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

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(C) 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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) 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 induced $emf$,$e = 200 \, V$. For self-inductance $(L)$ of the circuit,we use the relation for average $emf$: $e = L \frac{dI}{dt}$ Rearranging for $L$: $L = \frac{e}{(dI/dt)}$ Substituting the values: $L = \frac{200}{(5.0 / 0.1)} = \frac{200}{50} = 4 \, H$. 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 $Henry$.

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