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1. Derive an expression for equivalent capacitance of three capacitors when connected in series. 2. Derive an expression for equivalent capacitance of three capacitors when connected in parallel.

Vidyadip · Accepted Answer

In fig. (a) Three capacitors of capacitances C₁, C₂, C₃ are connected in series between points A and D.

In 'series’ first plate of each capacitor has charge +Q and second plate of each capacitor has charge -Q i.e., charge on each capacitor is Q.

Let the potential differences across the capacitors C₁, C₂, C₃ be V₁, V₂, V₃ respectively. As the second plate of first capacitor C₁ and first plate of second capacitor C₂ are connected together, their potentials are equal. Let this common potential be V_B. Similarly the common potential of second plate of C₂ and first plate of C₃ is V_C. The second plate of capacitor C₃ is connected to earth, therefore its potential V_D = 0. As charge flows from higher potential to lower potential, therefore V_A > V_B > V_C > V_D.

For the frist capacitor, $\text{V}_1=\text{V}_\text{A}-\text{V}_\text{B}=\frac{\text{Q}}{\text{C}_1}\ ...(\text{i)}$

For the second capacitor, $\text{V}_1=\text{V}_\text{B}-\text{V}_\text{C}=\frac{\text{Q}}{\text{C}_2}\ ...(\text{ii)}$

For the third capacitor, $\text{V}_3=\text{V}_\text{C}-\text{V}_\text{D}=\frac{\text{Q}}{\text{C}_3}\ ...(\text{iii)}$

Adding (i), (ii) and (iii), we get,

$\text{V}_1+\text{V}_2+\text{V}_3=\text{V}_\text{A}-\text{V}_\text{D}=\text{Q}\Big[\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}\Big]\ ...(\text{iv)}$

If V be the potential difference between A and D, then,

$\text{V}_\text{A}-\text{V}_\text{D}=\text{V}$

$\therefore$ From (iv), we get,

$\text{V}=\big(\text{V}_1+\text{V}_2+\text{V}_3=\text{Q}\Big[\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}\Big]\ ...(\text{v})$

If in place of all the three capacitors, only one capacitor is placed between A and D such that on giving it charge Q, the potential difference between its plates become V, then it will be called equivalent capacitor. If its capacitance is C, then,

$\text{V}=\frac{\text{Q}}{\text{C}}\ ...(\text{vi})$

Comparing (v) and (vi), we get,

$\frac{\text{Q}}{\text{C}}=\text{Q}\Big[\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}\Big]\ \text{Or}\ \frac{1}{\text{C}}=\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}$

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