Question

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

• A. 50Ω
• B. 100Ω
• C. 200Ω
• D. 288Ω

Answer 1

The Correct Option is C

Given:

• Bulb power: \( P = 60 \, \text{W} \)
• Bulb rated voltage: \( V = 120 \, \text{V} \)
• Source voltage: \( V_s = 220 \, \text{V} \)
• We need to find resistance \( R \) to be connected in series

Step 1: Calculate Resistance of Bulb

Using the formula:

\[ R_{\text{bulb}} = \frac{V^2}{P} = \frac{120^2}{60} = \frac{14400}{60} = 240 \, \Omega \]

Step 2: Let Required Series Resistance Be \( R \)

Total resistance in circuit:

\[ R_{\text{total}} = R_{\text{bulb}} + R \]

Total current using source voltage:

\[ I = \frac{V_s}{R_{\text{total}}} = \frac{220}{240 + R} \]

But we want the bulb to get its rated power (60 W at 120 V), so current through bulb must be:

\[ I = \frac{P}{V} = \frac{60}{120} = 0.5 \, \text{A} \]

Step 3: Use Current to Solve for \( R \)

\[ 0.5 = \frac{220}{240 + R} \Rightarrow 240 + R = \frac{220}{0.5} = 440 \Rightarrow R = 440 - 240 = 200 \, \Omega \]

Conclusion:

The required resistance to be connected in series is \( {200 \, \Omega} \), so the correct answer is (C).

Answer 2

The power \( P \) of the bulb is given by: \[ P = \frac{V^2}{R_{\text{bulb}}} \] Where \( V = 120 \, \text{V} \), so the resistance of the bulb is: \[ R_{\text{bulb}} = \frac{V^2}{P} = \frac{120^2}{60} = 240 \, \Omega \] Now, the total voltage \( V_{\text{total}} = 220 \, \text{V} \) is to be divided between the series combination of the bulb and the resistor. The current \( I \) flowing through the circuit is: \[ I = \frac{V_{\text{bulb}}}{R_{\text{bulb}}} = \frac{120}{240} = 0.5 \, \text{A} \] The required resistance \( R_{\text{series}} \) is: \[ R_{\text{series}} = \frac{V_{\text{total}} - V_{\text{bulb}}}{I} = \frac{220 - 120}{0.5} = 200 \, \Omega \]