Question

In the circuit given below, the equivalent capacitance is ______________

(a) 5.43 μF

(b) 4.23 μF

(c) 3.65 μF

(d) 5.50 μF

Answer 1

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.