Q141 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
A bulb of resistance 280 Ohm is supplied with a voltage $\mathrm{V}=400 \sin \pi \mathrm{t}$. The peak current is
- A2.02 A
- B2.22 A
- Correct answerC1.43 A
- D1.11 A
Solution
The given voltage is $V = 400 \sin(\pi t)$.
Comparing this with the standard form $V = V_{peak} \sin(\omega t)$, the peak voltage is $V_{peak} = 400$ V.
The resistance of the bulb is $R = 280$ Ohm.
Using Ohm's law for the peak values, the peak current $I_{peak}$ is given by $I_{peak} = \dfrac{V_{peak}}{R}$.
$I_{peak} = \dfrac{400}{280} = \dfrac{40}{28} = \dfrac{10}{7}$ A.
$I_{peak} \approx 1.4285$ A.
Rounding to two decimal places, $I_{peak} \approx 1.43$ A.
Answer: 1.43 A
Comparing this with the standard form $V = V_{peak} \sin(\omega t)$, the peak voltage is $V_{peak} = 400$ V.
The resistance of the bulb is $R = 280$ Ohm.
Using Ohm's law for the peak values, the peak current $I_{peak}$ is given by $I_{peak} = \dfrac{V_{peak}}{R}$.
$I_{peak} = \dfrac{400}{280} = \dfrac{40}{28} = \dfrac{10}{7}$ A.
$I_{peak} \approx 1.4285$ A.
Rounding to two decimal places, $I_{peak} \approx 1.43$ A.
Answer: 1.43 A
Hence, the correct answer is option: 'C'.