Question

In the figure is shown a system of four capacitors connected across a 10V battery. Charge that will flow from switch S when it is closed is:

A. Zero
B. $20\mu C$ from a to b
C. $5\mu C$ from b to a
D. $5\mu C$ from a to b

Answer 1

Hint: To solve this problem, we will use the concept of charging and discharging of capacitors and also we know that the capacitor blocks the DC current. We will proceed by solving the series and parallel capacitor circuit for the cases when the switch is on and when the switch is closed.
Formula used- When capacitor are in series, the equivalent capacitance is
$
  \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} \\
  {C_{eq}} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} \\
$
And when capacitors are in parallel, the equivalent capacitance is
${C_{eq}} = {C_1} + {C_2}$

Complete step-by-step answer:

Case 1 When the switch was open, the equivalent capacitance is
For parallel path XaY
$
  {C_{aeq}} = \dfrac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}} \\
  {C_{aeq}} = \dfrac{{2 \times 3}}{{2 + 3}} \\
  {C_{aeq}} = 1.2\mu F \\
$
Similarly for the other parallel path XbY, the equivalent capacitance is
$
  {C_{beq}} = \dfrac{{{C_3} \times {C_4}}}{{{C_3} + {C_4}}} \\
  {C_{beq}} = \dfrac{{2 \times 3}}{{2 + 3}} \\
  {C_{beq}} = 1.2\mu F \\
$
The net equivalent capacitance is
$
  {C_{eq}} = {C_{aeq}} + {C_{beq}} \\
  {C_{eq}} = 1.2 + 1.2 \\
  {C_{eq}} = 2.4\mu F \\
$
The voltage of the circuit is 10V
Net charge is given by
$
  {Q_{net}} = {C_{eq}}V \\
  {Q_{net}} = 2.4\mu F \times 10V \\
  {Q_{net}} = 24\mu C \\
$
This net charge will distribute among the capacitors as shown in the figure.
$$

Case 2 When the switch is closed, the charge distribution is shown in the figure below. As the switch is closed the potential at point a and b will be the same.
Let the potential at the positive terminal i.e. at a be V and the potential at the negative terminal of the capacitor 0.
The charge across the capacitor ${C_1} = 2\mu F$ is ${Q_1}$
${Q_1} = \left( {10 - V} \right) \times 2\mu F........\left( 1 \right)$
The charge across the capacitor ${C_2} = 3\mu F$ is ${Q_2}$
${Q_2} = \left( {10 - V} \right) \times 3\mu F........\left( 2 \right)$
The charge across the capacitor ${C_3} = 3\mu F$ is ${Q_3}$
\[{Q_3} = V \times 3\mu F........\left( 3 \right)\]
The charge across the capacitor ${C_2} = 2\mu F$ is ${Q_4}$
${Q_4} = V \times 2\mu F........\left( 4 \right)$
Since, the capacitor ${C_1} = {C_4}{\text{ and }}{{\text{C}}_2} = {C_3}$
Therefore ${Q_1} = {Q_4}{\text{ and }}{{\text{Q}}_2} = {Q_3}$
From equation (1) and equation (4)
$
  \left( {10 - V} \right) \times 2 = 2V \\
  20 - 2V = 2V \\
  4V = 20 \\
  V = 5 \\
$
From equation (1) and equation (3)
$
  {Q_3} = 5 \times 3 \\
  {Q_3} = 15\mu F \\
  {Q_1} = \left( {10 - V} \right) \times 2 \\
  {Q_1} = \left( {10 - 5} \right) \times 2 \\
  {Q_1} = 10\mu F \\
$
Therefore
${Q_1} = {Q_4}{\text{ = 10}}\mu {\text{F and }}{{\text{Q}}_2} = {Q_3} = 15\mu F$
From it is clear that the charge will flow from point b towards point a
And the net charge flow is $5\mu F$
Hence, the correct option is C.

Note- In order to solve these types of problems, remember the relation between charge and voltage and learn how to find the equivalent capacitances. Also when the capacitors are in series, the total capacitance is always less than the minimum of them and if two capacitors are in parallel, then the total capacitance is the sum of them. Remember charge flows from high value to low value.