Science (086) — AI-generated practice question

Comparing Resistances of X and Y:

Using $R = \rho \dfrac{l}{A}$, let wire Y have length $l$ and area $A$.

Then wire X has length $2l$ and area $A/2$.

$$R_X = \rho \frac{2l}{A/2} = \frac{4\rho l}{A}$$

$$R_Y = \rho \frac{l}{A}$$

$$\therefore \frac{R_X}{R_Y} = 4 \quad \Rightarrow \quad R_X = 4R_Y$$

Voltage across wire Y in series:

In series, the same current flows, so voltage divides in proportion to resistance.

Total resistance = $R_X + R_Y = 4R_Y + R_Y = 5R_Y$

$$V_Y = \frac{R_Y}{R_X + R_Y} \times 6 = \frac{R_Y}{5R_Y} \times 6 = \frac{1}{5} \times 6 = 1.2 \text{ V}$$

Fraction of total voltage across Y $= \dfrac{1}{5}$.

Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends

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• Examiners expect the formula $R = \rho l/A$ to be stated first, then substituted for both wires to get the 4:1 ratio.
• For the series voltage part, use the voltage divider rule: $V \propto R$ in series. Since $R_X : R_Y = 4:1$, wire Y carries $\frac{1}{5}$ of the total voltage = 1.2 V.
• Write the ratio clearly; a common error is inverting X and Y.