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:
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.
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Approach Solution -2
The given circuit can be rearranged as Now the equivalent capacitance of two 1 µF capacitors in parallel is given by \(Ceq = 1 + 1 = 2 µF\)
