TS EAMCET · Physics · Alternating Current
The frequency of an alternating voltage is 50 Hz. The time taken for instantaneous voltage to increase from zero to half of its peak voltage is
- A \(\frac{1}{800} \mathrm{~s}\)
- B \(\frac{1}{600} \mathrm{~s}\)
- C \(\frac{1}{300} \mathrm{~s}\)
- D \(\frac{1}{200} \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{600} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(V = V_p \sin(\omega t)\) \(\frac{V_p}{2} = V_p \sin(\omega t)\) \(\sin(\omega t) = \frac{1}{2}\) \(\omega t = \frac{\pi}{6}\) \(2\pi f t = \frac{\pi}{6}\) \(t = \frac{1}{12f}\) \(t = \frac{1}{12 \cdot 50}\) \(t = \frac{1}{600} \mathrm{~s}\)
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