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Cambridge IGCSE0625

Electric circuits

Physics 0625 Chapter Notes

What this chapter covers

Electric circuits - Circuit diagrams and circuit componentsElectric circuits - Series and parallel circuitsElectric circuits - Action and use of circuit components
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Electric circuits notes

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In this chapter

  1. 1Ammeters and voltmeters
  2. 2Circuit symbols
  3. 3Thermistor (NTC)
  4. 4Light-dependent resistor (LDR)
  5. 5Potential divider
  6. 6Relay
  7. 7Diodes and LEDs
  8. 8Current in a series circuit, and meters
  9. 9E.m.f. and p.d. in series
  10. 10Resistance in series
  11. 11Current and p.d. in a parallel circuit
  12. 12Resistance in parallel
  13. 13Lamps in parallel
  14. 14More resistance, more p.d.
  15. 15The potential divider
  16. 16Calculating with a potential divider
  17. 17Thermistors and LDRs as dividers

1. Ammeters and voltmeters

An ammeter measures current. It is connected in series with the component.

A voltmeter measures potential difference (p.d.). It is connected in parallel with only that component.

A circuit diagram must show both meters in the right place.

Examiner insight

“A common error was placing the voltmeter in series with the ammeter.” “As expected, the most common error was interchanging of the symbols for the ammeter and the voltmeter.”

2. Circuit symbols

Learn the symbol for every component in the syllabus list. A cell, a battery, a d.c. power supply, an a.c. power supply and an unspecified power supply each have their own symbol.

A thermistor, a variable resistor and a fuse are often confused. A bell and a buzzer have similar symbols.

The symbol for a potential divider has two wire connectors and no gap between the arrow tip and the rectangle.

Examiner insight

“Candidates need to know the electrical symbols given in the syllabus including knowing the difference between symbols for a cell, a battery, a d.c. power supply, an a.c. power supply and an unspecified power supply.” “The most common error was to draw the symbol for a fuse or a variable resistor instead of a thermistor.”

3. Thermistor (NTC)

For an NTC thermistor, the resistance decreases as the temperature increases.

In a series circuit a lower resistance means a larger current. A lamp in series with a warm thermistor is brighter. A lamp in series with a cold thermistor is dimmer.

A good answer links the change in resistance to the change in current.

Examiner insight

“Many candidates incorrectly thought that the resistance of a thermistor increases as it gets hotter.” “Most candidates gained credit for stating that the current in the circuit increased, but a significant number did not state that this was due to the resistance of the thermistor decreasing.”

Worked example 13 marks

A lamp and a thermistor are connected in a series circuit. The lamp is glowing with normal brightness. The circuit is moved from a warm room to a cold room. Explain how the change in temperature affects the brightness of the lamp.

  1. 1

    brightness of the lamp decreases [1]

  2. 2

    temperature of the thermistor decreases, so the resistance of the thermistor increases [1]

  3. 3

    the current in the lamp decreases [1]

4. Light-dependent resistor (LDR)

The resistance of an LDR decreases as the brightness of the light on it increases. When the light is made less bright, its resistance increases.

A bright LDR in series with a lamp lets a larger current through. A dim LDR lets a smaller current through.

Examiner insight

“When the brightness is increased, the resistance of the LDR goes down.” “Many candidates gained credit for stating the current increased but failed to link this to a decrease in the resistance of the LDR.”

Worked example 12 marks

A light-dependent resistor X is connected in series with a 240 Ω fixed resistor and a 12 V battery. A voltmeter is connected across the 240 Ω resistor.

The brightness of the light shining on X is reduced. The resistance of X increases. State the effect of this change on the current in the fixed resistor and on the reading on the voltmeter.

  1. 1

    the current in the fixed resistor decreases [1]

  2. 2

    the reading on the voltmeter decreases [1]

5. Potential divider

A potential divider splits the e.m.f. across resistors in series. The p.d. is shared in proportion to their resistances.

The larger resistor takes the larger share. When the resistance of a thermistor or an LDR falls, its share of the p.d. falls.

The current is not divided: the same current is in every resistor in series.

V₁/V₂ = R₁/R₂

Examiner insight

“Many candidates incorrectly stated that current would be divided, and some candidates tried to answer by describing the diagram given in the question.” “Most candidates correctly identified the circuit symbol for an LDR, but not all of them could work out that increasing the light intensity decreases the LDR’s resistance, so that the LDR has a lower proportion of the total resistance and therefore a lower proportion of the total p.d. across it.”

Worked example 13 marks

A 12 V battery is connected in series with a 240 Ω fixed resistor and component X, a light-dependent resistor. A voltmeter is connected across the 240 Ω resistor.

The resistance of component X is 400 Ω when the light shines on it. Calculate the reading on the voltmeter.

  1. 1

    total resistance = 400 + 240 = 640 Ω, so I = V/R = 12/640 = 0.01875 A [1]

  2. 2

    V = IR = 0.01875 × 240 [1]

  3. 3

    = 4.5 V [1]

6. Relay

A relay has a coil. When the p.d. across the relay coil increases, the magnetic field produced by the coil gets stronger.

The relay coil then attracts the switch, which causes the lamp to light.

Examiner insight

“The majority of candidates struggled to apply their knowledge of how LEDs, relays, thermistors and LDRs behave in electrical circuits.”

7. Diodes and LEDs

A diode lets current pass in one direction only. The arrow in its symbol gives the direction of the current.

A light-emitting diode (LED) is a diode that glows when current passes. Connect it the other way round and it does not glow.

To see whether a lamp lights, trace the path of the current: it must be able to pass through every diode.

Examiner insight

“Stronger candidates stated that the LED was a diode which would only allow current in one direction. Weaker candidates just stated that the current did not flow.” “Fewer candidates realised that the direction of the arrow on the diode gave the correct direction for the current.”

Worked example 11 mark

A light-emitting diode (LED) is connected to a battery and glows.

The LED is then connected to the battery the other way round.

Which statement is correct?

A The LED will not glow. · B The LED will still glow but less brightly. · C The LED will glow more brightly. · D The current in the LED will be too large.

  1. 1

    A: the LED will not glow. [1]

Recap

  • Ammeter: in series with the component. Voltmeter: in parallel with only that component.
  • Learn the symbols. Do not confuse a thermistor, a variable resistor and a fuse.
  • NTC thermistor: resistance decreases as temperature increases.
  • LDR: resistance decreases as light brightness increases.
  • Potential divider: the e.m.f. is shared in proportion to the resistances. The current is not divided.
  • Relay: a stronger magnetic field in the coil attracts the switch.
  • Extended: a diode lets current pass in one direction only, so an LED connected the wrong way round does not glow.

8. Current in a series circuit, and meters

In a series circuit the current is the same at every point.

An ammeter is connected in series with the component. A voltmeter is connected in parallel with the lamp or other component it measures, and with only that component.

Examiner insight

“Candidates generally lacked the knowledge that the current at every point in a series circuit is the same.” “Weaker candidates did not have a good understanding that the voltmeter should be placed in parallel with only that component and the ammeter placed in series with the component.”

9. E.m.f. and p.d. in series

The combined e.m.f. of cells in series is the sum of their e.m.f.s.

In a series circuit the total p.d. across the components is equal to the sum of the individual p.d.s across each component.

E_total = E₁ + E₂

V_total = V₁ + V₂

Examiner insight

“Some candidates stated that the e.m.f. was equal to the voltmeter reading.” “Common incorrect units for e.m.f. were mA or J.”

Worked example 12 marks

Two identical cells are in series with a thermistor and a fixed resistor. A voltmeter is connected across the thermistor and the fixed resistor together.

The switch is closed. The reading on the voltmeter is 9.0 V. Determine the e.m.f. of one cell. Give the unit.

  1. 1

    e.m.f. of one cell = 9.0 ÷ 2 [1]

  2. 2

    = 4.5 V (with the unit) [1]

10. Resistance in series

The combined resistance of resistors in series is the sum of the resistances.

When the p.d. across the whole series circuit is used in R = V ÷ I, the answer is the total resistance of the circuit.

R_total = R₁ + R₂

Examiner insight

“Over 50% of candidates applied the formula (V = I R) without realising that the resistance used must be the total resistance in the series circuit.” “Some candidates attempted to calculate the total resistance without converting the current from mA to A.”

Worked example 13 marks

Two identical cells are in series with a thermistor, a fixed resistor and an ammeter. A voltmeter is connected across the thermistor and the fixed resistor together.

The reading on the voltmeter is 9.0 V. The reading on the ammeter is 4.6 mA. The resistance of the fixed resistor is 480 Ω. Calculate the resistance of the thermistor.

  1. 1

    total resistance = V ÷ I = 9.0 ÷ (4.6 × 10⁻³) = 1957 Ω [1]

  2. 2

    resistance of thermistor = total resistance - 480 = 1957 - 480 [1]

  3. 3

    = 1477 = 1500 Ω to 2 significant figures [1]

11. Current and p.d. in a parallel circuit

In a parallel circuit the current from the source is larger than the current in each branch.

The sum of the currents entering a junction equals the sum of the currents leaving it. So the current from the supply is the sum of the currents in the branches.

The p.d. across an arrangement of parallel resistances is the same as the p.d. across one branch.

I_total = I₁ + I₂

Examiner insight

“Most candidates did not recall the rules about current in parallel circuits to determine where the largest current would be in the circuit shown.” “Common misconceptions were that the current was split between resistors or that the current was the same in all branches of the circuit.”

12. Resistance in parallel

The combined resistance of two resistors in parallel is less than either resistor by itself. The charges have more than one path to take.

To calculate it, add the reciprocals, then invert the result.

1/R = 1/R₁ + 1/R₂

Examiner insight

“The logic is that the charge carriers now have a choice of paths to take, meaning that more charge carriers can pass per unit time.” “Many went on to calculate the parallel resistance correctly, but some did not invert their answers at the end.”

Worked example 11 mark

Two resistors, X and Y, are connected in parallel. The two resistors have a combined resistance of 24 Ω. Resistor X has a resistance of 40 Ω. What is the resistance of resistor Y?

A 16 Ω · B 32 Ω · C 60 Ω · D 64 Ω

  1. 1

    1/R_Y = 1/24 - 1/40 = 1/60, so C: 60 Ω. [1]

13. Lamps in parallel

In a lighting circuit the lamps are connected in parallel. If one lamp fails, the other branch is still complete, so the other lamps stay lit.

The lamps can be controlled independently.

Examiner insight

“The strongest answers stated a clear advantage of the parallel connection, e.g. that if one lamp breaks the other branch of the circuit is still complete.” “Candidates needed to avoid statements such as that lamps receive the same voltage which is true but is not an advantage.”

Recap

  • Series: the current is the same at every point.
  • Series: add the e.m.f.s, add the resistances, add the p.d.s.
  • Ammeter in series, voltmeter in parallel with the component.
  • Parallel: the supply current is larger than the current in each branch.
  • Parallel: the combined resistance is less than either resistor.
  • Lamps in parallel: if one fails the others stay lit, and each is controlled independently.
  • Extended: current into a junction = current out; p.d. across each branch is the same.
  • Extended: for two resistors in parallel, add the reciprocals, then invert.

14. More resistance, more p.d.

In a series circuit the current is the same in every component. A voltmeter is connected in parallel with the component it measures.

For a constant current, the p.d. across an electrical conductor increases as its resistance increases. So in a series circuit the resistor with the greater resistance has the greater p.d. across it.

V = IR

Examiner insight

“Most candidates correctly added a voltmeter symbol to the diagram, but weaker candidates drew the voltmeter in series with the LDR.” “Candidates who attempted to use V = IR often made mistakes, such as using the full p.d. across the resistance of a single component to find a value for I.”

Worked example 11 mark

The diagram shows a battery connected to two resistors, X and Y.

Two voltmeters are connected into the circuit. The reading on voltmeter 1 is greater than the reading on voltmeter 2.

Which row about the current in the two resistors and the resistance of the two resistors is correct?

currentsize of resistors
Acurrent is the same in both resistorsboth resistors have the same resistance
Bcurrent is the same in both resistorsresistance of X is greater than resistance of Y
Ccurrent in X is greater than current in Yboth resistors have the same resistance
Dcurrent in X is greater than current in Yresistance of X is greater than resistance of Y
  1. 1

    X and Y are in series, so the current is the same in both.

  2. 2

    Voltmeter 1 reads more, so X has the larger p.d. for the same current: X has the greater resistance.

  3. 3

    B: current is the same in both resistors; resistance of X is greater than resistance of Y [1]

15. The potential divider

A potential divider is two resistors in series across a supply. Each resistor takes a share of the e.m.f., so the two p.d.s add up to the e.m.f.

In a variable potential divider one resistance can be changed. This changes how the e.m.f. is shared. When one resistance increases, the p.d. across it increases and the p.d. across the other resistor decreases. The resistor with the larger resistance has the larger share.

V₁ + V₂ = e.m.f.

Examiner insight

“Complete answers showed an understanding that the electromotive force (e.m.f.) is shared between the fixed resistor and the LDR.” “The correct symbol for a potential divider includes two wire connectors and no gap between the arrow tip and the rectangle.”

16. Calculating with a potential divider

For two resistors in series, the p.d.s are in the same ratio as the resistances.

The two p.d.s also add up to the e.m.f., so the p.d. across one resistor is the e.m.f. minus the p.d. across the other. Use the ratio and the e.m.f. together to find each p.d.

Keep the subscripts matched: V₁ is the p.d. across R₁. You can also find the current from the e.m.f. and the total resistance, then use V = IR for one resistor.

R₁/R₂ = V₁/V₂

Examiner insight

“Use of the formula V1 / V2 = R1 / R2 gave stronger candidates the correct value for Vout.” “Common mistakes included inconsistent use of subscripts in the formula and substituting values to give the potential difference across the variable resistor.”

Worked example 13 marks

A student connects the circuit shown in Figure 7.1.

A bright light shines on component X.

The resistance of component X is 400 Ω when the light shines on it.

Calculate the reading on the voltmeter.

  1. 1

    X is a light-dependent resistor in series with the 240 Ω resistor, so the total resistance is 400 + 240 = 640 Ω.

  2. 2

    I = V/R = 12/640 = 0.01875 A [1]

  3. 3

    V = IR = 0.01875 × 240 [1]

  4. 4

    = 4.5 V [1]

17. Thermistors and LDRs as dividers

The resistance of a thermistor decreases as its temperature increases. The resistance of an LDR decreases as the light gets brighter.

Put either one in series with a fixed resistor and it takes a share of the e.m.f. When its resistance changes, its share of the p.d. changes, and the fixed resistor's share changes the other way.

Name the component each time you give a p.d.: write "the p.d. across the LDR", not "the p.d." alone.

Examiner insight

“Other candidates started off poorly by stating that the increase in the intensity of the light would result in an increase in the resistance of the (light-dependent resistor) LDR.” “It should be noted that only NTC thermistors are required by the syllabus so no credit could be given for an answer stating that the resistance (of the thermistor) increased.”

Worked example 11 mark

The circuit shown can be completed by inserting components at X and at Y.

The completed circuit is a potential divider in which the potential difference (p.d.) across component Y increases when the temperature increases.

Which row shows the components X and Y?

XY
Alight-dependent resistorresistor
Bresistorlight-dependent resistor
Cresistorthermistor
Dthermistorresistor
  1. 1

    The p.d. across Y increases, so the p.d. across X decreases: X loses resistance when hotter.

  2. 2

    D: X is a thermistor and Y is a resistor [1]

Recap

  • For the same current, the greater the resistance, the greater the p.d. across it: V = IR.
  • Extended: a potential divider shares the e.m.f. between two resistors in series, so V₁ + V₂ = e.m.f.
  • Extended: R₁/R₂ = V₁/V₂. Keep the subscripts matched.
  • Extended: a thermistor or LDR that loses resistance takes a smaller share of the e.m.f.

End-of-chapter exercise

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

  1. State whether an ammeter is connected in series or in parallel with the component it measures.1 mark
  2. A student wants to measure the current in a lamp and the p.d. across it. State how each meter is connected.2 marks
  3. A lamp is in series with an NTC thermistor. The thermistor is warmed. State what happens to the resistance of the thermistor and to the brightness of the lamp.2 marks
  4. A lamp is in series with a light-dependent resistor (LDR). The room is made darker. State what happens to the resistance of the LDR and to the brightness of the lamp.2 marks
  5. A 12 V battery is connected in series with a 20 Ω resistor and a 40 Ω resistor. A voltmeter is connected across the 40 Ω resistor. Calculate the reading on the voltmeter.3 marks
  6. A thermistor circuit operates a relay. The temperature rises and the p.d. across the relay coil increases. State what happens to the magnetic field produced by the coil and what the coil then attracts.2 marks
  7. A fire alarm circuit has a 9.0 V battery in series with an NTC thermistor and a 300 Ω fixed resistor. A voltmeter is connected across the fixed resistor. At room temperature the resistance of the thermistor is 600 Ω. In a fire it is 150 Ω. (a) Calculate the voltmeter reading at room temperature. (b) Calculate the voltmeter reading in a fire. (c) State how the reading changes as the temperature rises.5 marks
  8. An LED in a circuit with a cell and a resistor glows. The cell is then turned round. State what happens to the LED and explain why.2 marks
  9. A greenhouse warning light uses two circuits. In the first circuit, a battery is in series with an NTC thermistor and the coil of a relay. In the second circuit, the switch of the relay is in series with a cell, a resistor and an LED. (a) Explain why the LED glows when the greenhouse becomes hot. (b) A student builds the same two circuits. The switch closes when the greenhouse is hot, but the LED does not glow. Every component works. Suggest the mistake the student made and explain why the LED does not glow.5 marks
  10. A 6.0 V battery is connected across two branches in parallel. Branch 1 has a diode P in series with a 300 Ω fixed resistor. Branch 2 has a diode Q in series with an LDR and a 100 Ω fixed resistor. The current from the battery is in the direction of the arrow on P. It would be against the arrow on Q. Treat a diode that allows current as having no resistance. The LDR is in bright light and its resistance is 500 Ω. (a) Calculate the current in the battery. (b) The battery is turned round. Calculate the p.d. across the 100 Ω resistor. (c) The battery stays turned round. The light on the LDR is made less bright. State and explain what happens to the p.d. across the 100 Ω resistor.7 marks
  11. In a series circuit, an ammeter next to the cell reads 0.35 A. An identical ammeter is placed next to the lamp, at the other end of the circuit. State its reading and give a reason.2 marks
  12. Three cells, each of e.m.f. 1.5 V, are connected in series. State the combined e.m.f.1 mark
  13. A 15 Ω resistor and a 25 Ω resistor are connected in series. Calculate their combined resistance.2 marks
  14. A student wants to measure the current in a lamp and the p.d. across it. Describe where to connect the ammeter and the voltmeter.2 marks
  15. A 4.0 Ω resistor and a 12 Ω resistor are connected in parallel. A student says the combined resistance is 8.0 Ω, the average of the two. Explain why this cannot be correct.2 marks
  16. A lighting circuit has its lamps connected in parallel. State two advantages of this.2 marks
  17. At a junction in a parallel circuit, currents of 2.5 A and 1.7 A enter. One wire leaves the junction. Calculate the current in that wire.2 marks
  18. A 20 Ω resistor and a 30 Ω resistor are connected in parallel. Calculate their combined resistance.3 marks
  19. A caravan has a 12 V battery. Three branches are connected in parallel across the battery: a heater, a pump, and two identical lamps in series with each other. The current from the battery is 7.5 A. The current in the heater is 4.0 A and the current in the pump is 2.0 A. Calculate (a) the current in the lamp branch and (b) the resistance of one lamp.5 marks
  20. A technician tests a heater control unit. A 9.0 V battery is connected to an unknown resistor R in series with two heating elements. The two elements, of resistance 30 Ω and 60 Ω, are in parallel with each other. An ammeter in series with the 30 Ω element reads 0.20 A. Calculate (a) the p.d. across the two elements, (b) the current from the battery, (c) the combined resistance of the two elements and (d) the resistance of R.7 marks
  21. A conductor carries a constant current. State how the p.d. across it changes when its resistance increases.1 mark
  22. A 2.0 Ω resistor and a 6.0 Ω resistor are connected in series to a cell. State which resistor has the larger p.d. across it. Give a reason.2 marks
  23. The current in a resistor is 250 mA. The resistance of the resistor is 18 Ω. Calculate the p.d. across the resistor.3 marks
  24. A 15 Ω resistor and a 25 Ω resistor are connected in series. The current in the circuit is 0.30 A. Calculate the p.d. across each resistor. State which resistor has the larger p.d. across it.4 marks
  25. A fixed resistor and a variable resistor are connected in series to a cell. The resistance of the variable resistor is increased. State what happens to the p.d. across the variable resistor and to the p.d. across the fixed resistor.2 marks
  26. Two resistors, R₁ = 200 Ω and R₂ = 300 Ω, are connected in series across a 10 V supply. Calculate the p.d. across R₂.3 marks
  27. A thermistor is in series with a fixed resistor across a battery. A voltmeter is connected across the fixed resistor. The temperature of the thermistor increases. Explain what happens to the reading on the voltmeter.3 marks
  28. An LDR and a 500 Ω resistor are connected in series across a 6.0 V supply. In bright light the resistance of the LDR is 100 Ω. Calculate the p.d. across the LDR.3 marks
  29. An egg incubator uses a thermistor in series with a 1200 Ω fixed resistor across a 9.0 V battery. A voltmeter is connected across the thermistor. (a) The voltmeter reads 6.0 V. Calculate the resistance of the thermistor. (b) The incubator warms up and the voltmeter reading falls to 3.0 V. Calculate the new resistance of the thermistor.5 marks
  30. A street lamp is controlled by an LDR in series with a variable resistor across a 12 V supply. The lamp switches on when the p.d. across the LDR rises above 8.0 V. (a) The variable resistor is set to 1500 Ω. Calculate the resistance of the LDR at which the lamp switches on. (b) The lamp must switch on earlier in the evening, when the resistance of the LDR is only 2000 Ω. Calculate the new resistance needed for the variable resistor. (c) Explain why the lamp is off in bright daylight.7 marks

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