A-LevelPhysicsD.C. circuits9702/21 · Oct/Nov 2019 · Q611 Marks

Define electric potential difference (p.d.). [1]

(a)Define electric potential difference (p.d.). [1][1]
(b)The variation with potential difference V of the current I in a semiconductor diode is shown in Fig. 6.1. [Figure 6.1] Use Fig. 6.1 to describe qualitatively the variation of the resistance of the diode as V increases from 0 to 1.0 V. [2][2]
(c)The diode in (b) is part of the circuit shown in Fig. 6.2. [Figure 6.2] The cell of electromotive force (e.m.f.) 2.0V and negligible internal resistance is connected in series with the diode and resistors X and Y. The resistance of Y is 60Ω. The current in the cell is 15mA.
(c)(i)Use Fig. 6.1 to determine the resistance of the diode. resistance = .................................................... Ω [3][3]
(c)(ii)1Calculate: 1. the resistance of X resistance = .................................................... Ω [3][3]
(c)(ii)22. the ratio power dissipated in resistor Y / total power produced by the cell ratio = .................................................... [2][2]

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Mark scheme

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

[B1] work done / charge; or energy (transferred from electrical to other forms) / charge
[B1] for V < 0.25 V resistance is infinite/very high (as current is zero)
[B1] for V > 0.25 V resistance decreases (as V increases)
[C1] R = V/I
[C1] = 0.75 / (15 x 10⁻³)
[A1] = 50 Ω
[C1] Vy = 15 x 10⁻³ × 60 (= 0.90 V)
[C1] Vx = 2.0-0.90 – 0.75 (= 0.35 V)
[A1] Rx = 0.35 / (15 × 10⁻³); = 23 Ω
[(C1)] total R = 60 + 50 + Rx
[(C1)] 60 + 50 + Rx = 2.0 / (15 × 10⁻³)
[(A1)] Rx = 23 Ω
[C1] P = VI or P = EI or P = I²R or P = V²/R; ratio = (15×10⁻³)² × 60 / 2.0 × 15 × 10⁻³ or 0.90×15×10⁻³ / 2.0×15×10⁻³ or (0.90²/60) / 2.0×15×10⁻³
[A1] = 0.45

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