DC circuits and potential dividers – A-level Physics exam questions with mark scheme

1Standard4 marks

The diagram shows a circuit containing a battery of emf 6.0 V and negligible internal resistance.

Calculate the current in the 12 Ω resistor. [4]

resistance of parallel combination: $\left(\dfrac{1}{6.0} + \dfrac{1}{12}\right)^{-1} = 4.0$ Ω(1)
total resistance = 8.0 Ω so current from battery = $\dfrac{6.0}{8.0} = 0.75$ A(1)
p.d. across parallel section = $0.75 \times 4.0 = 3.0$ V or by ratio of resistances(1)
current in 12 Ω = $\dfrac{3.0}{12} = 0.25$ A(1)
Answer: 0.25 A
Examiner tip: The p.d. across parallel branches is the same, but the current splits in inverse ratio to resistance — the larger resistor gets the smaller current.

2Hard5 marks

The diagram shows a potential divider used as a temperature sensor. The supply has negligible internal resistance. At 20 °C the (negative temperature coefficient, NTC) thermistor has a resistance of 4.7 kΩ.

(a) Calculate the output p.d. $V_{out}$ at 20 °C. [2]

(b) The temperature of the thermistor rises. Explain what happens to $V_{out}$. [2]

(a) $V_{out} = 9.0 \times \dfrac{2.2}{2.2 + 4.7}$(1)
(a) 2.9 V(1)
(b) the resistance of the (NTC) thermistor decreases as temperature rises(1)
(b) the fixed resistor gets a larger share of the supply p.d. (current increases), so $V_{out}$ increases allow p.d. across thermistor decreases so $V_{out} = 9.0 - V_{th}$ increases(1)
(c) 2.2 kΩ (p.d. is shared equally, so resistances are equal)(1)
Answer: (a) 2.9 V (b) $V_{out}$ increases (c) 2.2 kΩ
Examiner tip: Always state the thermistor's resistance change first, then link it to the share of p.d. — the second mark depends on the first.

3Hard5 marks

Two 100 kΩ resistors are connected in series to a 12 V supply of negligible internal resistance. A student connects a voltmeter of resistance 200 kΩ across one of the resistors.

(a) Calculate the reading on the voltmeter. [3]

(b) The student expected a reading of 6.0 V. Explain why the reading is lower and how the measurement could be improved. [2]

(a) combined resistance of voltmeter and resistor: $\left(\dfrac{1}{100}+\dfrac{1}{200}\right)^{-1} = 66.7$ kΩ(1)
(a) $V = 12 \times \dfrac{66.7}{66.7 + 100}$(1)
(a) 4.8 V(1)
(b) the voltmeter in parallel reduces the resistance of that section, so it takes a smaller share of the 12 V(1)
(b) use a voltmeter with a much higher resistance (ideally infinite) compared with 100 kΩ, e.g. a digital voltmeter of ~10 MΩ(1)
Answer: (a) 4.8 V (b) voltmeter lowers resistance of the section; use a much higher-resistance voltmeter
Examiner tip: An ideal voltmeter has infinite resistance; real ones only behave ideally when their resistance is much larger than the component they measure.

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