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