A 10, V battery connected to 5,Omega resistan | vedclass

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(D) The initial current $I$ in the circuit is given by Ohm's law: $I = \frac{V}{R} = \frac{10\, V}{5\,\Omega} = 2\, A$.When the switch is opened,the c

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$1	imes10^4\, V$

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(D) The initial current $I$ in the circuit is given by Ohm's law: $I = \frac{V}{R} = \frac{10\, V}{5\,\Omega} = 2\, A$.When the switch is opened,the current drops from $2\, A$ to $0\, A$ in a time interval $\Delta t = 2\, ms = 2 	imes 10^{-3}\, s$.The rate of change of current is $\frac{\Delta I}{\Delta t} = \frac{0 - 2}{2 	imes 10^{-3}} = -10^3\, A/s$.The average induced $emf$ $(\varepsilon)$ across the coil is given by the formula $\varepsilon = -L \frac{\Delta I}{\Delta t}$.Substituting the values: $\varepsilon = -10\, H 	imes (-10^3\, A/s) = 10^4\, V$.

$A$ $10\, V$ battery connected to $5\,\Omega$ resistance coil having inductance $10\, H$ through a switch drives a constant current in the circuit. The switch is suddenly opened and the time taken to open it is $2\, ms$. The average $emf$ induced across the coil is

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