In the circuit shown in the figure,the switch | vedclass

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(C) Just after the switch $S$ is closed,the capacitors act as short circuits (wires) because they are initially uncharged. However,looking at the circ

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Long time after the switch $S$ is closed,$I_1 = 0.6\,A, I_2 = 0$ and $I_3 = 0$.

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(C) Just after the switch $S$ is closed,the capacitors act as short circuits (wires) because they are initially uncharged. However,looking at the circuit,the capacitors are in series with the $4\,\Omega$ resistor and in parallel with the $8\,\Omega$ resistor. At $t=0$,the capacitor acts as a wire,so the $4\,\Omega$ branch is effectively shorted. The circuit simplifies to the $6\,V$ battery in series with the $2\,\Omega$ resistor and the $8\,\Omega$ resistor. Thus,$I_1 = I_3 = 6\,V / (2\,\Omega + 8\,\Omega) = 0.6\,A$ and $I_2 = 0$.Long time after the switch $S$ is closed,the capacitors are fully charged and act as open circuits. The branch containing the $4\,\Omega$ resistor and the capacitor will have no current $(I_2 = 0)$. The circuit simplifies to the $6\,V$ battery in series with the $2\,\Omega$ and $8\,\Omega$ resistors. Thus,$I_1 = I_3 = 6\,V / (10\,\Omega) = 0.6\,A$ and $I_2 = 0$.

In the circuit shown in the figure,the switch $S$ is initially open and the capacitor is initially uncharged. $I_1, I_2$ and $I_3$ represent the current in the resistance $2\,\Omega, 8\,\Omega$ and $4\,\Omega$ respectively.

Similar Questions

Two identical capacitors $A$ and $B$,charged to the same potential $V$,are connected in two different circuits as shown below at time $t=0$. If the charge on capacitors $A$ and $B$ at time $t=CR$ is $Q_{A}$ and $Q_{B}$ respectively,then (Here $e$ is the base of natural logarithm):

The charge stored by the capacitor $C$ in the given circuit in the steady state is . . . . . . $\mu C$.

$A$ $2\ \mu F$ capacitor and an $R$ resistor are connected in series to a $200\ V$ $DC$ supply. $A$ neon bulb is connected across the capacitor,which glows at $120\ V$. Calculate the value of $R$ so that the bulb glows for $5\ s$ after the switch is closed. (Given: $\log_{10} 2.5 = 0.4$)

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