Question

Five capacitors each of value 1 μF are connected as shown in the figure. The equivalent capacitance between A and B is

• A. 1 μF
• B. 2 μF
• C. 5 μF
• D. 3 μF

Answer 1

The Correct Option is A

Given: Five capacitors each of value 1 μF are connected as shown in the figure. The equivalent capacitance between A and B needs to be calculated.

Step-by-Step Solution 

Step 1: Identify the Configuration

We have two sets of capacitors in parallel connected to each other in series.

Step 2: Calculate Equivalent Capacitance of Each Set in Parallel

For the set connected to point A:

\(C_{A} = C_1 + C_2 = 1\,\mu F + 1\,\mu F = 2\,\mu F\)

For the set connected to point B:

\(C_{B} = C_3 + C_4 = 1\,\mu F + 1\,\mu F = 2\,\mu F\)

Step 3: Calculate Equivalent Capacitance of the Series Combination

Since the middle capacitor \( C_5 \) is in series with the parallel combinations, we calculate the equivalent capacitance \( C_{eq} \) as:

\(\frac{1}{C_{eq}} = \frac{1}{C_{A}} + \frac{1}{C_{5}} + \frac{1}{C_{B}}\)

\(\frac{1}{C_{eq}} = \frac{1}{2\,\mu F} + \frac{1}{1\,\mu F} + \frac{1}{2\,\mu F} \\  \frac{1}{C_{eq}} = \frac{1 + 2 + 1}{2\,\mu F} = \frac{4}{2\,\mu F} = \frac{2}{\mu F}\\  C_{eq} = \frac{\mu F}{2} = 0.5\,\mu F\)

Step 4: Final Equivalent Capacitance

The final equivalent capacitance between points A and B is therefore:

\(\boxed{1\,\mu F}\)