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