May/June 2026 Paper 22

2026 · 6 questions · 32 parts · 60 marks

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1(a)1(b)(i)1(b)(ii)1(b)(iii)1(b)(iv)1(b)(v)

–/11 mks

2(a)2(b)(i)2(b)(ii)2(c)(i)2(c)(ii)

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3(a)3(b)(i)3(b)(ii)3(c)(i)3(c)(ii)

–/10 mks

4(a)4(b)(i)4(b)(ii)4(b)(iii)

–/10 mks

5(a)5(b)(i)5(b)(ii)5(c)(i)5(c)(ii)5(c)(iii)

–/12 mks

6(a)6(b)(i)6(b)(ii)6(b)(iii)6(c)6(d)

–/9 mks

5(a)Series Parallel Circuit AnalysisEasy2 marks

Three resistors of resistances

R_{1}

R_{2}

and

R_{3}

are connected in a network in parallel, as shown in Figure 5.1.

Figure 5.1, three resistors R1, R2 and R3 connected in parallel between two junctions, with the potential difference V marked across the whole combination; the total current IT enters the left junction and splits into currents I1, I2 and I3 through R1, R2 and R3 respectively, recombining at the right junction

The currents in the resistors are

I_{1}

I_{2}

and

I_{3}

respectively. The total current entering the network is

I_{T}

. The potential difference (p.d.) across the resistors is

V

Derive the equation

\frac{1}{R _{T}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

where

R_{T}

is the total resistance of the network.

Draw with the mouse or a finger.

5(b)(i)Series Parallel Circuit AnalysisEasy2 marks

Two identical resistance wires, each of resistance 2.8 Ω, are connected in parallel to a battery with negligible internal resistance and electromotive force (e.m.f.) 12 V, as shown in Figure 5.2.

Figure 5.2, a circuit diagram: a 12 V battery connected across two identical resistance wires wired in parallel with each other

Calculate the current in the battery.

5(b)(ii)Current Potential DifferenceEasy2 marks

Each resistance wire has a cross-sectional area of

6.1 \times 1 0^{- 6}

^{2}

The wires are made from a metal with a number density of charge carriers (free electrons) of

8.5 \times 1 0^{28}

^{- 3}

Calculate the average drift speed of the charge carriers in the wires.

5(c)(i)Circuit Components I V CharacteristicsMedium1 mark

A component X is added to the circuit in 5(b), as shown in Figure 5.3.

Figure 5.3, a circuit diagram: a 12 V battery, with the battery current of 5.0 A shown flowing into the top wire, connected across two branches in parallel — one branch is the resistance wire from Figure 5.2, the other branch is a component X drawn as a rectangular resistor symbol with two arrows pointing in from the upper left, the symbol for a light-dependent resistor

The current in the battery is now 5.0 A.

The resistance of each resistance wire remains 2.8 Ω.

State the name of component X.

5(c)(ii)Series Parallel Circuit AnalysisMedium3 marks

Determine the resistance of component X.

Show your working

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5(c)(iii)Circuit Components I V CharacteristicsMedium2 marks

A change in the environment of component X causes the current in the battery to decrease. The resistance of the resistance wires remains constant.

State and explain how the environment of X has changed.

Answer

0 words