MHT CET · Physics · Capacitance
The equivalent capacity between terminal A and \(B\) is

- A \(\frac{\mathrm{C}}{4}\)
- B \(\frac{3 \mathrm{C}}{4}\)
- C \(\frac{\mathrm{C}}{3}\)
- D \(\frac{4 \mathrm{C}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{4 \mathrm{C}}{3}\)
Step-by-step Solution
Detailed explanation

\(
\begin{aligned}
\frac{1}{C_5} & =\frac{1}{C}+\frac{1}{C}+\frac{1}{C} \\
\therefore \quad C_5 & =\frac{C}{3}
\end{aligned}
\)
Now, \(\mathrm{C}_5\) and \(\mathrm{C}\) are connected in parallel,
\(
\therefore \quad \mathrm{C}_{\text {net }}=\mathrm{C}_{\mathrm{s}}+\mathrm{C}=\frac{\mathrm{C}}{3}+\mathrm{C}=\frac{4 \mathrm{C}}{3}
\)
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