Question

A wire when connected to 220 V mains supply has power dissipation $P_1$. Now, the wire is cut into two equal pieces, which are connected in parallel to the same supply. Power dissipation in this case is $P_2$. Then $P_2:P_1$ is

• A. 1
• B. 4
• C. 2
• D. 3

Answer 1

Answer: (B)

4

Here, $P_1=\frac{V^2}{R}$ where $R=$ resistance of wire.

we know that $R=\frac{\rho L}{A}\Rightarrow R\propto L$

When the wire is cut into two piece so resistance of each part becomes

$R/2$.

Now these part are in parallel so $R_{eq}=\frac{\left(R/2\right)\left(R/2\right)}{R/2+R/2}=R/4$

Thus, $P_2=\frac{V^2}{R_{eq}}=\frac{4V^2}{R}$

So $\frac{P_2}{P_1}=4$