Current in a circuit falls from 5.0 , A to 0. | vedclass

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(B) Initial current,$I_{1} = 5.0 \, A$.Final current,$I_{2} = 0.0 \, A$.Change in current,$di = I_{1} - I_{2} = 5.0 \, A$.Time taken for the change,$d

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$4$

Vedclass · Answer

(B) Initial current,$I_{1} = 5.0 \, A$.Final current,$I_{2} = 0.0 \, A$.Change in current,$di = I_{1} - I_{2} = 5.0 \, A$.Time taken for the change,$dt = 0.1 \, s$.Average $emf$,$e = 200 \, V$.For self-inductance $(L)$ of the circuit,we use the relation for induced $emf$:$e = L \frac{di}{dt}$$L = e \cdot \frac{dt}{di}$$L = 200 \cdot \frac{0.1}{5.0} = 200 \cdot 0.02 = 4 \, H$.Hence,the self-inductance of the circuit is $4 \, H$.

Current in a circuit falls from $5.0 \, A$ to $0.0 \, A$ in $0.1 \, s$. If an average $emf$ of $200 \, V$ is induced,give an estimate of the self-inductance of the circuit in $H$.

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