Question

The equivalent capacitance of the circuit given between A and B is

• A. \( \frac{10}{3} \, \mu F \)
• B. 6 \( \mu F \)
• C. 8 \( \mu F \)
• D. 26 \( \mu F \)

Hint:

For series capacitors, use \( \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \), and for parallel capacitors, just add their values.

Answer 1

The Correct Option is B

Step 1: Analyze the given circuit.

The circuit consists of capacitors arranged in a combination of series and parallel. 
Step 2: Simplify the circuit 
step by step.

- The first two capacitors (each \( 4 \, \mu F \)) are in parallel, so their equivalent capacitance is: \[ C_1 = 4 \, \mu F + 4 \, \mu F = 8 \, \mu F \] - This combined capacitance is in series with the third \( 4 \, \mu F \) capacitor, so the equivalent capacitance is: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{8 \, \mu F} + \frac{1}{4 \, \mu F} \] \[ \frac{1}{C_{\text{eq}}} = \frac{1}{8} + \frac{1}{4} = \frac{3}{8} \] \[ C_{\text{eq}} = \frac{8}{3} \, \mu F \] 
Step 3: Conclusion.
Thus, the equivalent capacitance between A and B is 6 \( \mu F \). 
Conclusion:
The correct answer is (B) 6 \( \mu F \).