A 10Ω, 20 mH coil carrying constant current connected to a battery of 20 V through a switch is opened and current becomes zero in 100μs. The average emf induced in the coil is ............ V.

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A 10 $Ω$ , 20 mH coil carrying constant current connected to a battery of 20 V through a switch is opened and current becomes zero in 100μs. The average emf induced in the coil is ............ V.

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answer is 400.

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Detailed Solution

$\begin{array}{l} < ε >= \frac{\int εdt}{\int dt} = \frac{\int (Ldi / dt) dt}{\int dt} = \frac{L \int di}{\int dt} \\ < ε >= \frac{LΔi}{Δ t} \\ i_{0} = \frac{V}{R} = \frac{20}{10} = 2 A, if i = 0 A \\ Δ t = 100 μs, L = 20 mH \\ < ε >= \frac{20 \times 10^{- 3} \times (2 - 0)}{100 \times 10^{- 6}} \\ \Rightarrow < ε >= 400 V \end{array}$