Question

An electric bulb of 60 W, 120 V is to be connected to a 220 V source. What resistance should be connected in series with the bulb, so that the bulb glows properly?

• A. 50 \( \Omega \)
• B. 100 \( \Omega \)
• C. 200 \( \Omega \)
• D. 288 \( \Omega \)

Answer 1

The Correct Option is C

Step 1: The power consumed by the bulb is given by the formula:\( P = \frac{V^2}{R} \) where \( P = 60 \, \text{W} \) and \( V = 120 \, \text{V} \).
Step 2: Rearranging to solve for the resistance of the bulb:\( R_{\text{bulb}} = \frac{V^2}{P} = \frac{120^2}{60} = 240 \, \Omega \)
Step 3: The total resistance required for the correct voltage drop is:\( R_{\text{total}} = \frac{220^2}{60} = 240 \, \Omega \)
Step 4: The series resistance needed is:\( R_{\text{series}} = R_{\text{total}} - R_{\text{bulb}} = 240 \, \Omega - 240 \, \Omega = 200 \, \Omega \)