In the circuit shown in the figure,C_1 = 2C_2 | vedclass

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(B) The circuit consists of two $RC$ branches connected in parallel to a voltage source $V$. When the switch $S$ is closed at $t = 0$,each branch char

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increases with increase in time $t$

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(B) The circuit consists of two $RC$ branches connected in parallel to a voltage source $V$. When the switch $S$ is closed at $t = 0$,each branch charges independently.For the first branch with capacitor $C_1$ and resistor $R$,the current is given by $i_1(t) = \frac{V}{R} e^{-t / (R C_1)}$.For the second branch with capacitor $C_2$ and resistor $R$,the current is given by $i_2(t) = \frac{V}{R} e^{-t / (R C_2)}$.The ratio of the currents is $\frac{i_1}{i_2} = \frac{\frac{V}{R} e^{-t / (R C_1)}}{\frac{V}{R} e^{-t / (R C_2)}} = e^{-t / (R C_1) + t / (R C_2)} = e^{\frac{t}{R} \left( \frac{1}{C_2} - \frac{1}{C_1} \right)}$.Given $C_1 = 2C_2$,we have $\frac{1}{C_2} = \frac{2}{C_1}$.Substituting this into the ratio: $\frac{i_1}{i_2} = e^{\frac{t}{R} \left( \frac{2}{C_1} - \frac{1}{C_1} \right)} = e^{\frac{t}{R C_1}}$.Since $t$ increases,the exponent $\frac{t}{R C_1}$ increases,and therefore the ratio $i_1 / i_2 = e^{\frac{t}{R C_1}}$ increases with time $t$.

In the circuit shown in the figure,$C_1 = 2C_2$. Switch $S$ is closed at time $t = 0$. Let $i_1$ and $i_2$ be the currents flowing through the branches containing $C_1$ and $C_2$ respectively at any time $t$. Then the ratio $i_1 / i_2$:

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