Question

A light bulb is rated at 200 W for a 220 V supply. Find the resistance of the bulb.
OR
A light bulb is rated at 200 W for a 220 V supply. Find the rms current through the bulb.

Hint:

To find the rms current, use the relation \( I_{\text{rms}} = \sqrt{\frac{P}{R}} \), where \( P \) is power and \( R \) is resistance.

Answer 1

Finding the Resistance of the Bulb:
We know that the power \( P \) consumed by the bulb is given by the formula:
\[ P = \frac{V^2}{R} \] where:
- \( P = 200 \, \text{W} \) is the power,
- \( V = 220 \, \text{V} \) is the voltage,
- \( R \) is the resistance.
Rearranging the equation to solve for \( R \):
\[ R = \frac{V^2}{P} \]
Substitute the given values:
\[ R = \frac{(220)^2}{200} = \frac{48400}{200} = 242 \, \Omega \]
Thus, the resistance of the bulb is \( 242 \, \Omega \).
Finding the rms Current Through the Bulb:
The rms (root mean square) current \( I_{\text{rms}} \) is related to the power \( P \) and voltage \( V \) by the formula:
\[ P = I_{\text{rms}}^2 R \]
Rearranging the equation to solve for \( I_{\text{rms}} \):
\[ I_{\text{rms}} = \sqrt{\frac{P}{R}} \]
Substitute the known values:
\[ I_{\text{rms}} = \sqrt{\frac{200}{242}} \approx \sqrt{0.826} \approx 0.91 \, \text{A} \]
Thus, the rms current through the bulb is approximately \( 0.91 \, \text{A} \).