NEET · Physics · STD 12 -7. Alternating current
The peak value of an alternating current is 5 A and frequency is 60 Hz. How long will the current, starting from zero, take to reach the peak value ?
- A \(\frac{1}{120} s\)
- B \(\frac{1}{60} s\)
- C \(\frac{1}{30} s\)
- D \(\frac{1}{240} s\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{240} s\)
Step-by-step Solution
Detailed explanation
(D) \(\frac{1}{240} s\)
The alternating current is given by \(I=I_0 \sin (\omega t)\).
The time taken to reach the peak value from zero corresponds to a phase angle of \(\frac{\pi}{2}\).
\(\omega t=\frac{\pi}{2}\)
\(2 \pi f t=\frac{\pi}{2}\)
\(t=\frac{1}{4 f}\)
Given \(f=60 Hz\), we get:
\(t=\frac{1}{4 \times 60}=\frac{1}{240} s\)
The alternating current is given by \(I=I_0 \sin (\omega t)\).
The time taken to reach the peak value from zero corresponds to a phase angle of \(\frac{\pi}{2}\).
\(\omega t=\frac{\pi}{2}\)
\(2 \pi f t=\frac{\pi}{2}\)
\(t=\frac{1}{4 f}\)
Given \(f=60 Hz\), we get:
\(t=\frac{1}{4 \times 60}=\frac{1}{240} s\)
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