Question
Derive an expression for the effective capacitance of three parallel plate capacitors connected in series.
✓
Answer
i. Diagram:
ii. Explanation:
a. Capacitors are said to be connected in series if they are connected one after the other in the form of a chain.
b. Let capacitors of capacitances $C_1, C_2, C_3$ be connected in series as shown in the figure.
c. Let $V_1, V_2, V_3$ be the corresponding potential differences in the capacitors.
d. Suppose a potential difference ' $V$ ' is applied across the combination. The left plate of capacitor $C_1$ has a charge $+Q$. An equal but opposite charge $-Q$ is induced on the right plate of this capacitor. This charge $-Q$ induces a charge $+Q$ on the left plate of $C_2$ and so on.
e. Thus, each capacitor receives a magnitude of charge $Q$. Hence, when the capacitors are connected in series, the same current flows through them and all have the same charge $+Q$ induced on the plate. Thus, potential difference induced across capacitors is given by,
$V _1=\frac{ Q }{ C _1}, V _2=\frac{ Q }{ C _2}, V _3=\frac{ Q }{ C _3}$
f. Total potential difference ' $V$ ' across the combination is given by,$V=V_1+V_2+V_3$
$\therefore V =\frac{ Q }{ C _1}+\frac{ Q }{ C _2}+\frac{ Q }{ C _3} \ldots$...[From equation (1)]
$\therefore V = Q \left(\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}\right)$
$g$. If these capacitors are replaced by a single capacitor of capacity $C_S$, such that effective capacity remains the same then
$ C _{ s }=\frac{ Q }{ V }$
$\therefore V =\frac{ Q }{ C _{ S }} \ldots .(3) $
From equation (2) and (3),
$ \frac{ Q }{ C _{ s }}= Q \left(\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}\right)$
$\therefore \frac{1}{ C _{ s }}=\left(\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}\right)$
$\therefore C _{ s }=\frac{ C _1 C _2 C _3}{ C _1 C _2+ C _2 C _3+ C _3 C _1} $
ii. Explanation:
a. Capacitors are said to be connected in series if they are connected one after the other in the form of a chain.
b. Let capacitors of capacitances $C_1, C_2, C_3$ be connected in series as shown in the figure.
c. Let $V_1, V_2, V_3$ be the corresponding potential differences in the capacitors.
d. Suppose a potential difference ' $V$ ' is applied across the combination. The left plate of capacitor $C_1$ has a charge $+Q$. An equal but opposite charge $-Q$ is induced on the right plate of this capacitor. This charge $-Q$ induces a charge $+Q$ on the left plate of $C_2$ and so on.
e. Thus, each capacitor receives a magnitude of charge $Q$. Hence, when the capacitors are connected in series, the same current flows through them and all have the same charge $+Q$ induced on the plate. Thus, potential difference induced across capacitors is given by,
$V _1=\frac{ Q }{ C _1}, V _2=\frac{ Q }{ C _2}, V _3=\frac{ Q }{ C _3}$
f. Total potential difference ' $V$ ' across the combination is given by,$V=V_1+V_2+V_3$
$\therefore V =\frac{ Q }{ C _1}+\frac{ Q }{ C _2}+\frac{ Q }{ C _3} \ldots$...[From equation (1)]
$\therefore V = Q \left(\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}\right)$
$g$. If these capacitors are replaced by a single capacitor of capacity $C_S$, such that effective capacity remains the same then
$ C _{ s }=\frac{ Q }{ V }$
$\therefore V =\frac{ Q }{ C _{ S }} \ldots .(3) $
From equation (2) and (3),
$ \frac{ Q }{ C _{ s }}= Q \left(\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}\right)$
$\therefore \frac{1}{ C _{ s }}=\left(\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}\right)$
$\therefore C _{ s }=\frac{ C _1 C _2 C _3}{ C _1 C _2+ C _2 C _3+ C _3 C _1} $
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