Q196 / 240 · KCET 2026 · Physics
A light bulb rated $100$ W is connected to an AC source of $220$ V, $50$ Hz. The rms current through the bulb is
- Correct answerA$0.454$ A
- B$0.545$ A
- C$2.20$ A
- D$0.22$ A
Solution
The power of the light bulb is given as $P = 100$ W.
The rms voltage of the AC source is $V_{rms} = 220$ V.
For a purely resistive load like a light bulb, the power factor is $1$. The average power is given by the relation:
$P = V_{rms} I_{rms}$
Rearranging the formula to solve for the rms current $I_{rms}$:
$I_{rms} = \dfrac{P}{V_{rms}}$
Substituting the given values:
$I_{rms} = \dfrac{100}{220}$
$I_{rms} = \dfrac{5}{11} \approx 0.454$ A
Answer: $0.454$ A
The rms voltage of the AC source is $V_{rms} = 220$ V.
For a purely resistive load like a light bulb, the power factor is $1$. The average power is given by the relation:
$P = V_{rms} I_{rms}$
Rearranging the formula to solve for the rms current $I_{rms}$:
$I_{rms} = \dfrac{P}{V_{rms}}$
Substituting the given values:
$I_{rms} = \dfrac{100}{220}$
$I_{rms} = \dfrac{5}{11} \approx 0.454$ A
Answer: $0.454$ A
Hence, the correct answer is option: 'A'.