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

The Correct Option is A

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:
\(P = 12.5 \, \text{watts}\)