Two resistors of 10, Omega and 20, Omega and | vedclass

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(A) At $t = 0$,when the key is just closed,the inductor acts as an open circuit because it opposes any change in current. Therefore,the $10\, \Omega$

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$\frac{1}{15}\,A, \frac{1}{10}\,A$

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(A) At $t = 0$,when the key is just closed,the inductor acts as an open circuit because it opposes any change in current. Therefore,the $10\, \Omega$ resistor and $20\, \Omega$ resistor are in series.Total resistance $R_{eq} = 10\, \Omega + 20\, \Omega = 30\, \Omega$.The initial current $I_0 = \frac{V}{R_{eq}} = \frac{2\, V}{30\, \Omega} = \frac{1}{15}\, A$.As $t \rightarrow \infty$,the inductor acts as a short circuit (ideal inductor has zero resistance). The $10\, \Omega$ resistor is now in parallel with the short-circuited inductor,so all current flows through the inductor,bypassing the $10\, \Omega$ resistor.Now,only the $20\, \Omega$ resistor is in the circuit.The final current $I_{\infty} = \frac{V}{20\, \Omega} = \frac{2\, V}{20\, \Omega} = \frac{1}{10}\, A$.

Two resistors of $10\, \Omega$ and $20\, \Omega$ and an ideal inductor of $10\, H$ are connected to a $2\, V$ battery as shown. The key $K$ is inserted at time $t = 0$. The initial $(t = 0)$ and final $(t \rightarrow \infty)$ currents through the battery are

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