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105 questions
Physics/Paper 2/D.C. Circuits
CAIEAS Level9702-as · Paper 2

D.C. Circuits

105 questions· page 1 of 11

Q62025 May/Jun·P244 partsMedium-Easy
(a)

Determine RR.

RR = ______ Ω\Omega

(b)

Explain why the potential difference VV between any two points on wire XY is proportional to the distance LL between those points.

(c)(i)

Calculate EE.

EE = ______ V\text{V}

(c)(ii)

The value of RR is now decreased.

State and explain the change that must be made to the position of P on wire XY so that the galvanometer reads zero again.

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Q62024 May/Jun·P215 partsEasy
(a)

State Kirchhoff’s first law.

(b)

A cell with internal resistance rr is connected to two resistors of resistances R1R_1 and R2R_2 as shown in Fig. 6.1.

The potential differences (p.d.s) across R1R_1 and R2R_2 are V1V_1 and V2V_2 respectively.
The terminal p.d. across the cell is VV.
The current in the circuit is II.

Use Kirchhoff’s laws to show that the total resistance RTR_T of the external circuit is given by

RT=R1+R2R_T = R_1 + R_2
(c)

The electromotive force (e.m.f.) of the cell in Fig. 6.1 is 1.50 V1.50\text{ V}.

The values of R1R_1 and R2R_2 are 10 Ω10\ \Omega and 15 Ω15\ \Omega respectively. The terminal p.d. of the cell is 1.35 V1.35\text{ V}.

Calculate the internal resistance rr of the cell.

rr = ______ Ω\Omega

(d)(i)

the current in the cell

(d)(ii)

the terminal p.d. of the cell.

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Q52024 May/Jun·P224 partsEasy
(a)

State Kirchhoff’s second law.

(b)(i)

The current in the circuit is 1.1×102 A1.1 \times 10^{-2}\ \text{A}. The potential difference across YY is 4.0 V4.0\ \text{V}.

Calculate the resistance of XX.

resistance = ______ Ω\Omega

(b)(ii)

The temperature of YY is changed to 190C190^\circ\text{C}. The resistance of XX remains unchanged.

Determine the new potential difference across YY.

potential difference = ______ V\text{V}

(b)(iii)

The resistance of XX is increased. The temperature of YY remains at 190C190^\circ\text{C}.

By reference to the current in the circuit, state and explain the effect of this change, if any, on the potential difference across YY.

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Q72024 Oct/Nov·P215 partsMedium-Easy
(a)(i)

Show that the current II in the circuit is 0.50 A0.50\text{ A}.

(a)(ii)

Calculate the combined resistance of the two resistors.

resistance = ______ Ω\Omega

(a)(iii)

Use your answer in (a)(ii) to determine resistance RR.

RR = ______ Ω\Omega

(b)(i)

On Fig. 7.2, complete the circuit diagram to show this arrangement.

(b)(ii)

Explain, without calculation, whether the terminal p.d. across the cell is now less than, equal to or greater than 1.36 V1.36\text{ V}.

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Q72023 May/Jun·P223 partsMedium-Easy
(a)

A battery of electromotive force (e.m.f.) 9.0 V9.0\ \text{V} and negligible internal resistance is connected to a light-dependent resistor (LDR) and a fixed resistor, as shown in Fig. 7.1.

The LDR and fixed resistor have resistances of 1800 Ω1800\ \Omega and 1200 Ω1200\ \Omega respectively.

Calculate the potential difference across the LDR.

potential difference = ______ V\text{V}

(b)(i)

Calculate the length XZ along the resistance wire.

length XZ = ______ m\text{m}

(b)(ii)

The environmental conditions change causing a decrease in the resistance of the LDR. The temperature of the LDR remains constant.

State whether there is a decrease, increase or no change to:

  • the intensity of the light illuminating the LDR
  • the total power produced by the battery
  • the length XZ so that the galvanometer reads zero.
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Q62023 Oct/Nov·P214 partsEasy
(a)(i)

Show that the current in the battery is 0.050 A0.050\ \text{A}.

(a)(ii)

Calculate the resistance of the LDR.

resistance = ______ Ω\Omega

(a)(iii)

Determine the ratio

power dissipated in the LDRpower dissipated in the 240 Ω resistor\frac{\text{power dissipated in the LDR}}{\text{power dissipated in the 240}\ \Omega\ \text{resistor}}

ratio = ______

(b)

The intensity of the light incident on the LDR increases.

State and explain what happens to the voltmeter reading.

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Q62022 May/Jun·P224 partsMedium-Easy
(a)

A network of three resistors of resistances R1R_1, R2R_2 and R3R_3 is shown in Fig. 6.1.

The individual potential differences across the resistors are V1V_1, V2V_2 and V3V_3. The current in the combination of resistors is II and the total potential difference across the combination is VV.

Show that the combined resistance RR of the network is given by

R=R1+R2+R3R = R_1 + R_2 + R_3
(b)(i)

Switch X is open.

Calculate the potential difference across R1R_1.

potential difference = ______ V\text{V}

(b)(ii)

Switch X is now closed. The resistance of the thermistor is 12.0 kΩ12.0\ \text{k}\Omega.

Calculate the current in the battery.

current = ______ A\text{A}

(c)

The switch X in the circuit in (b) remains closed. The temperature of the thermistor decreases.

By reference to the current in the battery, state and explain the effect, if any, of the decrease in temperature on the power produced by the battery.

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Q52021 May/Jun·P236 partsEasy
(a)

Define the electromotive force (e.m.f.) of a source.

(b)

The circuit shown in Fig. 5.1 contains a battery of e.m.f. EE that has internal resistance rr, a variable resistor, a voltmeter and an ammeter.

Readings from the two meters are taken for different settings of the variable resistor. The variation with current II of the potential difference (p.d.) VV across the terminals XY of the battery is shown in Fig. 5.2.

Explain why VV is not constant.

(c)(i)

the e.m.f. EE

EE = ______ V\text{V}

(c)(ii)

the maximum current that the battery can supply

maximum current = ______ A\text{A}

(c)(iii)

the internal resistance rr.

rr = ______ Ω\Omega

(d)

On Fig. 5.2, sketch a line to show a possible variation with II of VV for a battery with a lower e.m.f. and a lower internal resistance than the battery in (b). Your line should extend over at least the same range of currents as the original line.

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Q62020 May/Jun·P224 partsMedium-Easy
(a)(i)

Explain, in terms of energy and without a calculation, why the potential difference across the lamp must be less than the e.m.f. of the battery.

(a)(ii)

Calculate:

  1. the current in the circuit

current = ______ A\text{A}

  1. the potential difference across the lamp

potential difference = ______ V\text{V}

  1. the internal resistance of the battery.

internal resistance = ______ Ω\Omega

(b)(i)

Sketch a circuit diagram showing how two of these three resistors may be connected together to give a combined resistance of 30 Ω30\ \Omega between the terminals shown. Label the values of the resistances on your diagram.

(b)(ii)

A potential divider circuit is produced by connecting the three resistors to a battery of e.m.f. 9.0 V9.0\ \text{V} and negligible internal resistance. The potential divider circuit provides an output potential difference VOUTV_{\text{OUT}} of 3.6 V3.6\ \text{V}. The circuit diagram is shown in Fig. 6.2.

On Fig. 6.2, label the resistances of all three resistors and the potential difference VOUTV_{\text{OUT}}.

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Q52019 May/Jun·P225 partsEasy
(a)

State Kirchhoff’s second law.

(b)(i)
  1. the combined resistance of the two resistors connected in parallel

combined resistance = ______ Ω\Omega

  1. the current in the battery.

current = ______ A\text{A}

(b)(ii)

Show that the internal resistance rr is 2.5 Ω2.5\ \Omega.

(b)(iii)

Determine the ratio

power dissipated by internal resistance rtotal power produced by battery\frac{\text{power dissipated by internal resistance } r}{\text{total power produced by battery}}

ratio = ______

(c)

The battery in (b) is now connected to a battery of e.m.f. 7.2 V7.2\text{ V} and internal resistance 3.5 Ω3.5\ \Omega. The new circuit is shown in Fig. 5.2.

Determine the current in the circuit.

current = ______ A\text{A}

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