Question

In the following circuit, the equivalent capacitance between terminal A and terminal B is :

• A. 2 µF
• B. 1 µF
• C. 0.5 µF
• D. 4 µF

Answer 1

The Correct Option is A

To determine the equivalent capacitance between terminals A and B in the provided circuit, a step-by-step analysis of its configuration is performed.

Circuit Configuration:

The circuit comprises four 2 μF capacitors arranged in a diamond configuration:

1. Two capacitors are connected in series on the left.
2. Two capacitors are connected in series on the right.
3. These two series arrangements are connected in parallel between terminals A and B.

1. Left-Side Series Arrangement:
Capacitors: Two 2 μF capacitors in series.
Equivalent series capacitance:

$ \frac{1}{C_{\text{series}}} = \frac{1}{2} + \frac{1}{2} = 1 $
$ C_{\text{series}} = 1 \mu F $

2. Right-Side Series Arrangement:
Capacitors: Two 2 μF capacitors in series.
Equivalent series capacitance:

$ \frac{1}{C_{\text{series}}} = \frac{1}{2} + \frac{1}{2} = 1 $
$ C_{\text{series}} = 1 \mu F $

3. Parallel Arrangement of Series Combinations:
The two resulting 1 μF equivalent capacitors are connected in parallel.
Equivalent parallel capacitance:

$ C_{\text{parallel}} = 1 + 1 = 2 \mu F $

Conclusion:

The equivalent capacitance between terminal A and terminal B is $ 2 \mu F $.