The correct option is
B
384 μJFirst of all let us simplify the given circuit and replace the whole circuit by a single equivalent capacitance
Ceq connected across
X &
Y.
Using point potential techique,
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1201480/original_3s.png)
Therefore we have
′3′ points named
1, 2 &
3 with different potentials.
Redrawing the circuit gives,
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1201483/original_3s-1.png)
For series connection for
1−2 and
2−3 , equivalent capacitance,
C13=2×12+1=23 μF
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1201486/original_3s-3.png)
Now, the final equivalent capacitance,
CXY=23+2=83 μF
Therefore, we have final equivalent circuit,
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1201487/original_3s-4.png)
Total charge supplied by the battery,
q=(CXY×V)=83×12=32 μC
Hence, the workdone by battery will be,
Wbattery=qV
⇒Wbattery=(32 μC)×(12 V)
∴Wbattery=384 μJ
Hence, option (b) is correct answer.
Why this question?:
Whenever workdone by battery is asked, it should immediately strike that we require total charge (q) supplied by battery in the circuit. In order to find q supplied, we need to simplify circuit and find Ceq first. |