A battery of EMF 24 V has an internal resistance of 2 Ω. Three resistors of 10 Ω, 15 Ω, and 30 Ω are connected in parallel and this parallel combination is then connected in series with the battery's internal resistance. (a) Find the equivalent resistance of the parallel combination. (b) Find the total current drawn from the battery. (c) Find the potential difference actually available across the parallel combination. (d) Hence, determine the current through the 30 Ω resistor. Show your reasoning at each step.
Generated by claude-sonnet-4-6 · 2026-06-26 01:11 · grounding rag
Model Answer
Given: EMF (E) = 24 V, internal resistance (r) = 2 Ω, R₁ = 10 Ω, R₂ = 15 Ω, R₃ = 30 Ω (in parallel)
(a) Equivalent resistance of parallel combination:
$$\frac{1}{R_p} = \frac{1}{10} + \frac{1}{15} + \frac{1}{30} = \frac{3+2+1}{30} = \frac{6}{30} = \frac{1}{5}$$
$$\therefore R_p = 5 \text{ Ω}$$
(b) Total current from battery:
Total resistance = $R_p + r = 5 + 2 = 7$ Ω
$$I = \frac{E}{R_p + r} = \frac{24}{7} \approx 3.43 \text{ A}$$
(c) Potential difference across parallel combination:
Voltage drop across internal resistance = $I \times r = 3.43 \times 2 = 6.86$ V
$$V_{parallel} = E - Ir = 24 - 6.86 \approx 17.14 \text{ V}$$
(d) Current through 30 Ω resistor:
Same potential difference (17.14 V) appears across each branch.
$$I_3 = \frac{V_{parallel}}{R_3} = \frac{17.14}{30} \approx 0.57 \text{ A}$$
Source: Chapter 11, sections 11.6.1 and 11.6.2
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Explanation
- Examiners award 1 mark each for parts (a), (b), (c), and (d), with 1 mark for correct formula/method shown at any step.
- Show the formula before substituting values — this earns method marks even if arithmetic slips.
- Key concept: the same PD acts across every branch of a parallel combination, so Ohm's law applies directly to each branch for part (d).
- Internal resistance acts as a series resistor; voltage is "lost" across it, so the terminal voltage (PD across the external circuit) is less than the EMF.