electricity and electronics electric fields and resistors ... · electric fields and resistors in...

ELECTRICITY AND ELECTRONICS The knowledge and understanding for this unit is given below. Electric fields and resistors in circuits 1. State that, in an electric field, an electric charge experiences a force. 2. State that an electric field applied to a conductor causes the free electric charges in it

to move. 3. State that work W is done when a charge Q is moved in an electric field. 4. State that the potential difference (V) between two points is a measure of the work

done in moving one coulomb of charge between the two points. 5. State that if one joule of work is done moving one coulomb of charge between two

points, the potential difference between the points is one volt. 6. State the relationship V = W/Q. 7. Carry out calculations involving the above relationship. 8. State that the e.m.f. of a source is the electrical potential energy supplied to each

coulomb of charge which passes through the source. 9. State that an electrical source is equivalent to a source of e.m.f. with a resistor in

series, the internal resistance. 10. Describe the principles of a method for measuring the e.m.f. and internal resistance of

a source 11. Explain why the e.m.f. of a source is equal to the open circuit p.d. across the terminals

of a source 12. Explain how the conservation of energy leads to the sum of the e.m.f.’s round a closed

circuit being equal to the sum of the p.d.’s round the circuit. 13. Derive the expression for the total resistance of any number of resistors in series, by

consideration of the conservation of energy. 14. Derive the expression for the total resistance of any number of resistors in parallel, by

consideration of the conservation of charge. 15. State the relationship among the resistors in a balanced Wheatstone bridge. 16. Carry out calculations involving the resistance in a balanced Wheatstone bridge. 17. State that for an initially balanced Wheatstone bridge, as the value of one resistor is

changed by a small amount, the out of balance p.d. is proportional to the change in resistance.

18. Use the following terms correctly in context: terminal p.d., load resistor, bridge circuit, lost volts, short circuit current.

Alternating Current and Voltage 1. Describe how to measure frequency using an oscilloscope. 2. State the relationship between peak and r.m.s. values for a sinusoidally varying

voltage and current. 3. Carry out calculations involving peak and r.m.s. values of voltage and current. 4. State the relationship between current and frequency in a resistive circuit.

Physics: Electricity (H) 1

Capacitance 1. State that the charge Q on two parallel conducting plates is directly proportional to the

p.d. V between the plates. 2. Describe the principles of a method to show that the p.d. across a capacitor is directly

proportional to the charge on the plates. 3. State that capacitance is the ratio of charge to p.d. 4. State that the unit of capacitance is the farad and that one farad is one coulomb per

volt. 5. Carry out calculations using C = Q/V 6. Explain why work must be done to charge a capacitor. 7. State that the work done to charge a capacitor is given by the area under the graph of

charge against p.d. 8. State that the energy stored in a capacitor is given by (charge × p.d.) and equivalent

expressions.

12

9. Carry out calculations using QV or equivalent expressions. 10. Draw qualitative graphs of current against time and of voltage against time for the

charge and discharge of a capacitor in a d.c. circuit containing a resistor and capacitor in series.

12

11. Carry out calculations involving voltage and current in CR circuits (calculus methods are not required).

12. State the relationship between current and frequency in a capacitive circuit. 13. Describe the principles of a method to show how the current varies with frequency in

a capacitive circuit. 14. Describe and explain the possible functions of a capacitor: storing energy, blocking

d.c. while passing a.c.


ELECTRIC FIELDS AND RESISTORS IN CIRCUITS Force Fields In Physics, a field means a region where an object experiences a force without being touched. For example, there is a gravitational field around the Earth. This attracts masses towards the earth’s centre. Magnets cause magnetic fields and electric charges have electric fields around them. Electric Fields In an electric field, a charged particle will experience a force. We use lines of force to show the strength and direction of the force. The closer the field lines the stronger the force. Field lines are continuous - they start on positive and finish on negative charge. The direction is taken as the same as the force on a positive “test” charge placed in the field. Electric Field Patterns

Positive point charge

+

Negative point charge

-

+ test chargehas a force ‘outwards’

+ test chargehas a force ‘inwards’

These are called radial fields. The lines are like the radii of a circle. The strength of the field decreases as we move away from the charge. Electric Field Patterns

Positive and negative point charges

Parallel charged plates

++++

----

+-

The field lines are equally spaced between the parallel plates. This means the field strength is constant. This is called a uniform field. Electric fields have certain similarities with gravitational fields.


Gravitational Fields If a mass is lifted or dropped through a height then work is done i.e. energy is changed.

h

If the mass is dropped then the energy will change to kinetic energy. If the mass is lifted then the energy will change to gravitational potential energy.

Change in gravitational potential energy = work done. Electric Fields Consider a negative charge moved through a distance in an electric field. If the charge moves in the direction of the electric force, the energy will appear as kinetic energy. If a positive charge is moved against the direction of the force as shown in the diagram, the energy will be stored as electric potential energy.

Change in electric potential energy = work done If the charge moved is one coulomb, then the work done is the potential difference or voltage. If one joule of work is done in moving one coulomb of charge between two points in an electric field, the potential difference, (p.d.) between the two points is one volt. 1 volt = 1 joule per coulomb

W = QV In this section W will be used for the work done i.e. energy transferred. Example: A positive charge of 3 µC is moved, from A to B, between a potential difference of 10 V. (a) Calculate the electric potential energy gained. (b) If the charge is now released, state the energy change.

++++

----

+ Q

AB(c) How much kinetic energy will be gained on reaching the negative plate? (a) W = QV= 3 × 10-6 × 10 = 3 × 10-5 J (b) Electric potential energy to kinetic energy (c) By conservation of energy the energy will be the same, i.e. 3 × 10-5 J.


Moving Charges in Electric Fields From the previous example, when the positive charge is released at plate B then the electical potential energy is converted to kinetic energy.

QV = 12

mv2

Example An electron is accelerated (from rest) through a potential difference of 200 V. Calculate (a) the kinetic energy, Ek gained. (b) the final speed of the electron. (Mass of an electron = 9.1 × 10-31 kg, charge on an electron = -1.6x10-19 C)

(a) Ek = 12

mv2 = QV = 1.6x10-19 × 200

= 3.2 × 10-17 J

(b) 12

mv2 = 3.2 × 10-17

v2 =× ×

=× ×

×

− −

−

3.2 10 2m

3.2 10 2

9.1 10

17 17

31

v = 8.4 × 106 m s-1

Applications of electric fields (for background interest) A television involves the use of electron guns. The electrons gain kinetic energy by accelerating through an electric field. Deflection of the electrons is usually done by electromagnetic coils, although flat screen tubes are now dependent on electrostatic deflection. An oscilloscope also depends on electric fields acting on electrons.

Electron Gun

+-

+

-

Electrostatic Spraying makes use of electric fields. Paint or powder particles are blown from a nozzle, where they acquire a charge. The object to be coated is earthed. The charged paint or powder particles follow the field lines and so reach the object, some reaching the back of the object as well as the front. Other applications include photocopiers, ink jet and laser printers.


Electric Current When S is closed, the free electrons in the conductor experience an electric field which cause them to move in one direction.

Wire magnified(free electrons)

S Note: The electron current will flow from the negative terminal to the positive terminal of the battery. The energy required to drive the electron current around the circuit is provided by a chemical reaction in the battery or by the mains power supply. The electrical energy which is supplied by the source is converted to other forms of energy in the components which make up the circuit. The lamp has resistance R. 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. This is Ohm’s Law which is summarised by the equation below.

V = IR A component which 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 p.d. across the component is altered, for example the transistor. Electric Power For a given component the power P = I V where I is the current through that component and V is the potential difference across the component. Resistive Heating The expression I2 R gives the energy transferred in one second due to resistive heating. Apart from obvious uses in electric fires, cookers, toasters, etc. consideration has to given to heating effects in resistors, transistors and integrated circuits and care taken not to exceed the maximum ratings for such components.

P = I R = VR

22

Electromotive Force (e.m.f) The energy given to a coulomb of charge by a source of electrical energy is called the e.m.f. of the source. This is measured in J C-1 or volts. Note: the potential difference between two points in the external circuit is also measured in volts, but this is concerned with electrical energy being transformed outside the source.


Sources of e.m.f. E.M.F’s can be generated in a great variety of ways e.g: chemical cells, thermocouple, piezo - electric generator, solar cell, electromagnetic generator. Resistors in Series - Conservation of Energy

E

IR1 IR2 IR3 IRS

E

R1 R2 R3

electron flow

RS

Applying the conservation of energy to resistors in series for one coulomb of charge. Energy supplied by source = energy converted by circuit components

Rs = equivalent series resistance

e.m.f. = IR1 + IR2 + IR3

IRS = IR1 + IR2 + IR3

RS = R1 + R2 + R3 Resistors in Parallel - Conservation of Charge The total charge per second (current) passing through R1, R2 and R3 must equal the charge per second (current) supplied from the cell i.e. passing through RP. Conservation of charge gives:

I = I1 + I2 + I3 (Since I = Qt

for each resistor)

ER

= ER

+ ER

+ ER

1 2 3 4

R1 R3R2ERP

I

EI

I1 I2 I3

1 = 1 + 1 + 1 R P R1 R2 R3

Rp = equivalent parallel resistance


Internal Resistance In choosing a suitable power supply for a circuit, you would have to ensure that it :

• gives the correct e.m.f • is able to supply the required maximum current

When a power supply is part of a closed circuit, it must itself be a conductor. All conductors have some resistance. A power supply has internal resistance, r. Energy will be wasted in getting the charges through the supply (the heat from the supply will be noticeable) and so the energy available at the output (the terminal potential difference) will fall. There will be “lost volts”. The lost volts = I r The greater the current, the more energy will be dissipated in the power supply until eventually all the available energy (the e.m.f.) is wasted and none is available outside the power supply. This maximum current is the short circuit current. This is the current which will flow when the terminals of the supply are joined with a short piece of thick wire. Open Circuit When no current is taken from the power supply, no energy is wasted. The terminal potential difference is therefore the maximum available and equals the e.m.f. General Circuit With S closed and using the conservation of energy e.m.f. = lost volts + t.p.d. e.m.f. = lost volts + output voltage E = Ir + V which is: E = Ir + IR Notice that the total resistance of the circuit is R + r, giving the equation:

E = I ( R + r)

Any power supply can be thought of as a source of constant e.m.f. E, in series with a small resistance, the internal resistance.

With S open, the voltmeter reading gives the e.m.f. (an open circuit).

With S closed, the voltmeter reading will fall (lost volts). The voltmeter now gives the output voltage, the terminal potential difference, t.p.d.

VR

r

SE

The short circuit current is the maximum which can be supplied by a source. This occurs when there is no external component and R = 0.


Example E r

V

R = 28 Ω

A cell of e.m.f. 1.5 V is connected in series with a 28Ω 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. (a) E = Ir + IR = Ir + V

Lost volts = Ir = E - V = 1.5 - 1.4 = 0.1 V

r = lost volts

I =

0.11

I = VR

= 1.428

= 0.05 A

r = 0.1

0.05 = 2 Ω

(b) A short circuit occurs when R = 0 (no external resistance)

IR + r

= Er

= 1.52

= 0.75 A

(c) Lost volts = Ir

IR + r

= E

28 + 2 = 1.5

= 0.05 A Lost volts = 0.05 × 2 = 0.1 V


Wheatstone Bridge Circuit Any method of measuring resistance using an ammeter or voltmeter necessarily involves some error unless the resistances of the meters themselves are taken into account. The use of digital voltmeters largely overcomes this as they tend to have very high resistances. Further error may be introduced if the meter is not correctly calibrated. The only situation where neither of these errors matter is if the meter reading is zero. The Wheatstone bridge circuit is one such example of using a meter as a null deflection indicator. Bridge Circuit If VOA = VOB there is no p.d. between A and B hence no current flows. The potentials at A and at B depend on the ratio of the resistors that make up each of the two voltage dividers. The voltmeter forms a ‘bridge’ between the two voltage dividers to make up a bridge circuit.

R4

A B

R 1 R3

R 2

5 V

0 V

V

O

Balanced Bridge No potential difference will exist across AB when RR

RR

1

2

3

4=

The bridge is balanced when the voltmeter or galvanometer (milliammeter) reads zero. Unbalanced Bridge If the bridge is initially balanced, and the resistor × is altered by a small amount × then the out of balance p.d. (reading on G) is directly proportional to the change in resistance, provided the change is small. Hence for small ×, reading on G ×

Alternative ‘diamond’ Representation with galvanometer

+1.5 V

G

R1 R2

R4X+Δ X

ΔX/Ω

Reading on G


ALTERNATING CURRENT AND VOLTAGE Peak and r.m.s. values The graph of a typical alternating voltage is shown below. The maximum voltage is called the peak value. From the graph it is obvious that the peak value would not be a very accurate measure of the voltage available from an alternating supply. In practice the value quoted is the root mean square (r.m.s.) voltage. The r.m.s. value of an alternating voltage or current is defined as being equal to the value of the direct voltage or current which gives rise to the same heating effect (same power output). Consider the following two circuits which contain identical lamps.

A

A The variable resistors are altered until the lamps are of equal brightness. As a result the direct current has the same value as the effective alternating current (i.e. 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 r.m.s. valves and to the peak values a similar equation for currents can be deduced.

IpeakI = 12

VpeakV = 12

andr.m.s. r.m.s. Note: a moving coil a.c. meter is calibrated to give r.m.s. values.


Graphical method to derive relationship between peak and r.m.s. values of alternating current The power produced by a current I in a resistor R is given by I2 R. 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 I

2 Peak 2

An identical heating effect (power output) for a d.c. supply = I2r.m.s R

[since I (d.c.) = Ir.m.s.] Average power output for a.c. = I2 r.m.s R = hence I2

r.m.s. = giving Ir.m.s. = I Peak Frequency of a.c. To describe the domestic supply voltage fully, we would have to include the frequency i.e. 230 V 50 Hz. An oscilloscope can be used to find the frequency of an a.c. supply as shown below.

I2Peak

2 R

⊕ I2 Peak 2

0.005timebase (s cm-1)

y input

R ⊕I2 Peak 2

⊕2

Time base = 0.005 s cm-1 Wavelength = 4 cm Time to produce one wave = 4 × 0.005 = 0.02 s

Frequency = 1

time to produce one wave

= 1

0 02. = 50 Hz


Mains supply The mains supply is usually quoted as 230 V a.c. This is of course 230 V r.m.s. The peak voltage rises to approximately 325 V. Insulation must be provided to withstand this peak voltage. Example A transformer is labelled with a primary of 230 Vr.m.s. and secondary of 12 Vr.m.s. What is the peak voltage which would occur in the secondary? V peak = √ 2 × V r.m.s. V peak = 1.41 × 12 V peak = 17.0 V


CAPACITANCE The ability of a component to store charge is known as capacitance. A device designed to store charge is called a capacitor. A typical capacitor consists of two conducting layers separated by an insulator.

Circuit symbol

Relationship between charge and p.d.

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 coulomb meter with the switch set to B. In this way pairs of readings of voltage and charge are obtained.

V C

A B

+-

Q

V0

Charge is directly proportional to voltage.

QV

= constant

For any capacitor the ratio Q/V is a constant and is called the capacitance.

farad (F) capacitance = chargevoltage

volts (V)

coulombs (C)

The farad is too large a unit for practical purposes. In practice the micro farad (µF) = 1 × 10-6 F and the nano farad (nF) = 1 × 10-9 F are used. Example A capacitor stores 4 × 10-4 C of charge when the potential difference across it is 100 V. What is the capacitance ?

C = QV

4 10

100

-6

=×

= 4 µF


Energy Stored in a Capacitor A charged capacitor can be used to light a bulb for a short time, therefore the capacitor must contain a store of energy. The charging of a parallel plate capacitor is considered below.

Vs Vs

Electroncurrent

Vs

Electroncurrent

+-+-

Vc Vc = Vs+- +-+-+-

There is an initial surge ofelectrons from the negativeterminal of the cell onto oneof the plates (and electronsout of the other plate towardsthe +ve terminal of the cell).

Once some charge is on theplate it will repel more chargeand so the current decreases.In order to further charge thecapacitor the electrons mustbe supplied with enoughenergy to overcome thepotential difference across theplates i.e. work is done incharging the capacitor.

Eventually the current ceasesto flow. This is when thep.d. across the plates of thecapacitor is equal to thesupply voltage.

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

The energy stored in the capacitor is given by the area undergraph

Area under graph= 1 Q x V2

Energy stored = 1 Q x V 2

Q

Charge

p.d.V

If the voltage across thecapacitor was constant workdone = Q x V, but since V isvarying, the work done =area under graph.

Q = C × V and substituting for Q and V in our equation for energy gives:

Energy stored in a capacitor = 12

QV =12

CV = 12

QC

22

Example A 403F capacitor is fully charged using a 50 V supply. How much energy is stored?

Energy = 12

CV = 12

40 10 25002 6× × ×

= 5 × 10-2 J


Capacitance in a d.c. Circuit Charging Consider the following circuit:-

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

time

current

0

Supply voltage

p.d. acrosscapacitor

0time

Consider the circuit at three different times.

As soon as the switch isclosed there is no charge onthe capacitor the current islimited only by theresistance in the circuit andcan be found using Ohm’slaw.

As the capacitor charges ap.d. develops across theplates which opposes thep.d. of the cell as a resultthe supply currentdecreases.

The capacitor becomesfully charged and the p.d.across the plates is equaland opposite to that acrossthe cell and the chargingcurrent becomes zero.

0 0

- -- - - -+ +

+ ++ +

0

A A A

Physics: Electricity (H 14

Discharging

If the cell is taken out of the circuit and theswitch is set to A, the capacitor willdischarge

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

While the capacitor is discharging the current flowing in the circuit and the voltage across the capacitor behaves 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 p.d. across the plates decreases as charge leaves them. Factors affecting the rate of charge/discharge of a capacitor •

• When a capacitor is charged to a given voltage the timetaken depends on the value of the resistance in the circuit. The larger the resistance the smaller the initial charging current, hence the longer it takes to charge the capacitor as Q = It

When a capacitor is charged to a given voltage the time taken depends on the value of the capacitor. The larger the capacitor the longer the charging time, since a larger capacitor requires more charge to raise it to the same p.d. as a smaller capacitor as V= Q

AA

- - - -

+ + + +BA A B

p.d. acrosscapacitor

time

Current

0 0 time

Supply voltage

(The area under this I/t graph = charge. Both curves will have the same area since Q is the same for both.)

Current

Time

Current

Time

large capacitorsmall capacitor

small resistorlarge resistor

C


Example The switch in the following circuit is closed at time t = 0

1 MΩ

1 µF

Vs

10 V Immediately after closing the switch what is: (a) the charge on C (b) the p.d. across C (c) the p.d. across R (d) the current through R. When the capacitor is fully charged what is: (e) the p.d. across the capacitor (f) the charge stored. (a) Initial charge on capacitor is zero. (b) Initial p.d . is zero since charge is zero. (c) p.d. is 10 V = Vs - Vc = 10 - 0 = 10 V

(d) IR

= V = 1010

= 10 A 63

(e) Final p.d. across the capacitor equals the supply voltage = 10 V. (f) Q = CV = 2 × 10-6 × 10 = 2 × 10-5 C.


Resistors and Capacitors in a.c. Circuits Frequency response of resistor The following circuit is used to investigate the relationship between current and frequency in a resistive circuit.

ACurrent

(A)

frequency (Hz)

Signalgenerator(constante.m.f.)

0

The results show that the current flowing through a resistor is independent of the frequency of the supply. Frequency response of capacitor The following circuit is used to investigate the relationship between current and frequency in a capacitive circuit.

ACurrent

(A)

frequency (Hz)

Signalgenerator(constante.m.f.)

0

The results show that the current is directly proportional to the frequency of the supply. To understand the relationship between the current and frequency consider the two halves of the a.c. cycle.

Electrons

-

+

+

-Electrons

The electrons move back and forth around the circuit passing through the lamp and charging the capacitor one way and then the other (the electrons do not pass through the capacitor). The higher the frequency the less time there is for charge to build up on the plates of the capacitor and oppose further charges from flowing in the circuit More charge is transferred in one second so the current is larger.


Applications of Capacitors (for background interest) Blocking capacitor A capacitor will stop the flow of a steady d.c. current . This is made use of in the a.c./d.c. switch in an oscilloscope. In the a.c. position a series capacitor is switched in allowing passage of a.c. components of the signal, but blocking any steady d.c. signals. Flashing indicators A low value capacitor is charged through a resistor until it acquires sufficient voltage to fire a neon lamp. The neon lamp lights when the p.d. reaches 100 V. The capacitor is quickly discharged and the lamp goes out when the p.d. falls below 80V.

120 V

1 - 2 MΩ

Crossover networks in loudspeakers In a typical crossover network in low cost loudspeaker systems, the high frequencies are routed to LS-2 by the capacitor.

LS 1 LS 2 Smoothing The capacitor in this simple rectifier circuit is storing charge during the half cycle that the diode conducts. This charge is given up during the half cycle that the diode does not conduct. This helps to smooth out the waveform.

ICapacitorcharges Capacitor

discharges

t

Capacitor as a transducer A parallel plate capacitor can be used to convert mechanical movements or vibration of one of its plates into changes in voltage. This idea forms the basis of many measuring systems, e.g. by allowing a force to compress the plates we have a pressure transducer.


PROBLEMS Revision of Circuits 1. If a current of 40 mA passes through a lamp for 16 s, how much charge has passed any

point in the circuit? 2. A lightening flash lasted for 1 ms. If 5 C of charge was transferred during this time,

what was the current? 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. What is the p.d. across a 2 kΩ resistor if there is a current of 3 mA flowing through it? 5. Find the readings on the meters in the following circuits.

A10 V

70 Ω

10 Ω

20 Ω

(a)

(b) A12 V

15 Ω

9 Ω

V

A

3 Ω

5 Ω

V

6 V

(c)

R = ?

A40 V

5 Ω

10 Ω

I = 2 AA

40 V

4 Ω

10 Ω

R = ?

6. Find the unknown values of the following resistors. (a) (b)

20 V


7. Find the total resistance of the following combinations. (a) (b)

5 Ω

25 Ω

8 Ω

10 Ω 10 Ω

20 Ω10 Ω 10 Ω

10 Ω

10 Ω

10 Ω 10 Ω

1 Ω

20 Ω

4 Ω

5 Ω

10 Ω

20 Ω

10 Ω

10 Ω

25 Ω

5 Ω10 Ω

12 Ω

3 Ω

6 Ω5 Ω

A

3 kΩ

5 kΩ

V

(c) (d) (e) (f) 8. If the ammeter reads 2 mA,

find the voltmeter reading.


9. Calculate the power in each of the following cases. (a) A 12 V accumulator delivering 5 A. (b) A 60 Ω heater with a 140 V supply. (c) A 5 A current in a 20 Ω heater coil.

10. An electric kettle has a resistance of 30 Ω.

(a) What current will flow when it is connected to a 230 V supply? (b) Find the power rating of the kettle.

A

6 V

6 Ω 6 ΩV6 Ω

6 V

A

5 Ω 20 Ω2 Ω

11. A 15 V supply produces a current of 2 A for 6 minutes. How much energy is supplied in this time?

12. Find the readings on the

ammeters and the voltmeter. 13. Assuming each of the four cells

cell to be identical, find: (a) the reading on the ammeter (b) the current through the 20 Ω resistor (c) the voltage across the 2 Ω resistor.

14. A coil has a current of 50 mA flowing through it when the applied voltage is 12 V.

Find the resistance of the coil. 15. Write down the rules which connect the (a) potential differences and (b) the currents in

series and parallel circuits. 16. Draw the symbol for a fuse, diode, capacitor, variable resistor, battery and a d.c. power

supply.

10 V

R2

R1 V1

V2

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

18. Find the values of V1 and V2 of the circuit in question 17 if:

(a) R1 = 1 kΩ R2 = 49 kΩ (b) R1 = 5 kΩ R2 = 15 kΩ.


V1

V2

10 VR119. Explain what would happen to the

readings on V1 and V2 if light was shone onto the L.D.R. Suppose the L.D.R. was replaced with a thermistor which was then heated. Explain the effect on the readings.

20. (a) What would be the polarity of A and B when connected to a 5 V supply, so

that the LED would light? A

B

R

(b) What is the purpose of R in the circuit shown above? (c) If the L.E.D. rating is 200 mA at 1.5 V, find the value of R.


ELECTRICY AND ELECTRONICS PROBLEMS Electric fields and resistors in circuits 1. Draw the electric field pattern for the following charges:

(a) (b) (c) 2. Describe the motion of the small test charges in each of the following fields.

+ test charge

+ -

+ test charge (a) (b)

+ -+++

---

(c) (d)

- Q

-+ -+

+ Q --

+ + 3. An electron volt is a unit of energy. It represents the change in potential energy of an

electron which moves through a potential difference of 1 volt. If the charge on an electron is 1.6x10-19 C, what is the equivalent energy in joules?

4. Mass of an electron = 9.1 × 10-31 kg Charge on an electron = 1.6 × 10-19 C

The electron shown opposite is accelerated across a p.d. of 500 V. (a) How much electrical work is done? (b) How much kinetic energy has it gained? (c) What is its final speed?

----

- e

+500 V++++

5. Electrons are ‘fired’ from an electron gun at a screen. The p.d. across the gun is

2000 V. After leaving the positive plate the electrons travel at a constant speed to the screen. Assuming the apparatus is in a vacuum, at what speed will the electrons hit the screen?

----

++++

- e

Electron gun Screen


6. What would be the increase in speed of an electron accelerated from rest by a p.d. of 400 V?

7. An ×-ray tube is operated at 25 kV and draws a current of 3 mA. (a) Calculate

(i) the kinetic energy of each electron as it hits the target (ii) the velocity of impact of the electron as it hits the target (iii) the number of electrons hitting the target each second.

(mass of electron = 9.1 × 10-31 kg charge on electron = 1.6 × 10-19 C) (b) What happens to the kinetic energy of the electrons?

8. Sketch the paths which (a) an a-particle,

(b) a b-particle, and (c) a neutron,

would follow if each particle entered the given electric fields with the same velocity. (Students only studying this unit should ask for information on these particles).

- - - -

Path of particle----

++++

Path of particle 9. State what is meant by (a) the e.m.f. of a cell

(b) the p.d. between 2 points in the circuit. 10. Prove the expressions for the total resistance of resistors in (a) a series and (b) a

parallel circuit. 11. In the circuit below:

(a) what is the total resistance of the circuit (b) what is the resistance between X and Y (c) find the readings on the ammeters (d) calculate the p.d. between X and Y (e) what power is supplied by the battery ?

A1 A2

3 Ω

12 Ω 4 Ω

X

Y

12 V

12 Ω

8 Ω 24 Ω

S

230 V

12. The circuit opposite uses the 230 V

alternating mains supply. Find the current flowing in each resistor when: (a) switch S is open (b) switch S is closed.


13. An electric cooker has two settings, high and low. It takes 1 A at the low setting and 3 A at the high setting. (a) Find the resistance of R1 and R2. (b) What is the power consumption at each setting ?

R1230 V R2

12 V36 WR

24 V 14. (a) Find the value of the series resistor which

would allow the bulb to operate at its normal rating.

(b) Calculate the power dissipated in the resistor. 15. In the circuit below, r represents the internal resistance of the cell and R represents the

external resistance of the circuit. When S is open, the voltmeter reads 2.0 V. When S is closed, it reads 1.6 V and the ammeter reads 0.8 A. (a) What is the e.m.f. of the cell ? (b) What is the terminal potential difference

when S is closed? (c) Calculate the values of r and R. (d) If R was halved in value, calculate the new readings on

the ammeter and voltmeter.

RS

V

r

A

16. The cell in the diagram has an e.m.f. of 5 V. The current through the lamp is 0.2 A

and the voltmeter reads 3 V. Calculate the internal resistance of the cell.

V

r

A

17. A cell of e.m.f. 4 V is connected to a load resistor of 15 W. If 0.2 A flows round the

circuit, what must be the internal resistance of the circuit?


18. A signal generator has an e.m.f. of 8 V and internal resistance of 4 Ω. A load resistor is connected to its terminals and draws a current of 0.5 A. Calculate the load resistance.

19. (a) What will be the terminal p.d. across the cell in the circuit below.

R

rE = 1.5 V lost volts = 0.2 V

(b) Will the current increase or decrease as R is increased? (c) Will the terminal p.d. then increase or decrease ? Explain your answer.

20. A cell with e.m.f. 1.5 V and internal resistance 2 Ω is connected to a 3 Ω resistor.

What is the current? 21. A pupil is given a voltmeter and a torch battery. When he connects the voltmeter

across the terminals of the battery it registers 4.5 V, but when he connects the battery across a 6 Ω resistor, the voltmeter reading decreases to 3.0 V. (a) Calculate the internal resistance of the battery. (b) What value of resistor would have to be connected across the battery to reduce

the voltage reading to 2.5 V. 22. In the circuit shown, the cell has an e.m.f. of

6.0 V and internal resistance of 1 Ω. When the switch is closed, the reading on the ammeter is 2 A. What is the corresponding reading on the voltmeter ? 23. In order to find the internal resistance of a cell, the following sets of results were

taken.

Voltage (V) 1.02 0.94 0.85 0.78 0.69 0.60

Current (A) 0.02 0.04 0.06 0.08 0.10 0.12

(a) Draw the circuit diagram used. (b) Plot a graph of these results and from it determine

(i) the e.m.f. (ii) the internal resistance of the cell.

(c) Use the e.m.f. from part (b) to calculate the lost volts for each set of readings and hence calculate 6 values for the internal resistance.

(d) Calculate the mean value of internal resistance and the approximate random uncertainty.


24. The voltage across a cell is varied and the corresponding current noted. The results are shown in the table below.

Voltage (V) 5.5 5.6 5.7 5.8 5.9

Current (A) 5 4 3 2 1

Plot a graph of V against I. (a) What is the open circuit p.d? (b) Calculate the internal resistance. (c) Calculate the short circuit current. (d) A lamp of resistance 1.5 Ω is connected across the terminals of this supply. Calculate (i) the terminal p.d.

and (ii) the power delivered to the lamp. 25. Calculate the p.d. across R2 in each case.

R2

R1

+5 V

0 V

2 kΩ

8 kΩ R2

+5 V

0 V

4 kΩ

1 kΩ

+5 V

R2

0 V

500 Ω

750 Ω

(a) (b) (c)

t 26. Calculate the p.d. across AB (voltmeter reading) in each case.

B B

+12 V

0V

3 kΩ

3 kΩ

V

9 kΩ

3 kΩ

A

+10 V

0V

3 kΩ

2 kΩ

V6 kΩ

4 kΩ

A B

+5 V

0V

5 kΩ

2 kΩ

V

10 kΩ

8 kΩ

A

+9 V0 V

6 kΩ

3 kΩ

V

9 kΩ

6 kΩ

A

B

(a) (b) (c) 27. (a) Calculate the reading on the voltmeter. (b) What alteration could be made to balance the bridge circuit ?


28. Three pupils are asked to construct balanced Wheatstone bridges. Their attempts are shown.

+1.5 V

5 Ω

10 Ω

G

10 Ω

5 Ω

A

B

+1.5 V

7 Ω

10 Ω

G

14 Ω

20 Ω

A

B

+1.5 V

9 Ω

12 Ω

12 Ω

16 Ω

A

B

G

Pupil A Pupil B Pupil C

One of the circuits gives a balanced Wheatstone bridge, one gives an off - balance Wheatstone bridge and one is not a Wheatstone bridge.

(a) Identify each circuit. (b) How would you test that balance has been obtained ? (c) In the off – balance Wheatstone bridge ; (i) calculate the potential difference across the galvanometer. (ii) in which direction will electron current flow through the galvanometer. 29. Calculate the value of the unknown resistor × in each case.

5 Ω

10 Ω

G

20 Ω

X

+1.5 V

R

9 Ω

2 V

120 Ω

2 V

4 kΩ

12 kΩ

G

x

15 k

120 Ω

G

x Ω10 kΩ 25 kΩ

G

3.6 kΩ x

2 V

30. The circuit shown opposite is balanced. (a) What is the value of resistance ×? (b) Will the bridge be unbalanced if

(i) a 5 Ω resistor is inserted next to the 10 Ω resistor

(ii) a 3 V supply is used. (c) What is the function of resistor R and what is the

disadvantage of using it as shown ?


31. The following Wheatstone bridge circuit is used to monitor the mechanical strain on a girder in an oil rig.

(a) Explain how the circuit can

be used to monitor the strain. (b) Sketch the graph of current

through the galvanometer against the strain.

R3

G

+1.5 V

R4

Strain gauges

32. 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. The two meter readings were:

mA

10 2

1.71.8

OO1.3 V

(a) What are the readings? (b) What is the nominal resistance calculated from these readings? (c) Which reading is likely to cause the greatest uncertainty? (d) What is the smallest division on the milliammeter? (e) What is the absolute uncertainty on the milliammeter? (f) What is the absolute uncertainty on the voltmeter? (g) What is the percentage uncertainty on the milliammeter? (h) What is the percentage uncertainty on the voltmeter? (i) Which is the greatest percentage uncertainty? (j) What is the percentage uncertainty in the resistance? (k) What is the absolute uncertainty in the resistance? (l) Express the final result as “(resistance ± uncertainty)Ω” (m) Round both the result and the uncertainty to the relevant number of significant

figures or decimal places.


Alternating Current and Voltage 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 one second.

2. The circuit below is used to compare the a.c. and d.c. supplies when the lamp is at the

same brightness with each supply. The variable resistor is used to adjust the brightness of the lamp.

BA

(a) Explain how the brightness of the lamp is changed using the variable resistor. (b) What additional apparatus would you use to ensure the brightness of the lamp

was the same for each supply? (c) In the oscilloscope traces shown below diagram 1 shows the voltage across the

lamp when the switch is in position B and diagram 2 shows the voltage when the switch is in position A.

3.6 cm

Diagram 1

10.2 cm

Diagram 2

Y Gain set to 1 V cm-1

From the oscilloscope traces, how is the root mean square voltage numerically related to the peak voltage. (d) Redraw diagrams 1 and 2 to show what would happen to the traces if the time

base was switched on. 3. The root mean square voltage produced by a low voltage power supply is 10 V r.m.s.

(a) Calculate the peak voltage of the supply. (b) If the supply was connected to an oscilloscope, Y-gain set to 5 V cm-1 with the

time base switch off, describe what you would see on the screen.


4. (a) A transformer has a peak output voltage of 12 V. What is the r.m.s. output voltage?

(b) A vertical line 6 cm long appears on an oscilloscope screen when the Y gain is set to 20 V cm-1. Calculate: (i) the peak voltage of the input (ii) the r.m.s. voltage of the input.

6 ± 0.1 cm

5. The following trace appears on an oscilloscope screen when the time base is set at 2.5 ms cm-1.

(a) What is the frequency of the input including the uncertainty to the nearest Hz? (b) Sketch what you would see on the screen if the time base was changed to

(i) 5 ms cm-1 (ii) 1.25 ms cm-1

6. An a.c. input of frequency 20 Hz is connected to an oscilloscope with time base set at

0.01 s cm-1. What would be the wavelength of the waves appearing on the screen? Capacitance 7. A 50 µF capacitor is charged till the voltage across its plates reaches 100 V.

(a) How much charge has been transferred from one plate to the other? (b) If it was discharged in 4 milliseconds, what would be the average current?

8. A capacitor holds a charge of 3 × 10-4 C when it is charged to 600 V.

What is the value of the capacitor? 9. A 30 µF capacitor holds a charge of 12 × 10-4 C.

(a) What is the voltage that it is charged to? (b) If the tolerance of the capacitor is ± 5 µF, express this uncertainty as a

percentage. (c) What is the greatest voltage which could occur across the plates of the

capacitor? 10. A 15 µF capacitor is charged from a 1.5 V battery. What charge will be stored on the

plates? 11. (a) A capacitor has a voltage of 12 V across its plates and stores a charge of

1.2 × 10-5 C. Calculate its capacitance. (b) A 0.1 µF capacitor is connected to a 8 volt direct supply. How much charge

will it store?


12. Using the circuit below a capacitor is charged at a constant charging current of 2.0x10-5 A.

A

The time taken to charge the capacitor is 30 s and during this time the voltage across the capacitor rises from 0 V to 12 V. What is the capacitance of the capacitor?

13. A 100 µF capacitor is charged from a 20 V supply.

(a) How much charge is stored? (b) How much energy is stored in the capacitor?

14. A 30 µF capacitor stores 6 × 10-3 C of charge. How much energy is stored in the

capacitor? 15. The circuit below is used to investigate the charging of a capacitor.

A

10 kΩ 12 V

2000 µF

(a) What is the response of the ammeter when switch S is closed? (b) How can you tell when the capacitor is fully charged? (c) What would be a suitable range for the ammeter? (d) If the 10 kΩ resistor is replaced by a larger resistor, what will be the effect on

the maximum voltage across the capacitor?


16. In the circuit below the neon lamp flashes at regular intervals. The neon lamp requires a potential difference of 100 V across it before it will conduct and flash. It continues to glow until the potential difference across it drops to 80 V. While lit, its resistance is very small compared with R.

120 V dc

R

C

(i) Explain why the neon bulb flashes. (ii) Suggest two methods of decreasing the flash rate.

17. In the circuit below the capacitor C is charged with a steady current of 1mA by

carefully adjusting the variable resistor R.

9 V dc C

A

V

The voltmeter reading is taken every 10 seconds. The results are shown in the table.

Time 0 10 20 30 40

p.d.(V) 0 1.9 4 6.2 8.1

(a) Plot a graph of charge against voltage for the capacitor and hence find its capacitance (use graph paper).

(b) Calculate the capacitance for each of the readings (ignoring readings for t = 0). (c) Calculate the mean capacitance and the approximate random uncertainty in the

mean to two decimal places.


18. The circuit below is used to charge and discharge the capacitor

VR

VC

100 V

1 2

(a) What position should the switch be set (i) to charge and (ii) to discharge the capacitor?

(b) Draw graphs of VR against time for the capacitor charging and discharging, and of VC against time for the capacitor charging and discharging.

(c) If the capacitor has capacitance of 4.0 µF and the resistor has resistance of 2.5 MΩ calculate: (i) the maximum charging current in the circuit above (ii) the maximum charge stored by the capacitor when fully charged in the

above circuit.

3 V

3 MΩ

3 µF

19.

(a) For the above circuit draw graphs of (i) VC against time during charging and (ii) VA against time during charging.

(b) Calculate the final voltage across the capacitor and the final charged stored by it.

20. For each of the circuits below state what happens to the current flowing when the

frequency is (i) increased and (ii) decreased.

Low voltage variable frequency supply

Circuit 1 Circuit 2


21. In the circuit below the signal generator is set at 6.0 Vr.m.s., 1000 Hz. The lamp operates normally.

Signal generator

(a) Explain why the lamp can operate normally when the plates of the capacitor are separated by an insulator.

(b) What happens to the brightness of the lamp when the frequency of the signal generator is increased. Why does this happen?

22. For each of the following circuits sketch a graph of current against frequency.

Signal generator

Diagram A Diagram B

Signal generator

A B

A A

23.

The supply frequency to the above circuit is increased from a very low frequency, while the supply voltage remains constant. What will happen to the brightness of lamp A and B?


NUMERICAL ANSWERS Revision of circuits 1. Q = 0.64 C 2. Q = 5 × 103 A 3. t = 2 × 104 s. 4. V = 6 V 5. (a) 0.1 A (b) 0.5 A, 4.5 V (c) 2 A, 10 V. 6. (a) 5 Ω (b) 6 Ω 7. (a) 15 Ω (b) 25 Ω (c) 24.2 Ω (d) 13.3 Ω (e) 22.9 Ω (f) 14.7 Ω. 8. 3.75 × 10-3 V 9. (a) 60 W (b) 327 W (c) 500 W 10. (a) 7.67 A, (b) 1763 W 11. 9000 J 12. 0.67 A, 4 V 13. (a) 0.67 A (b) I (20 Ω) = 0.13 A, I (5 Ω) = 0.54 A (c) 13.4 V 14. 240 Ω 18. (a) V1 = 0.2 V V2 = 9.8 V (b) V1 = 2.5 V V2 = 7.5 V 20 (c) 17.5 Ω Electric fields and resistors in circuits 3. 1.6 × 10-19 J 4. (a) 8 × 10-17 J (b) 8 × 10-17 (c) v = 1.33 × 106 m s-1

5. v = 2.65 × 107 m s-1. 6. v = 1.2 × 107 m s-1 7. (a) (i) 4 × 10-15 J (ii) 9.4 × 107 m s-1 (iii) 1.875 × 1016 electrons 11. (a) 6 Ω (b) 3 Ω (c) 1.5 A (ammeter 2), 2 A (ammeter 1) (d) 6 V (e) 24 W 12. (a) 11.5 A (b) 12.8 A (12 Ω), 9.6 A (8 Ω), 3.2 A (24 Ω) 13. (a) R1 = 240 Ω R2 = 120 Ω (b) Low - 240 W High - 720 W 14. (a) 4 Ω (b) 36 W 15. (a) 2 V (b) 1.6 V (c) r = 0.5 Ω, R = 2 Ω, 1.3 A, 1.3 V 17. 10 Ω 18. 5 Ω 19. 12 Ω 20. 19.(a) 1.3 V 21. 0.3 A 22. 3 Ω 23. 4 V 24. 23. (b) 1.1 V (Intercept), 4.2 Ω (-gradient) (c) 4.1 ± 0.02 Ω 25. 24. (a) 6 V (b) 0.1 Ω (c) 60 A (d)(i) 5.63 V (ii) 21.1 W 25. (a) 4 V (b) 1 V (c) 3 V 26. (a) 3 V (b) -0.8 V (c) 0 V 27. (a) 0.6 V 28. (c) (i) 0.5 V (ii) from B to A. 29. (a) 4 Ω (b) 45 Ω (c) 9 Ω 30. (a) 10 Ω 32. (a,e)1.76±0.01 mA, (b,f) 1.3±0.1V, (m) 740 ± 60 Ω


NUMERICAL ANSWERS Alternating current and voltage 1. (a) Vr.m.s. = 325 V (b) 100 times per second Vr.m.s. = 0.71 peak 3 (a) Vpeak = 14 V (b) A vertical line of height 5.7 cm is seen 4 (a) Vr.m.s. = 8.4 cm (b) (i) Vpeak = 60 V (ii) Vr.m.s. = 42 V 5. (a) frequency = 100 ±2 Hz 6. wavelength = 5 cm Capacitance 7. (a) Q = 0.005 C (b) I = 1.25 A 8. C = 50 µF 9. (a) V = 40 V, (b) 0.167 %, (c) 46.7 V 10 Q = 22 µF 11. (a) C = 1 µF (b) Q = 0.8 µC 12. C = 50 µF 13. (a) 0.002 C (b) E = 0.02 J 14. E = 0.6 J 15. (a) ammeter reading starts at 1.2 mA and falls steadily to zero (b) Ammeter reads zero (c) Range 0 to 2 mA 17. (a) C= 0.005 F 18. (a) (i) position 1 (ii) position 2 (c) (i) I = 40 µA (ii) Q = 400µC 19. (b) V = 3 V, Q = 9 µC This publication may be reproduced in whole or in part for educational purposes provided that no profit is derived from the reproduction and that, if reproduced in part, the source is acknowledged. First published 1999 Higher Still Development Unit PO Box 12754 Ladywell House Ladywell Road Edinburgh EH12 7YH


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