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 terminal A and terminal B in the given circuit, let's analyze the configuration step by step.

Circuit Analysis:

The circuit consists of four capacitors, each with a capacitance of 2 μF, arranged in a diamond shape as follows:

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

1. Series Combination on the Left Side:
Capacitors: 2 μF and 2 μF in series.
Equivalent capacitance for series:

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

2. Series Combination on the Right Side:
Capacitors: 2 μF and 2 μF in series.
Equivalent capacitance for series:

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

3. Parallel Combination of the Two Series Pairs:
The two 1 μF equivalent capacitors are now in parallel.
Equivalent capacitance for parallel:

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

Final Answer:

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