Question 517: Electrostatics & Capacitance

Question

In the given capacitive network the resultant capacitance between point $\mathrm{A}$ and $\mathrm{B}$ is

Accepted Answer

In the given circuit, $C_3$ and $C_4$ are in series and $\mathrm{C}_3=\mathrm{C}_4=8 \mu \mathrm{F}$
$	herefore \quad \frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}_3}+\frac{1}{\mathrm{C}_4}$
$\begin{aligned}
	herefore \quad \mathrm{C}_{\mathrm{S}} & =\frac{\mathrm{C}_3^2}{2 \mathrm{C}_3} \
\mathrm{C}_{\mathrm{S}} & =\frac{\mathrm{C}_3}{2} \
	herefore \quad \mathrm{C}_{\mathrm{S}} & =4 \mu \mathrm{F} \quad 	ext{.... (i)}
\end{aligned}$
$\mathrm{C}_5$ and $\mathrm{C}_6$ are in parallel and $\mathrm{C}_5=\mathrm{C}_6=4 ~\mu \mathrm{F}$
$\mathrm{C}_{\mathrm{P}}=\mathrm{C}_5+\mathrm{C}_6$
$	herefore \quad \mathrm{C}_{\mathrm{P}}=8 \mu \mathrm{F}$
$	herefore \quad$ Equivalent circuit is as shown in figure.
Now, $\mathrm{C}_2$ and $\mathrm{C}_{\mathrm{P}}$ are in series and their combination in parallel with $\mathrm{C}_{\mathrm{S}}$
$\begin{aligned}
	herefore \quad \mathrm{C}_{\mathrm{E}} & =\frac{\mathrm{C}_2 \mathrm{C}_{\mathrm{p}}}{\mathrm{C}_2 \mathrm{C}_{\mathrm{p}}}+\mathrm{C}_{\mathrm{s}} \
& \mathrm{C}_{\mathrm{E}}=\frac{(8)(8)}{16}+4 \
	herefore \quad & \mathrm{C}_{\mathrm{E}}=8 \mu \mathrm{F}
\end{aligned}$
Now, $\mathrm{C}_1$ and $\mathrm{C}_{\mathrm{E}}$ are in series,
$\begin{aligned}
& 	herefore \quad C=\frac{C_1 C_E}{C_1+C_E} \
& 	herefore \quad C=\frac{(8)(8)}{16} \
& 	herefore \quad C=4 \mu F
\end{aligned}$

Answer

8 $\mu$F

Answer

4 $\mu$F

Answer

2 $\mu$F

Answer

16 $\mu$F

Electrostatics & Capacitance

In the given capacitive network the resultant capacitance between point \mathrm{A} and \mathrm{B} is

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