A 10 ; Omega, 20 ; mH coil carrying constant | vedclass

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(D) The steady current $i_0$ flowing through the coil before the switch is opened is given by Ohm's law: $i_0 = \frac{V}{R} = \frac{20 \; V}{10 \; \Om

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) The steady current $i_0$ flowing through the coil before the switch is opened is given by Ohm's law: $i_0 = \frac{V}{R} = \frac{20 \; V}{10 \; \Omega} = 2 \; A$.The average induced emf $\langle \varepsilon angle$ in the coil is defined as the change in magnetic flux linkage divided by the time interval, which is equivalent to the self-induced emf formula: $\langle \varepsilon angle = L \frac{\Delta i}{\Delta t}$.Given values are:Inductance $L = 20 \; mH = 20 	imes 10^{-3} \; H$Change in current $\Delta i = i_{final} - i_{initial} = 0 - 2 = -2 \; A$Time interval $\Delta t = 100 \; \mu s = 100 	imes 10^{-6} \; s$Substituting these values into the formula (considering the magnitude of emf):$|\langle \varepsilon angle| = L \frac{|\Delta i|}{\Delta t} = \frac{20 	imes 10^{-3} 	imes 2}{100 	imes 10^{-6}}$$|\langle \varepsilon angle| = \frac{40 	imes 10^{-3}}{10^{-4}} = 40 	imes 10^1 = 400 \; V$.

$A$ $10 \; \Omega, 20 \; mH$ coil carrying constant current is connected to a battery of $20 \; V$ through a switch. When the switch is opened, the current becomes zero in $100 \; \mu s$. The average emf induced in the coil is $\dots \; V$.

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