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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Physics
Chapter
Alternating Current

BITSAT 2024 · Paper 1 · Physics · Alternating Current

An alternating voltage V(t)=220 100 π t volt is applied to a purely...

An alternating voltage V(t)=220sin⁡100πtV(t)=220 \sin 100 \pi t volt is applied to a purely resistive load of 50Ω50 \Omega. The time taken for the current to rise from half of the peak value to the peak value is:

Options

  1. A5 ms
  2. B3.3 ms
  3. C7.2 ms
  4. D2.2 ms

Correct answer

B. 3.3 ms

Step-by-step solution

Rising half to peak t=T/6t=2π6ω=π3ω=π300π=1300=3.3mst=T / 6\mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.3 \mathrm{ ms} Here ω=100πrad/s\omega=100 \pi \mathrm{rad} / \mathrm{s}
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