Question

What is the net capacitance of the circuit given below? (C1 = 4$\mu$F, C2 = 8$\mu$F in series, C3 = 6$\mu$F in parallel)

• A. 8.6 $\mu$F
• B. 4.6 $\mu$F
• C. 6.6 $\mu$F
• D. 3.0 $\mu$F

Hint:

For two capacitors in series, use "Product over Sum" to find the equivalent value quickly.

Answer 1

The Correct Option is A

Step 1: Concept
Capacitors in series follow the reciprocal rule ($1/C_s = 1/C_1 + 1/C_2$), while capacitors in parallel add directly ($C_p = C_s + C_3$).

Step 2: Meaning
First, find the equivalent capacitance of the top branch (C1 and C2 in series).

Step 3: Analysis
For the series part: $C_s = (4 \times 8) / (4 + 8) = 32 / 12 = 2.66 \mu$F. Now, add the parallel capacitor C3: $C_{net} = 2.66 + 6 = 8.66 \mu$F.

Step 4: Conclusion
The net capacitance of the circuit is approximately 8.6 $\mu$F. Final Answer: (A)