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.....\(ms\)
- A \(2.2\)
- B \(3.3\)
- C \(5\)
- D \(7.2\)
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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