question_answer

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 \[\mu F\]

Correct Answer: A

Solution :

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                 \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{C}+\frac{1}{{{C}_{2}}}\]                 \[=\frac{1}{1}+\frac{1}{2}+\frac{1}{1}\]                 \[=2+\frac{1}{2}=\frac{2}{5}\] \[\therefore \] \[C=\frac{2}{5}=0.4\,\mu F\]