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: Determine Resistance (\(R\))

The formula for resistance, given rated voltage (\(V\)) and rated power (\(P\)), is:
\(R = \frac{V^2}{P}\)

Provided values:
\(V = 200 \, \text{volts}, \quad P = 50 \, \text{watts}\)

Calculation:
\(R = \frac{200^2}{50} = \frac{40000}{50} = 800 \, \Omega\)

Resulting resistance:
\(R = 800 \, \Omega\)

Step 2: Calculate Power (\(P\)) at a Different Voltage

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

Provided values:
\(V_{\text{applied}} = 100 \, \text{volts}, \quad R = 800 \, \Omega\)

Calculation:
\(P = \frac{100^2}{800} = \frac{10000}{800} = 12.5 \, \text{watts}\)

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