Physics - Current Electricity Question with Solution | TestHub

Rate of heat i.e., Power developed in the $$\text{ wire }=P=\frac{V^2}{R}$$ Resistance of the wire of length, $L$ $$R_1=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}$$ $\therefore$ Power, $P_1=\frac{V^2}{R_1}$ Resistance of the wire when length is halved i.e., $L/2$ $$R_2=\frac{\rho \frac{L}{2}}{\pi (2r{)}^2}=\frac{\rho L}{\pi 8r^2}=\frac{R_1}{8}$$ $\therefore$ Power, $P_2=\frac{V}{\frac{R_1}{8}}=\frac{8V}{R_1}$ or, ${\mathrm{P}}_2=8{\mathrm{P}}_1$ i.e., power increased 8 times of previous or original wire.