question_answer

Four capacitors are connected in a circuit as shown in figure. The effective capacitance in \[\mu F\] between P and Q will be :

A)  \[5\,\,\mu F\]

B)  \[10\,\mu F\]

C)  \[2\mu F\]

D)  \[7.5\mu F\]

Correct Answer: A

Solution :

 Key Idea: Capacitors in series have same amount of charge, and in parallel have same potential difference. In the given circuit, the two \[2\,\mu F\] capacitors are connected in parallel, hence resultant circuit is shown Equivalent capacitance \[C=2+2=4\,\mu \] (in parallel) \[\therefore \] \[C=2+2=4\mu F\] This is connected in series with 12 up capacitor, hence equivalent capacitance is \[C=\frac{C\times 12}{C+12}=\frac{4\times 12}{12+4}=\frac{48}{16}=3\,\mu F\] Now, effective capacitance, \[C=3+2=5\mu F\]