Define the ohm.
- (a)Define the ohm.[1]
- (b)A wire has a resistance of 1.8Ω. The wire has a uniform cross-sectional area of 0.38 mm² and is made of metal of resistivity 9.6 × 10⁻⁷Ωm. Calculate the length of the wire.[3]
- (c)(i)A resistor X of resistance 1.8Ω is connected to a resistor Y of resistance 0.60Ω and a battery P, as shown in Fig. 5.1. [Figure 5.1] The battery P has an electromotive force (e.m.f.) of 1.2V and negligible internal resistance. Explain, in terms of energy, why the potential difference (p.d.) across resistor X is less than the e.m.f. of the battery.[1]
- (c)(ii)Calculate the potential difference across resistor X.[2]
- (d)(i)Another battery Q of e.m.f. 1.2V and negligible internal resistance is now connected into the circuit of Fig. 5.1 to produce the new circuit shown in Fig. 5.2. [Figure 5.2] State whether the addition of battery Q causes the current to decrease, increase or remain the same in: resistor X[1]
- (d)(ii)battery P.[1]
- (e)(i)The circuit shown in Fig. 5.2 is modified to produce the new circuit shown in Fig. 5.3. [Figure 5.3] Calculate: the total resistance of the two resistors connected in parallel[1]
- (e)(ii)the current in resistor Y.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.
- B1 volt / ampere
- C1 R = ρL/A
- C1 L = (1.8 × 0.38 × 10⁻⁶) / 9.6 × 10⁻⁷
- A1 = 0.71 m
- B1 thermal energy is dissipated in resistor Y
- C1 V/1.2 = 1.8/(1.8 + 0.6)
- A1 V = 0.90 V
- (C1) or
- (C1) I = 1.2/(1.8 + 0.6) (= 0.50)
- (A1) V= 0.50 x 1.8
- (A1) = 0.90 V
- B1 remain the same
- B1 decrease
- A1 1/R = 1/1.8 + 1/3.6
- A1 R = 1.2 Ω
- C1 I = 1.2/(1.2 + 0.60)
- A1 = 0.67 A
- (C1) or
- (C1) Vy = 1.2 × 0.60 / (1.2 + 0.60) (= 0.40)
- (A1) I = 0.40/0.60
- (A1) = 0.67 A
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