Questions on electricity
1. (i) M marked at the end of the graph B1
(ii) current is 5 (A) and p.d is 6 (V) C1
P = VI \ p = 6.0 5.0
(Allow p = I2 R or p = V2 \ R) C1
power = 30 (W) A1
(iii) 1. VL = 1.0 (V) (From the I/V graph) \ RL = 1.0/2.0 or 0.5 () M1
VR = 1.2 2.0 \ RT = 1.2 0.5 M1
V = 1.0 2.4 \ V = 1.7 2.0 A1
voltmeter reading = 3.4 (V) A0
2. Vr = 4.5 3.4 (= 1.1 V) \ 4.5 = 2.0r 3.4 (Possible ecf) C1
r = 0.55 () (1.05 scores 0/2 since the lamp is ignored) A1
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2. (i) p.d across 1.5 k resistor = 5.0 1.2 = 3.8 (V) B1
(ii) C1
C1
R = 474 () 470 () A1
(Using 3.8 V instead of 1.2 V gives 4.75 k - allow 2/3)
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3. (i) Any four from:
1. The resistance of the thermistor decreases (as temperature is increased) B1
2. The total resistance (of circuit) decreases B1
3. The voltmeter reading increases B1
4. Explanation of 3. above in terms of ‘sharing voltage’
/ B1
5. The current increases / ammeter reading increases B1
6. Explanation of current increase in terms of B1
(Allow ecf for statements 3. and 5. if statement 1. is incorrect maximum
score of 2/4)
(ii) C1
C1
R = 467 () 470 () A1
(When 1.4 V and 3.6 V are interchanged, then R = 3.1 103 () can score2/3)
(Calculation of total circuit resistance of 1.67 103 () can score 2/3)
(Use of scores 0/3)
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4. (i) (NTC) thermistor B1
(ii) Resistance decreases when temperature is increased. (ora) B1
(Allow correct credit for a PTC thermistor)
(iii) 1 I = (0.032-0.006 =) 0.026 (A) B1
2 (V200 = 0.026 200 =) 5.2 (V) / (V720 = 0.006 700 =) 4.2 (V) C1
E = 5.2 – 4.2 (Allow E = 4.2 – 5.2) C1
E = 1.0 (V) (Allow 1 sf answer) A1
(9.4 (V) scores 1/3)
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5. (i) The resistance of LDR/circuit changes (as light intensity changes) B1
When blade blocks light, resistance of LDR/circuit is large(r) (ora) B1
Correct statement about p.d (Possible ecf) B1
(ii) 1. (V = 5.0 – 3.0)
2.0 (V) (Allow 1 sf answer) B1
2. V = I = 2.0/2200 / 9.1 10–4 (A) C1
(3.0 = ) (R = 3.0 / 9.1 10–4)
R = 3300 () R = 3300 () Possible ecf A1
(For VLDR = 2.0 V, R = 1.47 k. This scores 1/2)
(If 3.5 V given in (ii)1., then R = 940 . This scores 2/2)
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