Q.
In the adjoining figure, four capacitors are shown with their respective capacities and the P.D. applied. The charge and the P.D. across the $4\mu F$ capacitor will be
A) $600\mu C;150\ volts$
B) $300\mu C;75\ volts$
C) $800\mu C;200\ volts$
D) $580\mu C;145\ volts$
Explanation
(d) Total capacitance $\frac{1}{C} = \frac{1}{20} + \frac{1}{8} + \frac{1}{12} \Rightarrow C = \frac{120}{31}\mu F$
Total charge $Q = CV = \frac{120}{31} \times 300 = 1161\mu C$
Charge, through $4\ \mu F$ condenser $= \frac{1161}{2} = 580\mu C$
And potential difference across it $= \frac{580}{4} = 145V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.