Question

A circuit is wired up as shown below. The capacitor is initially uncharged and switches S1 and S2 are initially open.

1) What is the voltage across the capacitor immediately after switch S1 is closed?

V_c = 0

V_c = V

V_c = 2V/3

2) What is the voltage across the capacitor after switch S1 has been closed for a very long time?

V_c = 0

V_c = V

V_c = 2V/3

3) After being closed a long time, switch 1 is opened and switch 2 is closed. What is the current through the right resistor immediately after switch 2 is closed?

I_R = 0

I_R = V/3R

I_R = V/2R

I_R = V/R

4) Now suppose both switches are closed. What is the voltage across the capacitor after a very long time?

V_c = 0

V_c = V

V_c = 2V/3

Answer 1

Concepts and reason

The concept involved to solve the above problem is charging and discharging of capacitor.

The capacitor gets charge when the switch 1 is closed and it discharges as switch 2 is closed. As the capacitor charges, potential across it increases with time, whereas when it discharges, potential across it decreases with time.

Fundamentals

A capacitor is an electrical device that stores electric charge. It consists of two electric conductors separated by an insulator.

Capacitance if the capacity of a capacitor to store charge. It is given by

$C = \frac{Q}{V}$

Here, $C$ is the capacitance of the capacitor, $Q$ is the charge and $V$ is the voltage.

Charging of capacitor:

The equation of potential across the capacitor during charging is given by

${V_{\rm{C}}} = V\left( {1 - {e^{\frac{{ - t}}{{RC}}}}} \right)$

Here, ${V_C}$ is the potential across the capacitor, $V$ is the potential across the source and $t$ is time.

Therefore, potential across capacitor increases with time when it is charging.

Discharging of Capacitor:

The equation of potential across the capacitor during discharging is given by

${V_{\rm{C}}} = V\left( {{e^{\frac{{ - t}}{{RC}}}}} \right)$

Therefore, potential across capacitor decreases with time when it is discharging.

(1)

Just as the switch S1 is closed, the circuit is completed. Since the charging time is very short,

${\rm{t = 0}}$

The voltage across the capacitor immediately after closing of switch S1

${V_{\rm{C}}} = V\left( {1 - {e^{\frac{{ - t}}{{RC}}}}} \right)$

Substitute 0 second for t

$\begin{array}{l}\\{V_{\rm{C}}} = V\left( {1 - {e^0}} \right)\\\\{V_{\rm{C}}} = V\left( {1 - 1} \right)\\\\{V_{\rm{C}}} = 0\\\end{array}$

(2)

As the switch S1 is closed, the circuit is completed. Since the charging time is very long,

$t = \infty$ .

The voltage across the capacitor after switch S1 has been closed for a long time

${V_{\rm{C}}} = V\left( {1 - {e^{\frac{{ - t}}{{RC}}}}} \right)$

Substitute $\infty$ for $t$

$\begin{array}{l}\\{V_{\rm{C}}} = V\left( {1 - {e^{ - \infty }}} \right)\\\\{V_{\rm{C}}} = V\left( {1 - \frac{1}{{{e^\infty }}}} \right)\\\\{V_{\rm{C}}} = V\left( {1 - 0} \right)\\\\{V_{\rm{C}}} = V\\\end{array}$

(3)

Just as switch 2 is closed, the circuit 2 gets completed. The current across the circuit is due to current to discharging of capacitor.

The potential across Resistor = ${\rm{V}}$

Resistance of the resistor = $2{\rm{R}}$

Therefore, current I through the circuit,

Using Ohm’s law

$\begin{array}{l}\\{I_{\rm{R}}} = \frac{V}{R}\\\\{I_{\rm{R}}} = \frac{V}{{2R}}\\\end{array}$

Ans: Part 1

Voltage across the capacitor immediately after switch S1 is closed is ${V_{\rm{C}}} = 0$

Part 2

Voltage across the capacitor after switch S1 is closed for a long time is ${V_{\rm{C}}} = V$

Part 3

The current across the resistor immediately after switch 2 is closed is ${I_{\rm{R}}} = \frac{V}{{2R}}$