06 0·08 0·10 0·12

54

(a) Draw a diagram of the circuit used to produce these results.

(b) Plot a graph of the results and from it determine:

(i) the emf of the cell

(ii) the internal resistance of the cell

(iii) the short circuit current of the cell.

15. A variable resistor is connected across a power supply. A voltmeter is used to measure Vtpd, the voltage measured across the terminals of the supply. An ammeter is used to measure I, the current in the variable resistor. The table below shows the results obtained as the resistance of the variable resistor is changed.

Vt.p.d. (V) 5·5 5·6 5·7 5·8 5·9

I (A) 5·0 4·0 3·0 2·0 1·0

Plot a graph of Vtpd against I.

(a) What is the value of the open circuit p.d.?

(b) Calculate the internal resistance of the power supply.

(c) Calculate the short circuit current of the power supply.

(d) The variable resistor is now removed from the circuit and a lamp of resistance 1·5 is connected across the terminals of the supply.

Calculate:

(i) the terminal p.d.

(ii)the power delivered to the lamp.

55

RS

V

r

A

E

16. A circuit is set up as shown to investigate the properties of a battery.

The variable resistor provides known values of resistance R.

For each value of resistance R, the switch is closed and the current I noted.

The table shows the results obtained.

R (Ω) 0 2 4 6 8 10 12

I (A) 6·80 3·78 2·62 2·00 1·62 1·36 1·17

1/I (A–1)

(a) Show that the relationship E = I(R + r) can be put in the form:

R=EI−r

(b) Complete the third row in the table.

(c) Use the values of R and 1/I to plot a graph.

(d) Use the information in the graph to find:

(i) the internal resistance of the battery

(ii) the e.m.f. of the battery.

(e) The battery is now short circuited. Calculate the current in the battery when this happens.

56

RS

V

r

A

12 V

4

17. A student uses the following circuit to investigate the conditions for transferring the maximum power into a load resistor.

For each setting of the variable resistor the current in the circuit is recorded.

The table below shows the results obtained.

R (Ω) 1 2 3 4 5 6

I (A) 2·40 2·00 1·71 1·50 1·33 1·20

Power in R (W)

(a) Complete the table by calculating the power in the load for each value of R.

(b) Sketch a graph to show how the power in the load resistor R varies with R.

(c) In order to achieve maximum transfer of power, what is the relationship between the internal resistance of the power source and the resistance of the load resistor?

18. An automotive electrician needed to accurately measure the resistance of a resistor.

She set up a circuit using an analogue milliammeter and a digital voltmeter.

57

The two meter readings were as shown in the diagram below.

(a) What are the readings on the ammeter and the voltmeter?

(b) What is the nominal resistance calculated from these readings?

(c) What is the smallest division on the milliammeter?

(d) What is the absolute uncertainty on the milliammeter?

(e) What is the absolute uncertainty on the voltmeter?

(f) What is the percentage uncertainty on the milliammeter reading?

(g) What is the percentage uncertainty on the voltmeter reading?

(h) Which is the greatest percentage uncertainty?

(i) What is the percentage uncertainty in the resistance?

(j) What is the absolute uncertainty in the resistance?

(k) Express the final result as (resistance ± uncertainty) Ω

(l) Round both the result and the uncertainty to the relevant number of significant figures or decimal places.

58

Capacitors

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 milliseconds.

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 30 µF capacitor stores a charge of 12 × 10–4 C.

(a) What is the p.d. across its terminals?

(b) The tolerance of the capacitor is ± 0·5 µF. Express this uncertainty as a percentage.

4. A 15 µF capacitor is charged using a 1·5 V battery.

Calculate the charge stored on the capacitor when it is fully charged.

5. (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.

59

6. A circuit is set up as shown.

The capacitor is initially uncharged. The switch is now closed.

The capacitor is charged with a constant charging current of

2·0 × 10–5 A for 30 s.

At the end of this time the p.d. across the capacitor is 12 V.

(a) What has to be done to the value of the variable resistor in order to keep the current constant for 20 s?

(b) Calculate the capacitance of the capacitor.

7. A 100 µF capacitor is charged using a 20 V supply.

(a) How much charge is stored on the capacitor when it is fully charged?

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

60

A

8. A 30 µF capacitor stores 6·0 × 10–3 C of charge. How much energy is stored in the capacitor?

9. The circuit below is used to investigate the charging of a capacitor.

The battery has negligible internal resistance.

The capacitor is initially uncharged. The switch is now closed.

(a) Describe what happens to the reading on the ammeter from the instant the switch is closed.

(b) How can you tell when the capacitor is fully charged?

(c) What would be a suitable range for the ammeter?

(d) The 10 k Ω resistor is now replaced by a larger resistor and the investigation repeated.

What is the maximum voltage across the capacitor now?

10. In the circuit below the neon lamp flashes at regular intervals.

61

9 V C

A

V

S+

–

The neon lamp requires a potential difference of 100 V across it before it conducts and flashes. It continues to glow until the potential difference across it drops to 80 V. While lit, its resistance is very small compared with the resistance of R.

(a) Explain why the neon bulb flashes.

(b) Suggest two methods of decreasing the flash rate.

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

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 seconds. The results are shown in the table below.

Time (s) 0 10 20 30 40

V (V) 0 1·9 4·0 6·2 8·1

62

100 V

VR

VC

1 2A B

(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.

12. The circuit below is used to charge and discharge a capacitor.

The battery has negligible internal resistance.

The capacitor is initially uncharged.

VR is the p.d. across the resistor and VC is the p.d. across the capacitor.

(a) What is the position of the switch: (i) to charge the capacitor(ii) to discharge the capacitor?

(b) Sketch graphs of VR against time for the capacitor charging and discharging. Show numerical values for the maximum and minimum values of VR.

(c) Sketch graphs of VC against time for the capacitor charging and discharging. Show numerical values for the maximum and minimum values of VC.

(d) (i) When the capacitor is charging what is the direction of the electrons between points A and B in the wire?

(ii) When the capacitor is discharging what is the direction of the electrons between points A and B in the wire?

(e) The capacitor has a capacitance of 4·0 µF. The resistor has resistance of 2·5 MΩ.

63

3 V+

–

3 M

3 F

S

Calculate:

(i) the maximum value of the charging current

(ii) the charge stored by the capacitor when the capacitor is fully charged.

13. A capacitor is connected in a circuit as shown.

The power supply has negligible internal resistance. The capacitor is initially uncharged.

VR is the p.d. across the resistor and VC is the p.d. across the capacitor.

The switch S is now closed.

(a) Sketch graphs of:

(i) VC against time during charging. Show numerical values for the maximum and minimum values of VC.

(ii) VR against time during charging. Show numerical values for the maximum and minimum values of VR.

(b) (i) What is the p.d. across the capacitor when it is fully charged?

(ii) Calculate the charge stored by the capacitor when it is fully charged.

64

12 V+

–

6 k

20 F

S

(c) Calculate the maximum energy stored by the capacitor.

14. A capacitor is connected in a circuit as shown.

The power supply has negligible internal resistance.

The capacitor is initially uncharged. The switch S is now closed.

(a) Calculate the value of the initial current in the circuit.

(b) At a certain instant in time during charging the p.d. across the capacitor is 3 V. Calculate the current in the resistor at this time.

(c) Calculate the current in the circuit when the charge on the capacitor is 80 µC.

15. The circuit shown is used to investigate the charge and discharge of a capacitor.

65

12 V

VR

VC

2 1 1 k

10 mF

A

The switch is in position 1 and the capacitor is uncharged.

The switch is now moved to position 2 and the capacitor charges.

The graphs show how VC, the p.d. across the capacitor, VR, the p.d. across the resistor, and I, the current in the circuit, vary with time.

66

0 10 20 30 40 50 6002468

101214

Current versus time

time / s

I / m

A

0 10 20 30 40 50 600

4

8

12

VC versus time

time / s

VC /

V

0 10 20 30 40 50 600

4

8

12

VR versus time

time / s

VR /

V

12 V+

–

1 k

10 F

S

(a) The experiment is repeated with the resistance changed to 2 kΩ.

Sketch the graphs above and on each graph sketch the new lines which show how VC, VR and I vary with time.

(b) The experiment is repeated with the resistance again at 1 kΩ but the capacitor replaced with one of capacitance 20 mF. Sketch the original graphs again and on each graph sketch the new lines which show how VC, VR and I vary with time.

(c) (i) What does the area under the current against time graph represent?

(ii) Compare the areas under the current versus time graphs in the original graphs and in your answers to (a) and (b). Give reasons for any differences in these areas.

(d) At any instant in time during the charging what should be the value of (VC + VR) ?

(e) The original values of resistance and capacitance are now used again and the capacitor fully charged. The switch is now moved to position 1 and the capacitor discharges.

Sketch graphs of VC, VR and I from the instant the switch is moved until the capacitor is fully discharged.

16. A student uses the circuit shown to investigate the charging of a capacitor.

67

0 10 20 30 40 50 6002468

101214

Current versus time

time / s

I / m

A

The capacitor is initially uncharged.

The student makes the following statements:

(a) When switch S is closed the initial current in the circuit does not depend on the internal resistance of the power supply.

(b) When the capacitor has been fully charged the p.d. across the capacitor does not depend on the internal resistance of the power supply.

Use your knowledge of capacitors to comment on the truth or otherwise of these two statements.

68

Solutions

Monitoring and measuring a.c.

1. (a) 325 V

(b) 100 times

2 (c) Vr.m.s . = 0.71 Vpeak

3. (a) 14 V

4. (a) 8.5 V

(b) (i) 60 V

(ii) 42 V

5. (a) (i) 100 Hz

(ii) ±2 Hz

6. 5 cm

Current, voltage, power and resistance

1. 0.64 C

2. 5 × 103 A

3. 2·0 × 104 s

4. 6 V

69

5. (a) I = 0·1 A

(b) I = 0·5 A, V = 4·5 V

(c) I = 2 A, V = 10 V

6. (a) 5

(b) 6

7. (a) 25

(b) 25

(c) 24·2

(d) 13·3

(e) 22·9

(f) 14·7 .

8. 3·75 V

9. (a) 60 W

(b) 327 W

(c) 500 W

10. (a) 7·7 A

(b) 1763 W

11. 9000 J

12. I = 0·67 A, V = 4 V

13. (a) 0·67 A

(b) 0·13 A

(c) 1·34 V

14. 240

70

15. (a) A = +, B = -

(b) To reduce the current through/in the resistor to the correct value (and also to reduce the voltage across the resistor to the correct operating value.)

(c) 175

16. See notes.

17. R1/R2 = V1/V2

18. (a) V1 = 0·2 V, V2 = 9·8 V

(b) V1 = 2·5 V, V2 = 7·5 V

19. V2 decreases (as R_LDR decreases as light increases). V 1 increases (as V1+V2=Vsupply)

20. (a) 4 V

(b) 1 V

(c) 3 V

21. (a) 3 V

(b) -0·8 V

(c) 0 V

22. (a) 0·6 V

(b) (i) 12 k

(ii) 4 k

23. X = 9 Y = 45

71

Electrical sources and internal resistance1. See notes

2. (a) 6

(b) A1 = 2 A, A2 = 1·5 A

(c) 6 V

(d) 24 W

3. (a) In 12 Ω current is 11·5 A; in 8 Ω current is 11.5A; in 24 Ω current is 0 A

(b) In 12 Ω current is 12·8 A; in 8 Ω current is 3·2 A; in 24 Ω current is 9·6 A

4. (a) R1 = 230 R2 = 115

(b) Low 230 W high 690 W

5. (a) 4

(b) 36 W

6. (a) 2·0 V

(b) 1·6 V

(c) r = 0·5 R = 2

(d) 1·3 A, 1·3 V

7. 10

8. 5

9. 12

10. (a) 1·3 V

(b) (i) As R is increased, current decreases (since emf = I x total resistance and total resistance of circuit has increased)

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(b) (ii) As R is increased, the p.d. across the terminals INCREASES.

Since lost volts = Ir and I decreases, lost volts also decreases so

t.p.d = emf-lost volts increases (i.e. less energy is lost as heat through cell).

11. 0·30 A

12. (a) 3·0

(b) 3·75

13. 4·0 V

14. (a) Very similar to diagram for Q3.1.42 except battery is replaced by a cell and R is replaced by variable resistor. Switch not necessary .

(b) (i) 1·1 V, the intercept on the y-axis

(ii) 4·2 the gradient of the line

(iii) 0·26 A

15 (a) 6 V

(b) 0·1

(c) 60 A

(d) (i) 5·6 V

(ii) 21 W

16. (a) Divide through by I: E/I = R+r ==> rearrange R = E/I-r

(b) 0·147, 0·264,0·382, 0·500, 0·617, 0·735, 0·855

(d) (i) 2·5

(ii) 17 V

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(e) 6·8 A

18. (a) Ammeter 1·76 mA, voltmeter 1·3 V

(b) 740

(c) 0·02 mA

(d) ±0·01 mA

(e) ± 0.1 V

(f) 0·6 %

(g) 8%

(h) 8%

(i) 8%

(j) 59

(k) (740± 59)

(l) (740± 60)

Capacitors

1. (a) 5·0 × 10–3 C

(b) (i) 1·25 A

2. 0·5 F

3 (a) 40 V

(b) 1·7%

4. 2·25 × 10–5 C

5. (a) 1·0 F

(b) 0·8 C

6. (b) 50 F

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7. (a) 2·0 × 10–3 C

(b) 0·020 J

8. 0·60 J

9. (a) Initially the ammeter goes from 0 to a large (maximum) reading. This then rapidly decreases and eventually falls back to zero as the capacitor becomes fully charged

(b) Reading on ammeter is 0 A

(c) 0 to 2 mA (max. current 1·2 mA)

(d) 12 V

10. (a) The capacitor takes a short time to charge up. When the p.d. across the capacitor is greater than 100V, the neon lamp becomes conducting. Charge rapidly flows off the capacitor plates which causes the bulb to flash. When the p.d. drops below 80V the resistance of the neon bulb is again very high. No charge flows off the plates and the capacitor charges up again. This cycle repeats causing the bulb to flash

(b) Increase the capacitance of the capacitor or increase the resistance of the resistor.

11.

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(b) 4·9 mF

12. (a) (i) Charging Position 1

(ii) Discharging position 2

(b,c) See notes

(d) (i) While charging, electrons flow from B to A (attracted towards + terminal)

(ii) While discharging, electrons flow in the opposite direction, from A to B

(e) (i) 40 A

(ii) 4·0 × 102 C

13. (a) (i) 0V when switch is closed rises to maximum Vc = 3 V

(ii) Maximum VR = 3V (initial reading) falls to 0V

(b) (i) 3 V

(ii) 9 C

(c) 1·35 × 10–5 J

14. (a) 2 mA

(b) 1·5 mA

(c) First find voltage across capacitor = 4V VR = 8V I = 1.3 mA (1.3x10 -3 A)

15. (a) Vc will take longer to rise to the same value. V R will fall more slowly from the same initial value. The current will start from 6mA (half-height) but will take longer to fall to zero.

(b) Very similar to graphs in (a) V c will take longer to rise to the same value. VR will fall more slowly from the same initial value. The current will start from the same value 12mA but will take twice as long to fall to zero.

(c) (i) The charge that is stored on the capacitor.

(ii) Area under graphs for (a) is the same as the original experiment. (total capacitance has not changed). Area

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under graph (b) is twice that of the original graph (as the capacitance has doubled it has twice the capacity to store charge for the same applied voltage.)

(d) 12V

(e) Vc will start at 12 V and gradually fall to zero. The capacitor is now effectively acting as the power supply in a circuit with a resistor and so the voltage across the resistor will equal the voltage across the capacitor. (the graph of V R will be the same as the graph of Vc ). The current is now flowing in the opposite direction and so it will start from -12mA and gradually rise to zero!

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