Question

The effective capacitance between the points $A$ and $B$ in the circuit shown in $(\mu F)$ is (capacitance of each capacitor is$(\mu F)$

• A. 2
• B. 4
• C. 3
• D. 0.4

Answer 1

Answer: (D)

Capacitance of 2 capacitors in parallel = 1 + 1 = 2$\mu$F
Capacitance of 1 $\mu$F, 2$\mu$F and 1 $\mu$F in series is given by, $\frac{1}{C}=\frac{1}{1}+\frac{1}{2}+\frac{1}{1}=\frac{2+1+2}{2}=\frac{5}{2}$
$i.e.$ C = $\frac{2}{5}=0.4\mu F$