The switch shown in the figure is closed at $t=0$. The charge on the capacitor as a function of time is given by

Q. In an LCR circuit as shown below both switches are open initially. Now switch $S_1$ is closed, $S_2$ kept open. (q is charge on the capacitor and $\tau =RC$ is capacitive time constant). Which of the following statement is correct?

Q. It is given that when switch S is open, charge on one of the capacitors is $q^{\prime }=\frac{c\epsilon }{2}$ and another one is zero. At t = 0 switch S is closed, then

Q. Consider a simple RC circuit as shown in Figure 1


Process 1 : In the circuit the switch $S$ is closed at $t=0$ and the capacitor is fully charged to voltage $V_e$ (i.e. charging continues for time $T>>RC$ ). In the process some dissipation ($E_D$) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is $E_c$.

Process 2 :

In a different process the voltage is first set to $\frac{V_0}{3}$ and maintained for a charging time T>>RC. Then the voltage is raised to $\frac{2V_0}{3}$ without discharging the capacitor and again maintained for a time $T>>RC$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final voltage $V_0$ as in Process $1.$

These two processes are depicted in Figure 2.

In Process 1, the energy stored in the capacitor $E_c$ and heat dissipated across resistance $E_D$ are related by

Q. The electric current in a charging $R-C$ circuit is given by $i=i_0{e}^{-t/RC}$ where $i_0$, $R$ and $C$ are constant parameters of the circuit and $t$ is time. Find the rate of change of current at (a) $t=0$, (b) $t=RC$, (c) $t=10RC$.

Q.

Two capacitors of capacitance 2C and C are charged to potential difference 2V and V respectively. Now the charged capacitors are connected with a resistor R as shown in figure. Switch ‘S’ is closed at t = 0.

What is the current in circuit as function of time?