Cambridge IGCSE0625

Electrical quantities

Physics 0625 Chapter Notes

What this chapter covers

Electrical quantities - Electric chargeElectrical quantities - Electric currentElectrical quantities - Electromotive force and potential differenceElectrical quantities - ResistanceElectrical quantities - Electrical energy and electrical power

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Electrical quantities notes

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1. Positive and negative charges

There are two kinds of charge: positive and negative.

Charges of the same sign repel each other. Positive repels positive and negative repels negative.

Opposite charges attract: positive attracts negative.

Examiner insight

“These candidates did not understand that if one rod had gained electrons and the other rod had lost electrons, the rods would be oppositely charged and therefore attract.” “It should be noted that different charges are not equivalent to opposite or unlike charges.”

Worked example 11 mark

A student shows that one charged plastic rod repels another charged plastic rod.

Other students make the statements listed.

1 Both rods have gained electrons. 2 Both rods have lost electrons. 3 One rod has gained electrons and one rod has lost electrons.

Which statements are possible explanations of what the student shows?

A 1 and 2 · B 1 and 3 · C 2 and 3 · D 3 only

1
A: statements 1 and 2. The rods repel, so they have the same charge: both gained electrons or both lost electrons. [1]

2. Charging by rubbing and detecting charge

Rubbing a plastic rod or ruler with a cloth charges it by friction.

To detect the charge, bring the rod close to a charged sphere hanging from an insulating thread. If the two charges have the same sign, the sphere is pushed away, because like charges repel. If they have opposite signs, the sphere is attracted to the rod, because opposite charges attract.

Examiner insight

“Some candidates stated “the ruler repels the sphere” but this was stated in the question and so could not be credited.”

Worked example 11 mark

A metal sphere is suspended vertically by an insulating thread. The sphere is charged. A negatively charged metal rod is brought close to the sphere. Figure 6.1 shows the charged rod and charged sphere.

State and explain whether the charge on the metal sphere is positive or negative.

1
positive, because opposite charges attract (the sphere is attracted to the rod) [1]

3. Only electrons move

Charging by friction involves only a transfer of negative charge (electrons).

A solid that gains electrons becomes negatively charged. A solid that loses electrons is left positively charged.

Positive charges do not move.

Examiner insight

“Most candidates correctly described rubbing the ruler with a cloth but few went on to describe how electrons were moved from the ruler to the cloth.” “Many candidates gained partial credit by describing how to give the polythene strip a negative charge, but then failed to explain that the negative charge was the result of electrons moving onto the polythene strip.”

Worked example 11 mark

A balloon is charged by rubbing it with a cloth.

When a negatively charged plastic rod is held close to the charged balloon, the rod and balloon attract.

Which statement is correct?

A The balloon lost electrons to the cloth. B The balloon gained electrons from the cloth. C The balloon lost protons to the cloth. D The balloon gained protons from the cloth.

1
A: the balloon lost electrons to the cloth. The balloon attracts a negative rod, so it is positive. [1]

4. Conductors and insulators

In a metal such as copper, some electrons are free to move. These free electrons carry charge through the metal, so a metal is a conductor.

In an insulator, such as plastic, the electrons are not free to move.

To tell the two apart, join a charged metal sphere to earth through an ammeter, using the material as the lead. With a conductor the pointer deflects and then returns to zero as the charge flows away. With an insulator the pointer does not move.

Examiner insight

“Weaker candidates made statements about conductors having electrons and insulators not having electrons or conductors allowing current or electricity to flow.” “There was a common belief that positive ions can move around in a metal. It is just the electrons that do the moving.”

Worked example 11 mark

Which statement explains why copper is a good electrical conductor?

A Copper is a non-metal. B Electrons can flow freely in copper. C Particles in copper are closely packed. D The density of copper is very low.

1
B: electrons can flow freely in copper. [1]

5. Charge and the coulomb

Charge is measured in coulombs, symbol C.

The coulomb is the unit of charge only. Do not give C as the unit of current, and do not mix up an amount of charge with a current or with electrons.

Examiner insight

“Candidates needed to remember that the unit of charge is C, coulomb, to gain full credit.” “Some candidates confused amount of charge with either current or electrons.”

6. Electric fields

An electric field is a region in which an electric charge experiences a force.

The direction of the field at a point is the direction of the force on a positive charge at that point.

Examiner insight

“A clear definition of electric field includes the idea that it is a region or place where a charge experiences a force. Common insufficient answers described the region where any object experiences a force or the region around a charged object.”

Worked example 12 marks

Complete the definition of an electric field.

An electric field is a region where ........ experiences ........ .

1
an electric charge (or charged particle) [1]
2
experiences an electrostatic force [1]

7. Electric field patterns

Field lines show the direction of the field: away from a positive charge and towards a negative charge. Around a point charge or a charged sphere the lines are radial. They touch the surface and do not go inside.

Between two oppositely charged parallel plates the lines are parallel, evenly spaced and perpendicular to the plates. Draw them only in the gap, each one touching both plates.

Examiner insight

“Candidates need to remember that radial lines must point towards the centre of the sphere, even though the actual line does not extend inside the sphere.” “Some weaker candidates concentrated on edge effects or drew field lines outside the gap between the plates.”

Worked example 12 marks

Figure 8.1 shows two parallel, oppositely charged metal plates.

On Figure 8.1, draw four field lines to show the pattern and the direction of the electric field between the two parallel plates.

1
four field lines, evenly spaced, perpendicular to the plates, each touching both plates [1]
2
at least one arrow upwards, towards the negative plate, and no incorrect arrows [1]

Recap

Charges of the same sign repel. Opposite charges attract.
Charging by friction transfers only electrons. Gain electrons: negative. Lose electrons: positive.
A metal conducts because it has free electrons. In an insulator the electrons are not free to move.
Extended: charge is measured in coulombs.
Extended: an electric field is a region where a charge experiences a force.
Extended: the field direction is the direction of the force on a positive charge.
Extended: field lines point away from positive and towards negative.

8. Current as a flow of charge

Electric current is related to the flow of charge.

In a metal, the charges that flow are free electrons. An electric current in a metal is a flow of electrons, not ions.

Copper is a good electrical conductor because electrons can flow freely in it.

Examiner insight

“Some candidates confused the current in a metal with the current in an ionic solution and stated the current in a metal is a flow of ions.” “Current does not get ‘used up’. Current is a flow of charge, and charge is conserved.”

Worked example 11 mark

The connecting wires in the circuit are made of metal. State the name of the particles that flow in the metal wires.

1
electrons [1]

9. Ammeters

An ammeter measures current. It goes in series with the component. A voltmeter goes in parallel with it.

Ammeters have different ranges, and a reading may be in A or mA. Divide mA by 1000 to get A: 290 mA is 0.29 A.

On an analogue ammeter, work out what one division of the scale is worth before you read it. A digital ammeter is easier to read and has no parallax error.

Examiner insight

“A small number of candidates gave an incorrect ammeter reading, most frequently by misjudging the scale and recording 0.6 A rather than 0.12 A.” “A few candidates could not recall the correct circuit symbols for an ammeter or for a voltmeter.”

10. Direct and alternating current

The two types of current are direct current (d.c.) and alternating current (a.c.).

When a question asks what d.c. means, the full name is not enough: explain the meaning. d.c. does not mean ‘directly from the source or battery’. a.c. and d.c. do not have different kinds of charge.

Examiner insight

“Many of the answers seen merely stated that d.c. was the abbreviation for direct current instead of explaining its meaning.” “A common incorrect response was that d.c. was directly from the source or battery.”

11. Current, charge and time

Electric current is the amount of charge passing a point per unit time.

Charge is in coulombs (C). Time must be in seconds, so change minutes and hours first. Change mA to A as well.

I = Q/t

Q = It

Examiner insight

“Some candidates confused amount of charge with either current or electrons.” “Candidates needed to remember that the unit of charge is C, coulomb, to gain full credit.”

Worked example 12 marks

The battery moves 4300 C around the circuit in 1.8 h.

Complete the equation that defines electric current I. Show that the current in the motor is 0.66 A.

1
I = Q/t, charge ÷ time [1]
2
I = 4300/(1.8 × 60 × 60) = 0.66 A [1]

12. Conventional current and electron flow

Conventional current is from positive to negative.

The flow of free electrons is from negative to positive. The two arrows point in opposite directions.

Label each arrow, or it cannot be credited.

Examiner insight

“Although a large proportion of candidates could calculate the current in this question, a significant number chose the wrong direction for electron flow.” “The most common error was drawing unlabelled arrows that could not be given credit.”

Recap

Electric current is related to the flow of charge. Current is not used up.
In a metal the current is a flow of free electrons, not ions.
An ammeter goes in series. On an analogue scale, work out one division before you read it.
d.c. is direct current and a.c. is alternating current. d.c. does not mean ‘directly from the battery’.
Extended: current = charge passing a point per unit time, I = Q/t.
Extended: change minutes and hours to seconds, and mA to A, before using Q = It.
Extended: conventional current is + to –. Electrons flow – to +.

13. Electromotive force

The electromotive force (e.m.f.) of a source is the electrical work done by the source in moving a unit charge around a complete circuit.

E.m.f. is measured in volts (V). It is a property of the power source, not a force.

The work done is the energy transferred, so e.m.f. is the energy per unit charge. It is not the rate of energy transfer, and it is not the energy stored in a battery.

Examiner insight

“Just stating that e.m.f. is the voltage across a battery was insufficient to gain credit.” “A common error was to define e.m.f. as a force.”

Worked example 11 mark

Which quantity is defined as the energy transferred by a cell in driving unit charge around a complete circuit?

A current · B electromotive force (e.m.f.) · C power · D resistance

1
B: electromotive force (e.m.f.). [1]

14. Potential difference

The potential difference (p.d.) between two points is the work done by a unit charge passing through a component.

The p.d. between two points is measured in volts (V). E.m.f. is about the whole circuit. P.d. is about one component.

A full definition of p.d. states the work done (or energy), per unit charge, in a component. Stating only that it is the voltage gains no credit.

Examiner insight

“The most common mistakes were to refer to the whole circuit, confusing the definition with that of electromotive force (e.m.f), or to omit to state that it was the energy per unit charge.” “Weaker candidates stated “it is the voltage”.”

Worked example 11 mark

Which unit is equivalent to 1.0 V?

A 1.0 J / A · B 1.0 J / C · C 1.0 J / s · D 1.0 J / Ω

1
A p.d. is the work done per unit charge.
2
Work is in joules and charge is in coulombs, so B: 1.0 J / C. [1]

15. Using a voltmeter

A voltmeter measures the p.d. across a component. It is connected in parallel with the component. A voltmeter only is needed to measure the e.m.f. of a cell.

Voltmeters can be analogue or digital, and can have different ranges. Do not give an analogue reading to too great a resolution: 2.4 V, not 2.40 V. A digital meter is easier to read and avoids parallax error.

Examiner insight

“Most candidates correctly added a voltmeter symbol to the diagram, but weaker candidates drew the voltmeter in series with the LDR.” “A number of candidates missed out the word ‘more’ when referring to accuracy.”

Worked example 11 mark

Another student performs the same experiment but uses a digital ammeter. State one advantage of using a digital meter.

1
any one from: easier to read, no need to process divisions on the scale, no chance of parallax error, less or no zero error [1]

16. The e.m.f. equation

E.m.f. is the work done by the source per unit charge moved around the circuit.

Work in joules (J), charge in coulombs (C), e.m.f. in volts (V). So 1 V = 1 J/C.

From the equation below, the work done is W = EQ and the charge is Q = W/E.

E = W/Q

Examiner insight

“Symbols of C for Q and V for E were not acceptable.” “Incorrect answers included use of P = VI or Work done = Fd or an incorrect unit of W.”

Worked example 13 marks

A battery of e.m.f. 6.5 V is connected in series with a motor, a fixed resistor and a switch.

(a) Complete the equation that defines e.m.f. E.

(b) The switch is closed. The battery moves 4300 C around the circuit in 1.8 h. Calculate the work done by the battery.

1
(a) E = W/Q [1]
2
(b) W = EQ = 6.5 × 4300. The time is not needed. [1]
3
= 27 950 = 2.8 × 10⁴ J [1]

17. The p.d. equation

The p.d. across a component is the work done per unit charge passing through it.

It has the same form as the e.m.f. equation. Rearrange it to find the work done or the charge.

V = W/Q

Examiner insight

“The most common error was to forget to convert the time of 1 minute into seconds when using the equation Q = It.”

Worked example 11 mark

A smoke detector is connected to a 9.0 V cell. The cell transfers energy at a rate of 1.0 × 10⁻⁴ W. How much charge passes through the smoke detector in one day?

A 0.040 C · B 0.96 C · C 78 C · D 7.8 × 10⁹ C

1
B: 0.96 C. Energy in one day = 1.0 × 10⁻⁴ × 86 400 = 8.64 J, so Q = W ÷ V = 8.64 ÷ 9.0. [1]

Recap

E.m.f. = electrical work done by a source in moving a unit charge around a complete circuit. Unit: volt.
P.d. = work done by a unit charge passing through a component. Unit: volt.
E.m.f. is for the whole circuit. P.d. is for one component.
A voltmeter is connected in parallel. Digital meters are easier to read and avoid parallax error.
Extended: E = W/Q and V = W/Q.

18. Resistance and R = V / I

Resistance R is found from the p.d. V across a component and the current I in it. Its unit is the ohm, Ω.

Write the equation first: R = V ÷ I. Then substitute.

R = V/I

Examiner insight

“A common mistake was to use an incorrect form of the equation, such as R = I / V.” “Weaker candidates used an incorrect transformation of V = IR.”

Worked example 13 marks

A heater is connected to a 230 V mains supply. The current in the heater is 11.5 A. Calculate the resistance of the heater.

1
R = V ÷ I [1]
2
R = 230 ÷ 11.5 [1]
3
= 20 Ω [1]

19. Measuring resistance with a voltmeter and ammeter

To find the resistance of a component, measure the p.d. across it and the current in it. The voltmeter is connected in parallel with the component. The ammeter is connected in series with it.

Record V to at least 1 decimal place and I to at least 2 decimal places, with the units V and A. Then calculate R = V ÷ I and give it to 2 or 3 significant figures.

If the current is in mA, convert it to A first.

R = V/I

Examiner insight

“The potential difference had to be given to at least 1 decimal place, and the current to at least 2 decimal places.” “Some candidates attempted to calculate the total resistance without converting the current from mA to A.”

Worked example 13 marks

The potential difference (p.d.) across the lamp is 3.9 V. The current in the lamp is 0.60 A. Calculate the resistance of the lamp.

1
R = V ÷ I [1]
2
R = 3.9 ÷ 0.60 [1]
3
= 6.5 Ω [1]

20. Resistance, length and thickness of a wire

For a metallic wire, a longer wire has greater resistance.

A thinner wire, with a smaller cross-sectional area, also has greater resistance. A thicker wire has smaller resistance.

A greater resistance means a smaller current in the circuit.

Examiner insight

“A common misconception was that thinner wires have smaller resistance.” “The dependence of resistance on length was well known, although “shorter wires have bigger resistance” was a common incorrect response.”

Worked example 11 mark

A length of metal wire is used as a resistor in a circuit. Which change results in an increase in current in the circuit?

A connecting a second identical length of wire in parallel with the original wire · B connecting a second identical length of wire in series with the original wire · C decreasing the thickness of the wire · D increasing the length of the wire

1
A: connecting a second identical length of wire in parallel with the original wire. [1]

21. I–V graphs: resistor and filament lamp

A resistor of constant resistance gives a straight line through the origin.

For a filament lamp, the current rises more slowly as the p.d. increases. The filament is heated to white heat, so its resistance increases. The graph is a curve.

Examiner insight

“A considerable number thought that the current is proportional to the applied p.d. This is not so because as the current is increased, the lamp filament is heated to white heat and consequently its resistance increases.”

22. I–V graph of a diode

A diode lets current pass in one direction only.

With the diode connected in the other direction, the current is zero. So the graph is not a straight line through the origin, and current is not proportional to voltage.

The sketch shows the direction that lets current pass.

Examiner insight

“Some candidates suggested that it showed current is proportional to voltage.” “Stronger answers stated that the current would be zero as the diode was in the wrong direction to allow current to flow.”

23. Resistance, length and cross-sectional area

For a metallic electrical conductor, resistance is directly proportional to length. Double the length and the resistance doubles.

Resistance is inversely proportional to cross-sectional area. Double the area and the resistance halves.

The area is proportional to the diameter squared. Double the diameter and the area is 4 times as large, so the resistance is divided by 4.

For a change in both length and area, deal with each factor in turn.

Examiner insight

“Most candidates knew that resistance is directly proportional to the length of the wire, but a few candidates incorrectly thought that the resistance is inversely proportional to the diameter of the wire.” “Candidates needed to recognise that when the diameter is decreased by a factor of 2 the cross sectional area is decreased by a factor of 4 and the resistance is increased by a factor of 4.”

Recap

R = V ÷ I. The unit is the ohm, Ω.
To measure resistance: voltmeter in parallel with the component, read V and I, then R = V ÷ I.
Convert mA to A before dividing.
A longer wire has greater resistance. A thinner wire has greater resistance.
Extended: a resistor of constant resistance gives a straight line through the origin on an I–V graph.
Extended: a filament lamp gives a curve, because its resistance increases as the filament heats up.
Extended: a diode lets current pass in one direction only.
Extended: R is directly proportional to length and inversely proportional to cross-sectional area.
Extended: double the diameter and the area is 4 times as large, so R is divided by 4.

24. Energy transfer in a circuit

An electric circuit transfers energy from a source, such as a cell or the mains supply, to the components. The components then transfer energy to the surroundings.

A battery stores chemical energy. Energy that reaches the surroundings is thermal / internal energy.

Examiner insight

“Some candidates thought that electrical energy is stored in a battery and others that light energy is transferred to the surroundings.”

25. Electrical power

The power of a component is found from the current in it and the potential difference across it.

Write the equation down first. Then substitute. Give the unit, the watt (W).

P = IV

Examiner insight

“Some candidates rearranged the equation for calculating power output.” “A commonly seen approach was dividing the voltage output by the current output.”

Worked example 13 marks

The voltage output of the solar panel is 4.5 V. The current output from the solar panel is 0.62 A. Calculate the power output from the solar panel.

1
power = I × V [1]
2
= 4.5 × 0.62 [1]
3
= 2.8 W [1]

26. Electrical energy

The energy transferred in a component depends on the current, the potential difference and the time. Time is in seconds.

Some candidates work in two stages: P = IV, then E = Pt. This gives the same answer.

E = IVt

E = Pt

Examiner insight

“A few candidates gained partial credit for an energy calculated without the time conversion.” “There was some difficulty for a minority of candidates in recalling the formula E = IVt.”

Worked example 13 marks

The current in one lamp is 0.45 A and the potential difference across the lamp is 6.7 V. The lamp is switched on for a time of 50 s. Calculate the energy supplied to the lamp.

1
energy = I × t × V [1]
2
= 0.45 × 50 × 6.7 [1]
3
= 150 J [1]

27. The kilowatt-hour

The kilowatt-hour (kWh) is a unit of energy, not of power.

1 kW h is the energy transferred in one hour at the rate of 1 kW. It is the energy transferred by a device with a power rating of 1 kW in one hour.

To find the energy in kW h, use power in kW and time in hours.

energy (kW h) = power (kW) × time (h)

Examiner insight

“Only the strongest candidates correctly stated that 1 kW h is the amount of energy transferred in one hour at the rate of transfer of 1 kW or the amount of energy transferred by an electrical device with a power rating of 1 kW in one hour.” “A common incorrect answer was that it was the number of kW per hour.”

28. Calculating the cost of energy

The cost of using an appliance is the energy in kW h multiplied by the cost of 1 kW h. The energy is power × time.

Convert watts to kilowatts and minutes to hours first. Convert 1200 W to 1.2 kW before you multiply.

cost = power (kW) × time (h) × cost of 1 kW h

Examiner insight

“The omission or incorrect conversion of power to kilowatts was the most common reason for only partial credit being awarded.” “Some candidates added the unit J to their answer which was incorrect.”

Worked example 14 marks

The power of the lawnmower is 1200 W. A person uses the lawnmower for 0.75 h. One kilowatt-hour of energy costs $0.34.

(i) Calculate the energy transferred to the lawnmower in 0.75 h. Give your answer in kW h.

(ii) Calculate the cost of the energy calculated in (i).

1
energy = P × t [1]
2
convert 1200 W to 1.2 kW [1]
3
1.2 × 0.75 = 0.90 kW h [1]
4
cost = 0.90 × 0.34 = 0.31 dollars [1]

Recap

A circuit transfers energy from the source to the components, then to the surroundings.
A battery stores chemical energy. Energy reaching the surroundings is thermal.
Power: P = IV, in watts.
Energy: E = IVt, with time in seconds.
1 kW h is the energy transferred in one hour at the rate of 1 kW.
Cost = energy in kW h × cost of 1 kW h. Use kW and hours.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

Two charged balloons hang side by side from insulating threads and move apart. State what this shows about their charges.1 mark
A plastic rod is rubbed with a cloth and becomes negatively charged. State what is transferred and the direction of the transfer.2 marks
A rod becomes positively charged when rubbed with a cloth. Explain how the rod became positive.2 marks
Explain why a copper wire is a good electrical conductor but a plastic rod is a good insulator.2 marks
A student rubs a plastic ruler with a cloth. The ruler is brought near a charged sphere hanging from an insulating thread. The charges on the ruler and the sphere have the same sign. Describe and explain what happens to the sphere.2 marks
A positively charged rod is brought near a charged sphere on an insulating thread. The sphere is attracted to the rod. State the sign of the charge on the sphere and give a reason.2 marks
A charged metal sphere stands on an insulating stand. A copper wire joins it to earth through a sensitive ammeter. The pointer deflects briefly and then returns to zero. Explain this.3 marks
(a) State the unit in which electric charge is measured. (b) A small positive charge in an electric field experiences a force to the left. State the direction of the field at that point. (c) A small negative charge replaces the positive charge. State the direction of the force on the negative charge.3 marks
A hollow metal dome stands on an insulating column. A moving rubber belt rubs against the dome and the dome loses electrons to the belt. Point P is level with the centre of the dome and to the left of it. A small charged dust grain at P is pushed directly away from the dome. (a) State the sign of the charge on the dome and explain your answer. (b) State the direction of the electric field at P. (c) State the sign of the charge on the dust grain and explain your answer using the direction of the field.5 marks
In a dust filter, air flows between two parallel vertical metal plates. Plate X is on the left and plate Y is on the right. The plates are oppositely charged. On the way in, a dust grain rubs against a plastic pipe and gains electrons from it. Between the plates, the electric field pushes the grain towards plate Y. (a) State the sign of the charge on the grain. (b) State the direction of the electric field between the plates and explain your answer. (c) State the sign of the charge on each plate. (d) Describe the pattern of the field lines between the plates. (e) A second grain loses electrons to the pipe. State the direction of the force on this grain between the plates.7 marks
A copper wire carries an electric current. (a) State what is flowing in the wire. (b) State why copper is a good electrical conductor.2 marks
A student measures the current in a lamp. The ammeter reads 250 mA. (a) State how the ammeter is connected to the lamp. (b) Give the current in A.2 marks
The scale of an analogue ammeter is numbered 0, 0.2, 0.4, 0.6, 0.8 and 1.0 A. There are 10 small divisions between 0 and 0.2 A. The pointer is 6 small divisions past 0. Determine the current.2 marks
A student writes two statements. Statement 1: d.c. means the current comes directly from the battery. Statement 2: a.c. and d.c. are different kinds of charge. (a) State what the letters a.c. stand for. (b) State how many of the student’s statements are correct.2 marks
A student has two ammeters in series, one before a lamp and one after it. She says the lamp uses up some of the current, so the second ammeter reads less. Explain why she is wrong.2 marks
A charge of 90 C passes a point in a wire in 2.0 minutes. Calculate the current.3 marks
Free electrons in a metal wire move from the left end to the right end. State the direction of the conventional current. Explain your answer.2 marks
A sensor circuit has a steady current of 0.50 mA. Calculate the time taken for a charge of 3.6 C to pass through the sensor. Give your answer in hours.4 marks
A rechargeable cell is charged in two stages. The current is 450 mA for 1.5 hours, then 150 mA for 40 minutes. (a) Calculate the total charge that passes into the cell. (b) A second charger passes the same total charge in the same total time, using one constant current. Calculate this current.6 marks
An electric fence unit sends short pulses of current along a metal wire. In each pulse, a charge of 0.0030 C passes a point in the wire in 0.15 ms. During a pulse, the free electrons in the wire move away from the unit. (a) Calculate the current during one pulse. (b) The unit sends 40 pulses each minute. A steady current in a second wire moves the same charge in one minute. Calculate this steady current. Give your answer in mA. (c) State the direction of the conventional current in the fence wire during a pulse. State whether the wire is connected to the positive or the negative terminal of the unit.7 marks
State the unit of electromotive force.1 mark
Complete the definition. The e.m.f. of a source is the electrical ........ done by the source in moving a unit ........ around a complete ........ .3 marks
Define potential difference.2 marks
A cell has an e.m.f. of 1.5 V. A lamp has a p.d. of 1.5 V across it. Both are measured in volts. State one difference between what the two quantities describe.2 marks
A student wants to measure the p.d. across a lamp. State how the voltmeter is connected.1 mark
State one advantage of a digital voltmeter over an analogue voltmeter.1 mark
A cell does 18 J of work in moving 12 C of charge around a complete circuit. Calculate the e.m.f. of the cell.2 marks
The p.d. across a lamp is 6.0 V. A charge of 250 mC passes through the lamp. Calculate the work done on the charge.3 marks
An electric bicycle has a battery of e.m.f. 36 V. On one journey the battery does 0.27 MJ of work in moving charge around the complete circuit. (a) Calculate the charge moved around the circuit. Give your answer in kC. (b) On this journey, 240 kJ of work is done by this charge as it passes through the motor. Calculate the p.d. across the motor. (c) A student says the p.d. across the motor must be 36 V. State why the student is wrong.6 marks
A smoke alarm battery has an e.m.f. of 9.0 V. It can move 1.8 kC of charge around a complete circuit before it needs replacing. A torch cell has an e.m.f. of 1.5 V. It can move 12 kC of charge before it needs replacing. A student says the smoke alarm battery can do more work because its e.m.f. is larger. (a) Show by calculation whether the student is correct. (b) Calculate the charge the smoke alarm battery would have to move to do the same work as the torch cell. (c) The torch cell is used in a torch until it needs replacing. The work done by the charge passing through the lamp is 16.8 kJ. Calculate the p.d. across the lamp.7 marks
Write down the equation that links resistance, potential difference and current.1 mark
The potential difference (p.d.) across a lamp is 12 V. The current in the lamp is 0.50 A. Calculate the resistance of the lamp.2 marks
The p.d. across a resistor is 2.4 V. The resistance of the resistor is 480 Ω. Calculate the current in the resistor. Give your answer in mA.3 marks
A student wants to find the resistance of a lamp using a voltmeter and an ammeter. Describe what the student does.3 marks
A student measures the p.d. across a resistor and the current in it. The results are in the table. Calculate the resistance for each pair of readings and state whether the resistance is constant. | V / V | 1.0 | 2.0 | 3.0 | | --- | --- | --- | --- | | I / A | 0.20 | 0.40 | 0.60 |3 marks
A wire is replaced by a wire of the same metal that is longer and thinner. State the effect on the resistance and give a reason.2 marks
A diode is connected in a circuit in the direction that does not allow current to pass. State the current in the diode and explain your answer.2 marks
The graph shows the current–voltage graph for a filament lamp. (a) Calculate the resistance of the lamp when the p.d. is 8.0 V. (b) The resistance of the lamp when the p.d. is 2.0 V is 5.0 Ω. Explain why the resistance is different at 8.0 V.4 marks
A sensor contains a thin metal wire of length 0.60 m. When the p.d. across the wire is 1.5 V, the current in it is 12.5 mA. (a) Calculate the resistance of the wire. (b) A new sensor uses wire of the same metal with twice the cross-sectional area. The new wire must have a resistance of 180 Ω. Calculate the length of the new wire.5 marks
The heating wire in a plant propagator is a metal wire of length 3.0 m and diameter 0.20 mm. When the p.d. across the wire is 12 V, the current in it is 0.80 A. The wire is replaced by a wire of the same metal, of length 4.5 m and diameter 0.60 mm. The p.d. across the new wire is 6.0 V. Calculate the current in the new wire. The temperature of each wire is constant.6 marks
State the type of energy stored in a battery.1 mark
The current in a lamp is 0.50 A and the potential difference across it is 12 V. Calculate the power of the lamp.2 marks
A kettle has a power of 2.3 kW. It is connected to a 230 V supply. Calculate the current in the kettle.3 marks
The current in a heater is 3.0 A and the potential difference across it is 12 V. Calculate the energy transferred in 2.0 minutes.3 marks
A hair dryer is connected to a 230 V mains supply. It transfers 90 000 J of energy in 1.0 minute. Calculate the current in the hair dryer.3 marks
Define the kilowatt-hour.1 mark
A toaster has a power of 500 W. It is used for 12 minutes. Calculate the energy transferred in kW h.3 marks
A 750 W microwave oven is used for 20 minutes each day for 30 days. One kilowatt-hour of energy costs $0.34. Calculate the cost of using the oven for the 30 days.4 marks
An electric scooter is charged from a 230 V mains supply. Charging takes 4 hours 30 minutes and costs 81 cents. One kilowatt-hour of energy costs 30 cents. Calculate the current in the charger.5 marks
A filament lamp and an LED lamp are each connected to a 230 V mains supply. The current in the filament lamp is 0.26 A. The current in the LED lamp is 0.040 A. Each lamp is used for 5.0 hours a day for 365 days. One kilowatt-hour of energy costs 30 cents. Calculate, in dollars, how much less it costs to use the LED lamp for the 365 days.6 marks

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1. Positive and negative charges#

2. Charging by rubbing and detecting charge#

3. Only electrons move#

4. Conductors and insulators#

5. Charge and the coulomb#

6. Electric fields#

7. Electric field patterns#

8. Current as a flow of charge#

9. Ammeters#

10. Direct and alternating current#

11. Current, charge and time#

12. Conventional current and electron flow#

13. Electromotive force#

14. Potential difference#

15. Using a voltmeter#

16. The e.m.f. equation#

17. The p.d. equation#

18. Resistance and R = V / I#

19. Measuring resistance with a voltmeter and ammeter#

20. Resistance, length and thickness of a wire#

21. I–V graphs: resistor and filament lamp#

22. I–V graph of a diode#

23. Resistance, length and cross-sectional area#

24. Energy transfer in a circuit#

25. Electrical power#

26. Electrical energy#

27. The kilowatt-hour#

28. Calculating the cost of energy#

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Workbook

Chapter at a Glance

Free notes are just the start.

Related chapters

1. Positive and negative charges

2. Charging by rubbing and detecting charge

3. Only electrons move

4. Conductors and insulators

5. Charge and the coulomb

6. Electric fields

7. Electric field patterns

8. Current as a flow of charge

9. Ammeters

10. Direct and alternating current

11. Current, charge and time

12. Conventional current and electron flow

13. Electromotive force

14. Potential difference

15. Using a voltmeter

16. The e.m.f. equation

17. The p.d. equation

18. Resistance and R = V / I

19. Measuring resistance with a voltmeter and ammeter

20. Resistance, length and thickness of a wire

21. I–V graphs: resistor and filament lamp

22. I–V graph of a diode

23. Resistance, length and cross-sectional area

24. Energy transfer in a circuit

25. Electrical power

26. Electrical energy

27. The kilowatt-hour

28. Calculating the cost of energy