Question
Explain the parallel connection of capacitors.
✓
Answer
$\rightarrow$ As shown in the fig. $(a)$ two capacitors $C _1$ and $C _2$ are connected in parallel.
$\rightarrow$ For such connection, the $p.d$. across each capacitor is same, suppose its value is $V$ .
$\rightarrow$ The charge on the plate of capacitor $' 1\ '$ is $\left(+Q_1\right)$ and the charge on the plate of capacitor $' 2 '$ is $\left(+Q_2\right)$
$\rightarrow Q=C V$ for first capacitor $Q_1=C_1 V$ and for second capacitor $Q _2= C _2 V$
$\rightarrow$ Hence, the total charge $($ charge of the equivalent capacitor$)$
$Q =Q_1+Q_2$
$\therefore Q =C_1 V+C_2 V$
$\therefore Q =V\left(C_1+C_2\right)$
$\therefore \frac{Q}{V} =C_1+C_2$
$\rightarrow$ Suppose, the equivalent capacitance for the given parallel connection is $C$ .
$\therefore C =\frac{ Q }{ V } \ldots$
$\therefore C = C _1+ C _2 \ ($From $(1)$ and $(2))$
$\rightarrow $ As shown in fig.$ (b), C_1, C_2, C_3 \ldots . . C_n$ total $n$ capacitors are connected in parallel. $P.d$. for each is $V$ and the charge on each capacitor is $Q _1, Q _2, Q _3 \ldots . . Q _n$
$\rightarrow$ Hence, the total charge,
$Q=Q_1+Q_2+Q_3+\ldots .+Q_n$
$\therefore Q = C _1 V+ C _2 V+ C _3 V+\ldots .+ C _n V$
$\therefore \frac{ Q }{ V }= C _1+ C _2+ C _3+\ldots . .+ C _n$
$\rightarrow$ Suppose, the equivalent capacitance for the given parallel connection is $C$ .
Where $\therefore C =\frac{ Q }{ V }$
$\rightarrow$ So, from eq $(3)$ and $(4),$
$\rightarrow C = C _1+ C _2+ C _3+\ldots . .+ C _n$
$\rightarrow$ Suppose, the capacitance of each capacitor is same and it is ' $C ^{\prime}$
$\therefore$ Equivalent capacitance,
$C _{e q} = C + C + C +\ldots . .+ C \text { (n times) }$
$\therefore \left( C _{e q}\right. =n C )$
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