The current (i) at time t = 0 and t = infty r | vedclass

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(D) At $t = 0$,the inductor acts as an open circuit (infinite resistance). The circuit simplifies to two parallel branches: one with $(5 \Omega + 1 \O

Vedclass · Accepted Answer

$\frac{5 E }{18}, \frac{10 E }{33}$

Vedclass · Answer

(D) At $t = 0$,the inductor acts as an open circuit (infinite resistance). The circuit simplifies to two parallel branches: one with $(5 \Omega + 1 \Omega) = 6 \Omega$ and the other with $(5 \Omega + 4 \Omega) = 9 \Omega$. These two branches are in parallel,so the equivalent resistance is $R_{eq} = (6 	imes 9) / (6 + 9) = 54 / 15 = 18 / 5 \ \Omega$. The current $i$ at $t = 0$ is $i_1 = E / R_{eq} = E / (18 / 5) = 5E / 18$. At $t = \infty$,the inductor acts as a short circuit (zero resistance). The circuit becomes a bridge-like structure where the $5 \Omega$ and $5 \Omega$ resistors are in parallel,and the $1 \Omega$ and $4 \Omega$ resistors are in parallel. The equivalent resistance is $R_{eq} = (5 \parallel 5) + (1 \parallel 4) = (2.5) + (4 / 5) = 2.5 + 0.8 = 3.3 \ \Omega = 33 / 10 \ \Omega$. The current $i$ at $t = \infty$ is $i_2 = E / R_{eq} = E / (33 / 10) = 10E / 33$.

The current $(i)$ at time $t = 0$ and $t = \infty$ respectively for the given circuit is

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