Explain the parallel connection of capacitors. - Vidyadip

Question

Explain the parallel connection of capacitors.

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Answer

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

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