Question

An electric bulb is designed to draw power $P_0$ at voltage $V_0$. If the voltage is V, it draws a power P, then-

• A. $P=\left(\frac{V}{V_0}\right)P_0$
• B. $P=\left(\frac{V_0}{V}\right)P_0$
• C. $P={\left(\frac{V_0}{V}\right)}^2{P}_0$
• D. $P={\left(\frac{V}{V_0}\right)}^2{P}_0$

Answer 1

The correct option is D $P={\left(\frac{V}{V_0}\right)}^2{P}_0$

Power in resistor is given by

$P=\frac{V^2}{R}$ so $R=\frac{V^2}{P}$

Now since the bulb is operating initially at Power $P_0$ and Voltage $V_0$. So

$P_0=\frac{{V_0}^2}{R}$

Resistance of bulb is $R=\frac{V_0^2}{P_0}$------$\left(1\right)$

now the bulb is attached to voltage $V$ and it consumes power $P$ now. So Using value of $R$ from equation $\left(1\right)$

$P=\frac{V^2}{R}=\frac{V^2{P}_0}{V_0^2}={\left(\frac{V}{V_0}\right)}^2{P}_0$