JEE Mains · Physics · STD 12 -7. Alternating current
An alternating voltage \(\mathrm{V}(\mathrm{t})=220 \sin 100 \ \pi \mathrm{t}\) 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 to the peak value is _______.
- A \(5 \mathrm{~ms}\)
- B \(3.3 \mathrm{~ms}\)
- C \(7.2 \mathrm{~ms}\)
- D \(2.2 \mathrm{~ms}\)
Answer & Solution
Correct Answer
(B) \(3.3 \mathrm{~ms}\)
Step-by-step Solution
Detailed explanation
Rising half to peak \(\mathrm{t}=\mathrm{T} / 6\) \(\mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms}\)
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