The correct answer is D because:

Ohm's law states that:
- V = IR
Rearranging for current gives:
- I = $\frac{V}{R}$
From Kirchhoff's Second Law the p.d. is the same across P and Q
So, the ratio is:
- $\frac{I_{P}}{I_{Q}}$ = $\frac{R_{Q}}{R_{P}}$
The resistivity of a wire is given by:
- ρ = $\frac{R L}{A}$
Rearranging for the resistance gives:
- R = $\frac{ρ L}{A}$
Cross-sectional area is equal to A = $π r^{2}$ = $π {(\frac{d}{2})}^{2}$ = $\frac{π d^{2}}{4}$
- So, R = $\frac{ρ L}{A}$ = $ρ L (\frac{4}{π d^{2}})$
For wire Q, the length is L, and the diameter is 3d
- So, the resistance of wire Q is equal to:
- R_Q= $ρ L (\frac{4}{π {(3 d)}^{2}})$ = $\frac{4}{9} (\frac{ρ L}{π d^{2}})$
For wire P, the length is 2L, and the diameter is d
- So, the resistance of wire P is equal to :
- R_P = $ρ (2 L) (\frac{4}{π d^{2}})$ = $8 (\frac{ρ L}{π d^{2}})$
Substituting these values into the ratio gives:
- $\frac{I_{P}}{I_{Q}}$ = $\frac{R_{Q}}{R_{P}}$ = $\frac{4}{9} (\frac{ρ L}{π d^{2}})$ $\div 8 (\frac{ρ L}{π d^{2}})$ = $\frac{4}{9}$ × $\frac{1}{8}$ = $\frac{1}{18}$

Resistance & Resistivity (AQA AS Physics)

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