Question

Two wires of different metals but having the same length and area of cross- section are connected in series. If their specific resistances are $p_{1}and{p}_2$, then their equivalent specific resistance will be

• A. $p_1+p_2$
• B. $\sqrt{p_1{p}_2}$
• C. $\frac{p_1+p_2}{2}$
• D. $2(p_1+p_2)$

Answer 1

Answer: (C)

$\frac{p_1+p_2}{2}$

Resistance,

$R_1=\frac{p_{1}L}{A}$

$R_2=\frac{p_{2}L}{A}$

In series combination of resistance,

$R_n=R_1+R_2$

$\frac{p_n{L}_n}{A_n}=\frac{p_{1}L}{A}+\frac{p_{2}L}{A}$. . . . . . .(1)

We know that, $A_n=A$

$L_n=2L$

Substitute in equation (1), we get

$\frac{p_n.2L}{A}=\frac{p_{1}L}{A}+\frac{p_{2}L}{A}$

$p_n=\frac{p_1+p_2}{2}$

The correct option is C.