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(D) At $t = 0$,the capacitor acts as a short circuit. The equivalent resistance of the circuit is $R_{eq} = R_3 + (R_1 || R_2) = 1 + \frac{2 	imes 2}

Vedclass · Accepted Answer

All of the above.

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(D) At $t = 0$,the capacitor acts as a short circuit. The equivalent resistance of the circuit is $R_{eq} = R_3 + (R_1 || R_2) = 1 + \frac{2 	imes 2}{2 + 2} = 1 + 1 = 2 \, \Omega$. The total current from the battery is $I = \frac{\varepsilon}{R_{eq}} = \frac{10}{2} = 5 \, A$. This current flows through $R_3$,so option $A$ is correct.As $t 	o \infty$,the capacitor is fully charged and acts as an open circuit. The current flows through the series combination of $R_1 || R_2$ and $R_3$. The total resistance is $R_{eq} = 2 \, \Omega$. The current through $R_3$ is $I = \frac{10}{2} = 5 \, A$. Note: The original option $B$ was incorrect; the current is $5 \, A$ at $t 	o \infty$ as well.Since $R_1$ and $R_2$ are in parallel,the voltage across them is the same. Thus,$I_{R_1} R_1 = I_{R_2} R_2$,which implies $\frac{I_{R_1}}{I_{R_2}} = \frac{R_2}{R_1} = \frac{2}{2} = 1$. This ratio is constant. Thus,option $C$ is correct.Given the options,$D$ is the most appropriate choice as $A$ and $C$ are correct.

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