A resistor is connected across a potential difference V and carries current I. (i) If the potential difference is tripled while the resistance remains fixed, what happens to the current? (ii) The resistor is replaced by one made of the same material but with double the length and half the cross-sectional area. By what factor does the resistance change? Show your working. (iii) With this new resistor connected across the tripled potential difference, calculate the ratio of the new current to the original current I.
Generated by claude-sonnet-4-6 · 2026-06-26 01:12 · grounding rag
Model Answer
(i) By Ohm's law, $I = V/R$. If V is tripled and R is unchanged, the current also triples to 3I.
(ii) Resistance $R = \rho L/A$. New length = 2L, new area = A/2.
$$R_{\text{new}} = \rho \cdot \frac{2L}{A/2} = \frac{4\rho L}{A} = 4R$$
The resistance increases by a factor of 4.
(iii) Original current: $I = V/R$
New current: $I_{\text{new}} = \dfrac{3V}{4R} = \dfrac{3}{4} \cdot \dfrac{V}{R} = \dfrac{3I}{4}$
$$\frac{I_{\text{new}}}{I} = \boxed{\frac{3}{4}}$$
Source: Chapter 11, Sections 11.4 and 11.5 (Ohm's Law; Resistance of a Conductor)
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Explanation
- Part (i): Direct application of $I = V/R$ — if R is constant, I is proportional to V.
- Part (ii): Examiners expect the formula $R = \rho L/A$ with substitution shown step by step. Both changes (doubled L and halved A) each double the resistance, giving a combined factor of 4.
- Part (iii): Combine both changes: voltage ×3, resistance ×4. The net effect is $3/4$, so the new current is less than the original despite the higher voltage. Always show the ratio explicitly.