Question
Explain the series connection of capacitors.
✓
Answer
→As shown in fig. (a), two capacitors $C _1$ and $C _2$ are connected in series.
→The left plate of $C _1$ and the right plate of $C _2$ are connected to two terminals of a battery, and have charges $Q$ and $-Q$ on them, respectively.
→Consequently the right plate of $C _1$ has charge $-Q$ and left plate of $C_2$ has charge $Q$ induced on it.
→Like this, even though the capacitors may have different capacitance, the charge on them (the charge on each capacitor plate) is same.
→Suppose, the potential difference between two terminals of $C_1$ and $C_2$ is $V_1$ and $V_2$ respectively.
→The total potential drop V across the combination will be :
$\begin{aligned}
V & = V _1+ V _2 \\
\text { but } C _1 & =\frac{ Q }{ V _1} \\
\therefore V _1 & =\frac{ Q }{ C _1}
\end{aligned}$
Similarly we get $V_2=\frac{Q}{C_2}$
$\begin{array}{l}
\therefore \text { From eq }{ }^{ n } \text { (1) } \\
V =\frac{ Q }{ C _1}+\frac{ Q }{ C _2} \\
\therefore \quad \frac{ V }{ Q }=\frac{1}{ C _1}+\frac{1}{ C _2} \\
\end{array}$
→Suppose, the equivalent capacitance for the given combination of capacitors is C , then,
$\begin{array}{l}
\therefore C =\frac{ Q }{ V } \\
\therefore \quad \frac{1}{ C }=\frac{ V }{ Q }
\end{array}\$
$\therefore$ From equation (2) and (3)
$\therefore \frac{1}{ C }=\frac{1}{ C _1}+\frac{1}{ C _2}$
→As shown in fig. (b), n capacitors are connected in series.
→Their capacitance are $C _1, C _2, C _3$ $C _n$ respectively. Electric charge on each of those, is Q .
→Suppose the p.d. across these capacitors, are $V _1$, $V _2, V_3 \ldots . V _n$
→The total p.d. of the series combination will be :
$\begin{aligned}
V & = V _1+ V _2+ V _3+\ldots .+ V _n \\
\therefore V & =\frac{ Q }{ C _1}+\frac{ Q }{ C _2}+\frac{ Q }{ C _3}+\ldots .+\frac{ Q }{ C _n} \\
\therefore \frac{ V }{ Q } & =\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}+\ldots .+\frac{1}{ C _n}
\end{aligned}$
→Suppose, the equivalent (/ effective) capacitance for the given series combination of capacitors is C .
$\begin{array}{l}
\therefore C =\frac{ Q }{ V } \\
\therefore \frac{1}{ C }=\frac{ V }{ Q }
\end{array}$
→From equations (4) and (5),
$\therefore \frac{1}{ C }=\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}+\ldots .+\frac{1}{ C _n}$
→Suppose, the capacitance of each capacitor is same.
Then we get the equivalent capacitance.
$C _{\text {eq }}=\frac{ C }{n}$
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