Q. A $4\mu F$ condenser is connected in parallel to another condenser of $8\mu F.$ Both the condensers are then connected in series with a $12\mu F$ condenser and charged to $20\ volts.$ The charge on the plate of $4\mu F$ condenser is
A) $3.3\mu C$
B) $40\mu C$
C) $80\mu C$
D) $240\mu C$
Explanation
(b) Equivalent capacitance of the circuit $C_{eq} = 6\mu F$
Charge supplied from source $Q = 6 \times 20 = 120\mu C$
Hence charge on the plates of $4\ \mu F$ capacitor
$= Q^{'} = \frac{4}{(4 + 8)} \times 120 = 40\mu C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.