A-LevelPhysicsD.C. circuits9702/22 · Feb/Mar 2019 · Q613 Marks

Using energy transformations, describe the electromotive force (e.m.f.) of a battery and the potential difference (p.d.) across a resistor.

(a)Using energy transformations, describe the electromotive force (e.m.f.) of a battery and the potential difference (p.d.) across a resistor.[2]
(b)A battery of e.m.f. 6.0V and negligible internal resistance is connected to a network of resistors and a voltmeter, as shown in Fig. 6.1. Resistor Y has a resistance of 24 Ω and resistor Z has a resistance of 32 Ω.
(b)(i).1The resistance Rx of the variable resistor X is adjusted until the voltmeter reads 4.8 V. Calculate: the current in resistor Z[1]
(b)(i).2The resistance Rx of the variable resistor X is adjusted until the voltmeter reads 4.8 V. Calculate: the total power provided by the battery[2]
(b)(i).3The resistance Rx of the variable resistor X is adjusted until the voltmeter reads 4.8 V. Calculate: the number of conduction electrons that move through the battery in a time interval of 25 s[2]
(b)(i).4The resistance Rx of the variable resistor X is adjusted until the voltmeter reads 4.8 V. Calculate: the total resistance of X and Y connected in parallel[2]
(b)(i).5The resistance Rx of the variable resistor X is adjusted until the voltmeter reads 4.8 V. Calculate: the resistance Rx.[2]
(b)(ii)The resistance Rx is now decreased. State and explain the change, if any, to the reading on the voltmeter.[2]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2019). The mark or mark type is shown in brackets.

[1] (a) e.m.f.: energy transferred from chemical to electrical (per unit charge)
[1] (a) p.d.: energy transferred from electrical to thermal (per unit charge)
[1] (b)(i) 1 I = 4.8/32 = 0.15 A
[1] (b)(i) 2 P = ΕΙ or P = VI or P = I²R or P = V²/R
[1] (b)(i) 2 = 6.0 × 0.15 or 0.15² × 40 or 6.0²/40 = 0.90 W
[1] (b)(i) 3 number = It / e
[1] (b)(i) 3 = [0.15 × 25] / 1.6 × 10⁻¹⁹ = 2.3 x 10¹⁹
[1] (b)(i) 3 or Q = 0.15 x 25 (= 3.75) number = 3.75 / 1.6 × 10⁻¹⁹ = 2.3 x 10¹⁹
[1] (b)(i) 4 4.8/6.0 = 32 / (Rxy + 32) or 1.2/6.0 = Rxy / (Rxy + 32) or 4.8/1.2 = 32/RxY
[1] (b)(i) 4 Rxy = 8.0 Ω
[1] (b)(i) 4 Alternative methods: Rxy = (6.0-4.8)/0.15 or = 8.0 Ω
[1] (b)(i) 4 or 6.0 = 0.15 (32 + RxY) Rxy = 40-32 = 8.0 Ω
[1] (b)(i) 5 1/8.0 = 1/Rx + 1/24
[1] (b)(i) 5 Rx = 12 Ω
[1] (b)(i) 5 Alternative method: Iz = 4.8/32 = 0.15 and Iy = 1.2/24 = 0.05 Ix = 0.15-0.05 (= 0.10)
[1] (b)(i) 5 Rx 1.2/0.10 = 12 Ω
[1] (b)(ii) total resistance decreases
[1] (b)(ii) (so voltmeter) reading increases

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