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Question 106 of 132

Q.The alternating current in a circuit is given by the equation i=10sin⁡(100πt+π6)i = 10\sin\left(100\pi t + \dfrac{\pi}{6}\right). The current attains its first maximum at t is :

(a) 1600\dfrac{1}{600} s
(b) 150\dfrac{1}{50} s
(c) 1100\dfrac{1}{100} s
(d) 1300\dfrac{1}{300} s
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2019MCQ· 1mImportance★★★★★
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Setting the sine function's argument to π/2\pi/2 (its first maximum) and solving for tt gives t=1/300t=1/300 s.

The given alternating current is i=10sin⁡(100πt+π6)i = 10\sin\left(100\pi t+\dfrac{\pi}{6}\right)

The function sin⁡θ\sin\theta attains its first (and maximum) value of 1 when its argument θ=π2\theta = \dfrac{\pi}{2}. So the current ii is maximum when: 100πt+π6=π2100\pi t+\dfrac{\pi}{6} = \dfrac{\pi}{2}

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