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

Question

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

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

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

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In a series arrangement of capacitors the charges that is absorbed by the capacitors is identical however the potential difference is spread out across the capacitors. It is the reciprocal value of equivalent capacitance equals the total of the reciprocals of the individual capacitances

1⁄ C = 1⁄ C₁ + 1⁄ C₂ + 1⁄ C₃

In a combination of parallel capacitors the potential difference across all capacitors is the same however, the current within the circuit is distributed between the capacitors. Equivalent capacitance equals the total resistances of each capacitor:

C = C1 + C2 + C3

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Oscar Davis · Accepted Answer

Concepts and Reason

This problem can be solved using the series and parallel combination of capacitance and the charge across capacitance concepts.

First, calculate the equivalent capacitance by using the series or parallel combination of the capacitor. Next, calculate the charge across circuit using the expression of charge, which is a term that includes capacitance and potential.

Fundamentals

Capacitance refers to the conductor's ability to store an electrical charge. Farad is the SI unit for capacitance.

If [katex]N[/katex]

[katex]\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}..... + \frac{1}{{{C_N}}}[/katex]

The voltage division rule gives the voltage of the capacitor:

[katex]{V_X} = {V_S}\frac{{{C_T}}}{{{C_X}}}[/katex]

Here, [katex]{C_T}[/katex]

This is the equivalent circuit to the one shown.

(a) At [katex]a[/katex], the capacitors are identical

[katex]{V_a} = V[/katex]

The voltage drops when it reaches [katex]b[/katex].

[katex]{V_b} = V[/katex]

Potential at [katex]a[/katex]

[katex]\begin{array}{c}\{V_a} - {V_b} = 0\\\Delta V = 0\\end{array}[/katex]

(b) [katex]\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}..... + \frac{1}{{{C_N}}}[/katex]

Two capacitors are shown in the diagram as part A.

[katex]\begin{array}{c}\\frac{1}{{{C_{equ}}}} = \frac{1}{C} + \frac{1}{C}\\ = \frac{2}{C}\\{C_{equ}} = \frac{C}{2}\\end{array}[/katex]

If [katex]N[/katex]

In the diagram, part B, you can see three capacitors connected in parallel.

[katex]\begin{array}{c}\{C_{equ}} = \frac{C}{2} + \frac{C}{2} + \frac{C}{2}\\ = \frac{{3C}}{2}\\end{array}[/katex]

Ans:

Part A

The difference in points [katex]a[/katex] and b is 0 V

Part B

The equivalent capacitance for the six capacitors is [katex]\frac{{3C}}{2}[/katex]

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