Question

An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is :

• A. 12.5 W
• B. 25 W
• C. 50 W
• D. 100 W

Answer 1

Answer: (A)

12.5 W

Answer 2

Step 1: Finding Resistance ($R$)

We use the formula for resistance when the rated voltage ($V$) and rated power ($P$) are given:
$R = \frac{V^2}{P}$

Given:
$V = 200 \, \text{volts}, \quad P = 50 \, \text{watts}$

Substituting the values:
$R = \frac{200^2}{50} = \frac{40000}{50} = 800 \, \Omega$

Thus, the resistance is:
$R = 800 \, \Omega$

Step 2: Finding Power ($P$) for a Different Voltage

To calculate the power consumed for an applied voltage ($V_{\text{applied}}$) of 100 volts, we use the formula:
$P = \frac{V_{\text{applied}}^2}{R}$

Given:
$V_{\text{applied}} = 100 \, \text{volts}, \quad R = 800 \, \Omega$

Substituting the values:
$P = \frac{100^2}{800} = \frac{10000}{800} = 12.5 \, \text{watts}$

Thus, the power consumed is:
$P = 12.5 \, \text{watts}$