Question

The total capacity of the system of capacitors shown in the adjoining figure between the points A and B is

• A. 1μF
• B. 2μF
• C. 3μF
• D. 4μF

Answer 1

The Correct Option is B

The problem involves finding the equivalent capacitance of the given system of capacitors between points A and B. Let's analyze the configuration step-by-step:

1. **Identify Series and Parallel Combinations**: 

• The two capacitors each of \(2 \, \mu F\) on the top and bottom are in parallel.
• The \(1 \, \mu F\) capacitor in the middle is in series with the combination of the two \(2 \, \mu F\) capacitors.

2. **Calculate Equivalent Capacitance of Parallel Capacitors**:

The top two \(2 \, \mu F\) capacitors in parallel have an equivalent capacitance:

\(C_{top} = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F\)

The bottom two \(2 \, \mu F\) capacitors in parallel also have an equivalent capacitance:

\(C_{bottom} = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F\)

3. **Calculate Equivalent Capacitance of Series Combination**:

The \(1 \, \mu F\) capacitor in the middle is in series with the above configurations, and the \(1 \, \mu F\) from both sides is also connected, forming a combination:

\(\frac{1}{C_{eq}} = \frac{1}{4 \, \mu F} + \frac{1}{1 \, \mu F} + \frac{1}{4 \, \mu F}\) \(\frac{1}{C_{eq}} = \frac{1}{4} + 1 + \frac{1}{4} = \frac{1.5}{1}\) \(C_{eq} = \frac{1}{1.5} \approx 0.666 \, \mu F\)

4. **Final Series Combinations**:

To find the total capacitance between A and B, consider the two central \(1 \, \mu F\) capacitors are symmetrical around the common \(1 \, \mu F\), effectively making it \(1 \, \mu F \) in parallel, yielding:

Therefore, the total capacitance is \(2 \, \mu F\) as the configurations were symmetrical.

5. **Conclusion**:

The total equivalent capacitance between A and B is:

\(C_{total} = 2 \, \mu F\)

Hence, the correct answer is 2μF.

Answer 2

The given circuit can be rearranged as

Now the equivalent capacitance of two 1 µF capacitors in parallel is given by 
\(Ceq = 1 + 1 = 2 µF\)