Question
Derive an expression for equivalent capacitance of three capacitors when connected
i. in series and
ii. in parallel.
i. in series and
ii. in parallel.
✓
Answer
i. 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 first capacitor, $V _1= V _{ A }- V _{ B }=\frac{Q}{C_1} \ldots$ (i)
For the second capacitor, $V _2= V _{ B }- V _{ C }=\frac{Q}{C_2} \ldots$ (ii)
For the third capacitor, $V _3= V _{ C }- V _{ D }=\frac{q}{c_3} \ldots$ (iii)
Adding (i), (ii) and (iii), we get
$V _1+ V _2+ V _3= V _{ A }- V _{ D }=Q\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right] \ldots$ (iv)
If V be the potential difference between A and D, then
$V _{ A }- V _{ D }= V$
$\therefore$ From (iv), we get
$V =\left( V _1+ V _2+ V _3\right)=Q\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right] \ldots( v )$
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
$V =\frac{Q}{C} \ldots( vi )$
Comparing (v) and (vi), we get
$\frac{Q}{C}=Q\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right]$ or $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \ldots$ (vii)
Thus in series arrangement, "The reciprocal of equivalent capacitance is equal to the sum of the reciprocals of the individual capacitors."
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 first capacitor, $V _1= V _{ A }- V _{ B }=\frac{Q}{C_1} \ldots$ (i)
For the second capacitor, $V _2= V _{ B }- V _{ C }=\frac{Q}{C_2} \ldots$ (ii)
For the third capacitor, $V _3= V _{ C }- V _{ D }=\frac{q}{c_3} \ldots$ (iii)
Adding (i), (ii) and (iii), we get
$V _1+ V _2+ V _3= V _{ A }- V _{ D }=Q\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right] \ldots$ (iv)
If V be the potential difference between A and D, then
$V _{ A }- V _{ D }= V$
$\therefore$ From (iv), we get
$V =\left( V _1+ V _2+ V _3\right)=Q\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right] \ldots( v )$
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
$V =\frac{Q}{C} \ldots( vi )$
Comparing (v) and (vi), we get
$\frac{Q}{C}=Q\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right]$ or $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \ldots$ (vii)
Thus in series arrangement, "The reciprocal of equivalent capacitance is equal to the sum of the reciprocals of the individual capacitors."
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