Question

Six identical capacitors with capacitance C are connected as shown in the figure (Figure 1) .

Six identical capacitors with capacitance C are connected as shown in the figure (Figure 1) What is the potential difference between points a and b? What is the equivalent capacitance of these six capacitors?

1)What is the potential difference between points a and b?

2) What is the equivalent capacitance of these six capacitors?

Answer 1

Concepts and reason

The concepts used to solve this problem are series and parallel combination of capacitance, and the charge across the capacitance.

Firstly, calculate the equivalent capacitance of the circuit by using the series and parallel combination of the capacitance, after that calculate the charge across the circuit by using the expression of charge in term of capacitance and voltage.

Fundamentals

Capacitance is the ability of a conductor to store an electric charge. The SI unit of capacitance is Farad.

If $N$ number of capacitors are connected in parallel then the equivalent capacitance is, ${C_{{\rm{eq}}}} = {C_1} + {C_2} + ....... + {C_{\rm{N}}}$

If $N$ number of capacitors are in series then the equivalent capacitance is given by:

$\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}..... + \frac{1}{{{C_N}}}$

The voltage on the capacitor is given by the voltage division rule:

${V_X} = {V_S}\frac{{{C_T}}}{{{C_X}}}$

Here, ${C_T}$ is the total capacitance and ${V_S}$ is the source voltage.

The equivalent circuit for the circuit given is as follows,

(a)

The capacitors are same at point $a$ and $b$ .The voltage at $a$ is the voltage dropped due to one capacitor $C$ , as the voltage source has one capacitor $C$ between it and point $a$ .

${V_a} = V$ ……(1)

Similarly, the voltage is dropped, when it reaches $b$ , by the same amount as $a$ . Again there is one capacitor $C$ between the source and point $b$ .

${V_b} = V$ ……(2)

The potential at point $a$ is the same as points $b$ so the potential difference between points a and b is,

$\begin{array}{c}\\{V_a} - {V_b} = 0\\\\\Delta V = 0\\\end{array}$

[Part a]

Part a

Part a

(b)

If $N$ number of capacitors are in series then the equivalent capacitance is given by:

$\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}..... + \frac{1}{{{C_N}}}$

From the diagram part A, two capacitors are connected in series:

$\begin{array}{c}\\\frac{1}{{{C_{equ}}}} = \frac{1}{C} + \frac{1}{C}\\\\ = \frac{2}{C}\\\\{C_{equ}} = \frac{C}{2}\\\end{array}$

If $N$ number of capacitors are connected in parallel then the equivalent capacitance is, ${C_{{\rm{eq}}}} = {C_1} + {C_2} + ....... + {C_{\rm{N}}}$

Form the diagram part B, three capacitors are connected in parallel.

$\begin{array}{c}\\{C_{equ}} = \frac{C}{2} + \frac{C}{2} + \frac{C}{2}\\\\ = \frac{{3C}}{2}\\\end{array}$

[Part b]

Part b

Part b

Ans: Part a

The potential difference between points $a$ and $b$ is $0{\rm{ V}}$ .

Part b

The equivalent capacitance of the six capacitors is $\frac{{3C}}{2}$ .