Question
- Derive an expression for equivalent capacitance of three capacitors when connected in series.
- Derive an expression for equivalent capacitance of three capacitors when connected in parallel.
✓
Answer
In fig. $(a) $ Three capacitors of capacitances $C_1, C_2, C_3$ 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 $C_1, C_2, C_3$ be $V_1, V_2, V_3$ respectively.
As the second plate of first capacitor $C_1$ and first plate of second capacitor $C_2$ are connected together, their potentials are equal.
Let this common potential be $V_B$.
Similarly the common potential of second plate of $C_2$ and first plate of $C_3$ is $V_C$.
The second plate of capacitor $C_3$ 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, $\frac{\text{Q}}{\text{C}}=\text{Q}\Big[\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}\Big]$ Or $ \frac{1}{\text{C}}=\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}$
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 $C_1, C_2, C_3$ be $V_1, V_2, V_3$ respectively.
As the second plate of first capacitor $C_1$ and first plate of second capacitor $C_2$ are connected together, their potentials are equal.
Let this common potential be $V_B$.
Similarly the common potential of second plate of $C_2$ and first plate of $C_3$ is $V_C$.
The second plate of capacitor $C_3$ 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, $\frac{\text{Q}}{\text{C}}=\text{Q}\Big[\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}\Big]$ Or $ \frac{1}{\text{C}}=\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}$
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