Question

In the circuit shown in figure, charge stored in $6\mu F$ capacitor will be

• A. $18\mu C$
• B. $54\mu C$
• C. $36\mu C$
• D. $72\mu C$

Answer 1

The correct option is A $18\mu C$

Voltage across two parallel elements are equal.
Hence,
Voltage across $4\mu F=9V$.
Applying K.V.L in loop containing battery of voltage $\left(ϵ\right)=12V$, we can find voltage across $6\mu F$ capacitor.
So, the voltage across $6\mu F$ capacitor will be $12-9=3V$
Therefore, charge $\left(Q\right)=C\times V$
$=6\times 3=18\mu C$