In the circuit shown,the key (K) is closed at | vedclass

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(B) The circuit can be analyzed by splitting it into two independent loops connected to the battery. Loop $1$ (Left loop): Contains the battery $(20 \

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$2.5 \, A$

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(B) The circuit can be analyzed by splitting it into two independent loops connected to the battery. Loop $1$ (Left loop): Contains the battery $(20 \, V)$,inductor $(L = 5 \, mH)$,and resistors ($4 \, \Omega$ and $6 \, \Omega$). The total resistance is $R_1 = 4 + 6 = 10 \, \Omega$. The time constant is $	au_1 = L/R_1 = 5 	imes 10^{-3} / 10 = 5 	imes 10^{-4} \, s$. The current $I_1(t) = \frac{V}{R_1} (1 - e^{-t/	au_1}) = \frac{20}{10} (1 - e^{-t / (5 	imes 10^{-4})}) = 2 (1 - e^{-2000t})$. Loop $2$ (Right loop): Contains the battery $(20 \, V)$,capacitor $(C = 0.1 \, mF = 10^{-4} \, F)$,and resistors ($5 \, \Omega$ and $5 \, \Omega$). The total resistance is $R_2 = 5 + 5 = 10 \, \Omega$. The time constant is $	au_2 = R_2 C = 10 	imes 10^{-4} = 10^{-3} \, s$. The current $I_2(t) = \frac{V}{R_2} e^{-t/	au_2} = \frac{20}{10} e^{-t/10^{-3}} = 2 e^{-1000t}$. At $t = 10^{-3} \ln 2 \, s$: $I_1 = 2 (1 - e^{-2000 	imes 10^{-3} \ln 2}) = 2 (1 - e^{-2 \ln 2}) = 2 (1 - e^{\ln(2^{-2})}) = 2 (1 - 1/4) = 2(3/4) = 1.5 \, A$. $I_2 = 2 e^{-1000 	imes 10^{-3} \ln 2} = 2 e^{-\ln 2} = 2 (1/2) = 1 \, A$. The total current through the key $K$ is $I = I_1 + I_2 = 1.5 + 1 = 2.5 \, A$.

In the circuit shown,the key $(K)$ is closed at $t = 0$. Find the current through the key at the instant $t = 10^{-3} \ln 2 \, s$.

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