Question

The total capacitance across points 1 and 2 in the circuit shown is

• A. \(0.66\,\mu F\)
• B. \(4.66\,\mu F\)
• C. \(3.66\,\mu F\)
• D. \(2.66\,\mu F\)

Hint:

Always reduce complex capacitor networks step-by-step.

Answer 1

The Correct Option is D

Step 1: Combine the parallel capacitors in the network.
In the given capacitor network, two $1\,\mu F$ capacitors appear in parallel in one branch. Parallel capacitors add directly: $C_p = 1 + 1 = 2\,\mu F$.

Step 2: Find the series combination with another capacitor.
This $2\,\mu F$ is in series with a $1\,\mu F$ capacitor. Series capacitors: $C_s = \dfrac{2 \times 1}{2+1} = \dfrac{2}{3}\,\mu F \approx 0.667\,\mu F$.

Step 3: Add the final parallel branch.
The result $0.667\,\mu F$ is in parallel with another $2\,\mu F$ capacitor: $C_{eq} = 2 + \dfrac{2}{3} = \dfrac{8}{3} \approx 2.66\,\mu F$. \[ \boxed{2.66\,\mu F} \]