The current in an inductor of self-inductance | vedclass

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(A) Given:Self-inductance of the inductor,$L = 40 	ext{ mH} = 40 	imes 10^{-3} 	ext{ H}$.Initial current,$I_1 = 2 	ext{ A}$.Final current,$I_2 = 1

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) Given:Self-inductance of the inductor,$L = 40 	ext{ mH} = 40 	imes 10^{-3} 	ext{ H}$.Initial current,$I_1 = 2 	ext{ A}$.Final current,$I_2 = 12 	ext{ A}$.Time interval,$dt = 8 	ext{ ms} = 8 	imes 10^{-3} 	ext{ s}$.The magnitude of the induced emf in an inductor is given by the formula:$|\varepsilon| = L \frac{di}{dt}$Substituting the values:$|\varepsilon| = (40 	imes 10^{-3} 	ext{ H}) 	imes \frac{(12 	ext{ A} - 2 	ext{ A})}{8 	imes 10^{-3} 	ext{ s}}$$|\varepsilon| = 40 	imes 10^{-3} 	imes \frac{10}{8 	imes 10^{-3}}$$|\varepsilon| = 40 	imes \frac{10}{8} = 5 	imes 10 = 50 	ext{ V}$.

The current in an inductor of self-inductance $L=40 \text{ mH}$ is to be increased uniformly from $2 \text{ A}$ to $12 \text{ A}$ in $8 \text{ ms}$. The emf induced in the inductor during this process is (in $\text{ V}$)

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