The network of six capacitors is as shown in figure. The equivalent capacitance between $\mathrm{A}$ and…
- $\frac{2 C}{3}$
- $\frac{4 C}{3}$
- 2C
- 3C
Solution
Taking the similar nodes together the equivalent diagram can be drawn as:
Therefore,
the eqiuvalent capacitance between $\mathrm{AQ}=3 C+2 C+C=6 C$
the equivalent capacitance between $\mathrm{QB}=3 C+2 C+C=6 C$
Now, for equivalent capacitance between $A B$
$\begin{aligned}
& \frac{1}{C_{\mathrm{AB}}}=\frac{1}{6 C}+\frac{1}{6 C}=\frac{1}{3 C} \\
& \therefore C_{\mathrm{AB}}=3 C
\end{aligned}$Asked in: MHT CET 2022 (08 Aug Shift 1)