Question

Two copper wires A and B have the same length and are connected across the same battery. If RB=2RAR_{B}=2 R_{A} find the ratio of their cross-sectional areas, AB/AAA_{B} / A_{A}.

Solution

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Both of the wires are copper, so their resistivities ρ\rho will be the same. Thus, the ratio of the cross-sectional area for the wires is:

RaRb=12=ρaL/AaρbL/Ab=AbAa    Aa=2Ab\begin{gather*} \dfrac{R_a}{R_b}=\dfrac{1}{2}=\dfrac{\rho_aL/A_a}{\rho_bL/A_b}=\dfrac{A_b}{A_a}\implies \boxed{A_a=2A_b} \end{gather*}

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