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(D) The initial current $I$ in the circuit is given by Ohm's law: $I = \frac{V}{R} = \frac{12\,V}{6\,\Omega} = 2\,A$.When the switch is opened,the cur

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(D) The initial current $I$ in the circuit is given by Ohm's law: $I = \frac{V}{R} = \frac{12\,V}{6\,\Omega} = 2\,A$.When the switch is opened,the current drops from $2\,A$ to $0\,A$ in a time interval $\Delta t = 1\,ms = 10^{-3}\,s$.The magnitude of the induced emf is given by the formula: $|\varepsilon| = L \left| \frac{\Delta I}{\Delta t} ight|$.Substituting the given values: $20 = L 	imes \frac{2 - 0}{10^{-3}}$.$20 = L 	imes \frac{2}{10^{-3}}$.$20 = L 	imes 2000$.$L = \frac{20}{2000} = 0.01\,H$.Converting to millihenry $(mH)$: $L = 0.01 	imes 1000\,mH = 10\,mH$.

$A$ $12\,V$ battery connected to a coil of resistance $6\,\Omega$ through a switch,drives a constant current in the circuit. The switch is opened in $1\,ms$. The emf induced across the coil is $20\,V$. The inductance of the coil is ........... $mH$.

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