AS Physics: D.C. Circuits — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: D.C. Circuits revision notes

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Section A

1. State Kirchhoff’s two laws and name the conservation law each expresses. [4]

2. Distinguish between e.m.f. and potential difference. [2]

3. Explain why adding a resistor in parallel decreases the total resistance of a circuit. [2]

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Section B

4. A cell of e.m.f. 1.50 V and internal resistance 0.80 Ω is connected to a 4.20 Ω resistor.

(a) Calculate the current in the circuit. [2]

(b) Calculate the terminal potential difference. [2]

(c) Calculate the power dissipated inside the cell. [2]

(d) The external resistance is reduced. State and explain what happens to the terminal p.d. [3]

5. A student plots terminal p.d. V against current I for a cell and obtains a straight line with gradient −0.45 V A⁻¹ and intercept 1.62 V.

(a) State the values of the e.m.f. and the internal resistance. [2]

(b) Explain why the graph has a negative gradient. [2]

6. A 12 V supply is connected across two resistors in series: a fixed 2.0 kΩ resistor and a thermistor. The output is taken across the thermistor.

(a) At 20 °C the thermistor has resistance 6.0 kΩ. Calculate the output p.d. [2]

(b) The temperature rises. State and explain the effect on the output p.d. [3]

7. Two resistors, 6.0 Ω and 3.0 Ω, are connected in parallel. Calculate the combined resistance. [2]

8. A potential divider uses a fixed resistor and an LDR in series. Explain how this circuit could be used as a light sensor, describing what happens to the output voltage across the LDR as light intensity increases. [3]

9. Under what condition(s) are e.m.f. and terminal p.d. equal? [2]

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Answers

1. First law: the sum of currents into a junction equals the sum of currents out [1] — conservation of charge [1]. Second law: around any closed loop, the sum of e.m.f.s equals the sum of p.d.s [1] — conservation of energy [1].

2. E.m.f. is the energy transferred to each unit charge by the source [1]; p.d. is the energy transferred from each unit charge in a component [1]. Both are defined between two points regardless of whether current actually flows — a potential difference can exist across an open switch, an unconnected cell, or a charged capacitor with no current crossing it; p.d. describes the energy that would be transferred per unit charge, not only energy already transferred.

3. It provides an additional path for current [1], so a greater total current flows for the same potential difference, which means a lower total resistance [1]. The combined resistance of any parallel network is always smaller than its smallest individual resistor.

4. (a) I = E ÷ (R + r) = 1.50 ÷ (4.20 + 0.80) [1] = 0.30 A [1].

(b) V = IR = 0.30 × 4.20 [1] = 1.26 V [1]. (Or V = E − Ir = 1.50 − 0.24 = 1.26 V.)

(c) P = I²r = 0.30² × 0.80 [1] = 0.072 W [1].

(d) Terminal p.d. decreases [1]. Lower external resistance means a larger current [1], so the “lost volts” Ir inside the cell increase and less energy per coulomb is available at the terminals [1].

5. (a) E = 1.62 V (the intercept) [1]; r = 0.45 Ω (the magnitude of the gradient) [1].

(b) V = E − Ir, which is of the form y = c + mx with m = −r [1]. As current increases, more energy is dissipated inside the cell by its internal resistance, so the terminal p.d. falls [1].

6. (a) V_out = 12 × 6.0 ÷ (2.0 + 6.0) [1] = 9.0 V [1].

(b) Output p.d. decreases [1]. The thermistor’s resistance falls as temperature rises [1], so it takes a smaller share of the total resistance and therefore a smaller share of the supply p.d. [1].

7. 1/R = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0 [1]; R = 2.0 Ω [1]. Note the combined resistance (2.0 Ω) is smaller than the smallest individual resistor (3.0 Ω) — always true for resistors in parallel.

8. An LDR’s resistance falls as light intensity increases [1]. If the output is taken across the LDR, it takes a smaller share of the total resistance as light increases, so the output voltage falls [1] — the circuit can therefore signal changing light levels via a changing output voltage [1].

9. When the internal resistance is negligible, or when no current flows (open circuit) [1] [1] — otherwise, while the cell is discharging (supplying current), the “lost volts” Ir make terminal p.d. less than e.m.f.; while it is being charged, the applied terminal voltage instead exceeds the e.m.f. by Ir. This is why a genuinely fresh, low-resistance cell measured on open circuit gives a reading very close to its rated e.m.f.

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Where marks are usually lost

• Forgetting internal resistance when a question specifies a real cell.
• Confusing e.m.f. with terminal p.d.
• Using the wrong resistance in the potential divider ratio.
• Saying parallel resistors increase total resistance.
• Not naming the conservation law behind each of Kirchhoff’s laws.
• Adding parallel resistances directly instead of using the reciprocal formula.
• Assuming an LDR’s output voltage behaves the same way as a thermistor’s without checking which component the output is taken across.
• Treating e.m.f. and terminal p.d. as always equal, forgetting they only coincide when internal resistance is negligible or no current flows – and, more generally, treating terminal p.d. as always below e.m.f., forgetting this only holds while the cell is discharging (supplying current); while it is being charged, the terminal voltage exceeds the e.m.f. by Ir.