Question

The effective capacitance between the points P and Q in the circuit shown in figure, is (capacitance of each capacitor is $1\mu F$ ):

• A. $0.4\mu F$
• B. $3\mu F$
• C. $4\mu F$
• D. $2\mu F$

Answer 1

Answer: (A)

The circuit diagram is shown in figure

As shown, $C_3$ and $C_4$ are in parallel, hence their effective capacitance is
$C=C_3+C_4=1+1=2\mu F$
Now $C_1,C_2$ and $C$ are in series, hence their effective capacitance is

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C}+\frac{1}{C_2}$
$=\frac{1}{1}+\frac{1}{2}+\frac{1}{1}$
$=2+\frac{1}{2}=\frac{5}{2}$
$\therefore$ $C=\frac{2}{5}=0.4\mu F$