Q.1 (b) An alternating current of frequency 50 Hz has a maximum value of 12 A. Find (i) Equation for instantaneous value of current and (ii) Time taken to reach 9.6 A for the first time.
Best answer
Given: f = 50 Hz, Im = 12 A. \( \omega = 2\pi f = 100\pi = 314.16 \) rad/s.
(i) \( i = I_m\sin\omega t \Rightarrow \) \( i = 12\sin(314.16\,t) \) A (i.e. \( 12\sin 100\pi t \)).
(ii) \( 9.6 = 12\sin(314.16\,t) \Rightarrow \sin(314.16\,t) = 0.8 \Rightarrow 314.16\,t = \sin^{-1}0.8 = 0.9273 \) rad (53.13°).
\( t = \dfrac{0.9273}{314.16} = 2.952\times10^{-3} \) s. t ≈ 2.95 ms (measured from the instant the current is zero and increasing).
Check: 12 sin(314.16 × 0.002952) = 12 sin(0.9273 rad) = 12 × 0.8 = 9.6 A.
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