In the circuit given below, the equivalent capacitance is ______________(a) 5.43 μF(b) 4.23 μF(c) 3.65 μF(d) 5.50 μFThis question was addressed to me in semester exam.I need to ask this question from Advanced Problems Involving Complex Circuit Diagram topic in division Network Theory and Complex Circuit Diagram of Network Theory

In the circuit given below, the equivalent capacitance is ____________

1.	In the circuit given below, the equivalent capacitance is ______________(a) 5.43 μF(b) 4.23 μF(c) 3.65 μF(d) 5.50 μFThis question was addressed to me in semester exam.I need to ask this question from Advanced Problems Involving Complex Circuit Diagram topic in division Network Theory and Complex Circuit Diagram of Network Theory
Answer» CORRECT answer is (a) 5.43 μF For explanation: The 2 μF capacitor is in parallel with 1 μF capacitor and this combination is in SERIES with 0.5 μF. Hence, C1 = \(\frac{0.5(2+1)}{0.5+2+1}\) = \(\frac{1.5}{3.5}\) = 0.43 Now, C1 is in parallel with the 5 μF capacitor. ∴ CEQ = 0.43 + 5 = 5.43 μF.

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