A-LevelPhysicsD.C. circuits9702/23 · Oct/Nov 2019 · Q611 Marks

A battery of electromotive force (e.m.f.) 12V and negligible internal resistance is connected to a network of two lamps and two resistors, as shown in Fig. 6.1. [Figure 6.1] The two lamps in the circuit have equal resistances. The two resistors have resistances R and 28Ω. The lamps are connected at junction X and the resistors are connected at junction Y. The current in the battery is 0.50A and the current in the lamps is 0.20A.

(a)(i)the resistance of each lamp[2]
(a)(ii)resistance R.[2]
(b)Determine the potential difference VXY between points X and Y.[3]
(c)Calculate the ratio total power dissipated by the lamps total power produced by the battery[2]
(d)The resistor of resistance R is now replaced by another resistor of lower resistance. State and explain the effect, if any, of this change on the ratio in (c).[2]

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[1] R = V/I
[1] resistance = (12/0.20)/2 or 6/0.20 = 30 Ω
[1] I = 0.50 - 0.20 (= 0.30 A)
[1] R + 28 = 12/0.30 (= 40 Ω)
[1] R = 12 Ω
[1] p.d. across lamp = 0.20 × 30 (= 6.0 V)
[1] p.d. across R = 0.30 × 12 (= 3.6 V)
[1] Vxy = 6.0 – 3.6 = 2.4 V
or
[C1] p.d. across lamp = 0.20 × 30 (= 6.0 V)
[C1] p.d. across 28 Ω resistor = 0.30 × 28 (= 8.4 V)
[A1] Vxy = 8.4 – 6.0 = 2.4 V
[1] P = VI or P = EI or P = I2R or P = V2/R
[1] ratio = (6.0 × 0.20) × 2 / (12 × 0.50) or 0.20 / 0.50 = 0.40
[1] no change to V across lamps, so power in lamps unchanged or current in battery/total current increases (and e.m.f. the same) so power produced by battery increases
[1] both the above statements and so the ratio decreases

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