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JEE Mains · Physics · STD 12 -7. Alternating current

An alternating voltage \(v\left( t \right) = 220\,\sin \,100\pi l\,volt\) is applied to a purely resistive load of \(50\,\Omega \) . The time taken for the current to rise from half of the peak value of the peak value is.....\(ms\)

  1. A \(2.2\)
  2. B \(3.3\)
  3. C \(5\)
  4. D \(7.2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3.3\)

Step-by-step Solution

Detailed explanation

\(V(t)=220 \sin (100 \pi t)\) volt time taken, \(t=\frac{\theta}{\omega}=\frac{\frac{\pi}{3}}{100 \pi}=\frac{1}{300} \,\mathrm{sec}\) \(=3.3 \,\mathrm{ms}\)
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