A student connects a battery to two resistors. The circuit diagram is shown in Fig. 10.1. [Figure 10.1 shows a battery connected in series to a 10 Ω resistor, which is then connected in series to a 40 Ω resistor. The potential difference (p.d.) across the 40 Ω resistor is 9.6 V.] [Figure 10.2 shows the circuit from Figure 10.1, but with an additional resistor R connected in parallel with the 40 Ω resistor.] [Figure 10.3 is a table with columns "quantity", "increases, decreases or stays the same as resistor R is added", and "brief explanation of why the change occurs". The table has rows for "current in 10Ω resistor" (pre-filled with "increases"), "p.d. across 10Ω resistor", and "p.d. across 40Ω resistor".]
- (a)State what is meant by the potential difference across a resistor.[2]
- (b)(i)Calculate the current in the 40 Ω resistor. current = .....................................................[2][2]
- (b)(ii)Calculate the electromotive force (e.m.f.) of the battery. e.m.f. = .....................................................[2][2]
- (c)The student has three different voltmeters to measure the p.d. across the 40 Ω resistor. These are labelled 0–2 V, 0–20V and 0–200 V. Each has a pointer that shows the p.d. on a scale. State and explain which of the three voltmeters is best to measure this p.d.[2]
- (d)(i)Calculate the power P produced in the 10 Ω resistor. P = ..................................................... [2][2]
- (d)(ii)The student has available two 10 Ω resistors, with power ratings of ½ P and 2P. Suggest why a resistor with a power rating of ½ P is not suitable for the circuit in Fig. 10.1.[1]
- (e)The student adds a resistor R to the circuit, to make the circuit shown in Fig. 10.2. Complete the table in Fig. 10.3 to show what happens as resistor R is connected. You should state whether each quantity increases, decreases or stays the same and give a brief explanation of why any change occurs. Calculations are not required. [4][4]
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Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [1] (a) work done / energy transferred
- [1] (a) work done / energy transferred in taking unit charge (through resistor)
- [1] (b)(i) (I=) V/R in any form algebraic or numerical
- [1] (b)(i) 0.24 A
- [1] (b)(ii) (total resistance of) 50 (Ω) seen or 2.4 (V) p.d. across 10 Ω resistor
- [1] (b)(ii) 12 V
- [1] (c) voltmeter chosen is 0-20 V
- [1] (c) correct explanation of why 0–200 V meter is unsuitable OR correct explanation of why 0–2 V meter is unsuitable
- [1] (d)(i) (P)= VI or I²R or V² / R in any form
- [1] (d)(i) 0.58 W
- [1] (d)(ii) ½P rated resistor blows / too hot / melts
- [1] (e) current in 10 Ω increases
- [1] (e) resistance decreases (of parallel combination or total resistance)
- [1] (e) p.d. across 10 Ω increases
- [1] (e) p.d.is proportional to current or (p.d.) increases as current increases or potential divider argument
- [1] (e) p.d. across 40 Ω decreases
- [1] (e) sum of p.d.s constant / 12V / value in (b)(ii) or p.d. across 10 Ω increases or resistance (40 Ω and R) a smaller fraction of total
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