(a) Voltmeter connected in parallel with X B1
(b) Same reading / no effect / no change B1
(c)(i) LDR / light-dependent resistor B1
(c)(ii) The resistance decreases (as the intensity of light increases) B1
(c)(iii) 3.5 – 4.0 × 10-7 (m) (to) 6.5 – 7.5 × 10-7 (m) B1
(d)(i) IVR= / )10(8.48.13−×=R C1
resistance = 375 ≈ 380 (Ω) A1
(d)(ii)1 Q = It (Allow with or without the Δ notation) C1
Q = 4.8 × 10-3 × 30 C1
charge = 0.144 ≈ 0.14 (C) A1
(d)(ii)2 W = VQ / W = VIt C1
W = 1.8 × 0.144 / W = 1.8 × 4.8 × 10-3 × 30
energy = 0.259 ≈ 0.26 (Possible ecf) A1
unit: joule / J / VC /VAs B1
(Allow 1/3 if power is 0.0086 (W))
[Total: 13]
2
(a) Kirchhoff’s second B1
(b) Ohm’s B1
(c) Resistance B1
(d) Electronvolt (Allow eV) B1
[Total: 4]
3
(a)(i) 21111RRR+= / 2121RRRRR+= C1
3012011+=R / 30203020+×=R
resistance = 12 (Ω) A1
(a)(ii) R = 10 + 12
resistance = 22 (Ω) (Possible ecf) B1
(b) R = 10 (Ω) / Resistance between B and C = 0 M1
100.5=I
reading = 0.5 (A) A1
[Total: 5]
5
Any four from: B1 × 4
1 (As temperature increases) the resistance of the thermistor / T decreases
2 The total resistance decreases (Possible ecf)
3 The current increases (in the circuit) (Possible ecf)
4 The (voltmeter) reading increases / voltage across R increases (Possible ecf)
5 The voltage across the thermistor / T decreases (Possible ecf)
6 Correct use of the potential divider equation / comment on the ‘sharing’
of voltage / correct use of V = IR
[Total: 4]
6
(a) ALRρ= (Allow any subject) B1
(b) The resistance decreases M1
by a factor of four (because resistance is inversely proportional to radius2) A1
(c)(i) A25103.1105.32200−−×××= / RLAρ= C1
2200103.1105.3)(25−−×××=A C1
(A = ) 2.07 × 10-10 (m2) ≈ 2 × 10-10 (m2) A0
(c)(ii) RIP2= / VIP= and V = IR C1
0.50 = I2 × 2200 C1
current = 0.015 (A) A1
(2.23 × 10-4 scores 2/3 – answer not square rooted)
[Total: 8]
