The current in an inductor is given by i = 5 + 16 t , where i and t are in ampere and second respectively. The induced emf is 10 mV . Find out i self-inductance and ii the energy stored in the inductor at t = 1 second.

Given that di dt = 16 As - 1 The formula to calculate emf across the conductor is: ε = L di dt From this, the formula used to calculate the self inductance ( L ) is obtained as ε = L di dt We get, 10 × 10 − 3 = L d dt ( 5 + 16 t ) = L × 16 Or L = 10 16 × 10 − 3 H = 6 . 25 × 10 − 3 H So, The self-inductance = 6 . 25 × 10 − 3 H = 0 . 625 mH Now for the energy stored in the inductor at At t = 1 s , I = 21 A U = 1 2 LI 2 = 1 2 ( 0 . 625 × 10 − 3 ) ( 21 ) 2 = 0 . 137 J P = Ei = ( 10 × 10 − 3 ) ( 21 ) = 0 . 21 J / s

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The network shown in the figure is a part of a complete circuit. If at a certain instant, the current $' i'$ is $5 A$ and is decreasing at a rate of $10^{3} A s^{- 1},$ then the potential difference $V_{B A} =$

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Important Questions on Electromagnetic Induction and Alternating Current

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The network shown in the figure is a part of a complete circuit. If at a certain instant, the current $' i'$ is $5 A$ and is decreasing at a rate of $10^{3} A s^{- 1},$ then the potential difference $V_{B A} =$