Question

The areas of cross section of two wires of the same material are in the ratio $1:2$ and their lengths are in the ratio $2:1$. If their ends have the same potential differences then currents through them will be in the ratio of

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

Answer: (C)

$1:4$

The resistance of a wire of length $l$ and cross section area $A$ is given by,
$R=\rho \left(l/A\right)$.
So
$\frac{R_1}{R_2}=\frac{l_1}{l_2}\frac{A_2}{A_1}=\frac{4}{1}$.
Now current through a wire is $i\propto 1/R$.
$\therefore \frac{i_1}{i_2}=\frac{R_2}{R_1}=\frac{1}{4}$
If their ends have the same potential differences then currents through them
will be in the ratio of $1:4$.