Question

The figure shows three conductors I, II and III of same material, different lengths $l,2l$ and $3l$ and of different areas of cross-section $3A,A$ and $2A$ respectively. Arrange them in the increasing order of current drawn from battery.

• A. $i_1<i_2<i_3$
• B. $i_3<i_2<i_1$
• C. $i_2<i_1<i_3$
• D. $i_2<i_3<i_1$

Answer 1

Answer: (D)

As we know, resistance $R=\rho \frac{l}{A}$.
The resistance of conductor $l$ is given by
$R_l=\rho \frac{l}{3A}=\frac{R}{3}($ where $R=\rho \frac{l}{A})$
Similarly, $R_{II}=\rho \frac{2l}{A}=2R$
and $R_{lll}=\rho \frac{3l}{2A}=\frac{3}{2}R$
From this we conclude that $R_{II}>R_{III}>R_I$. Since in parallel combination of resistances current distributes in inverse ratio of resistances, therefore $i_2<i_3<i_1$.