Question

A wire when connected to 220V 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

Answer 2

In 1st case: Using the formula $P=\frac{{{V}^{2}}}{R}$ where $R$ is resistance of wire, V is voltage across wire and P is power dissipation in wire and $R=\frac{\rho l}{A}$ From eqs. (1) and (2) $=\frac{{{V}^{2}}}{\frac{\rho l}{A}}=\frac{{{V}^{2}}}{\rho l}.A$ ${{P}_{1}}=\frac{{{V}^{2}}}{\rho l}.A$ ...(1) In 2nd case: Let ${{R}_{2}}$ is net resistance ${{R}_{2}}=\frac{R.R}{R+R}=\frac{R}{2}$ ${{R}_{2}}=\frac{\rho .\left( \frac{l}{2} \right)}{A.2}$ $=\frac{\rho l}{4A}$ $\therefore$ ${{P}_{2}}=\frac{{{V}^{2}}}{\rho l}.4A$ ...(2) Hence, from eqs. (3) and (4) $\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{1}{4}$ $\Rightarrow$ $\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{4}{1}$