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