Question

Three capacitors each having capacitance $C=2\mu F$ are connected to battery of emf 30V as shown in the figure. When the switch is closed. If heat generated in the circuit is $\frac{x}{10}mJ$,then $x$ is :

• A. $6$
• B. $2$
• C. $3$
• D. $12$

Answer 1

The correct option is C $3$

In the initial condition,
$C_{net}=(2C\times C)/(2C+C)=2C/3$
Charge provided by battery $=2CV/3$
Charge on capacitors $=CV/3,CV/3,2CV/3$
Potential across capacitors $=V/3,V/3,2V/3$
Energy of capacitors $=C{V}^2/18,C{V}^2/18,2C{V}^2/9$
Total energy $=C{V}^2/3$
In the final condition,
$C_{net}=C$
Charge provided by battery $=CV$
Charge across capacitors $=0,0,CV$
Potential across capacitors $=0,0,V$
Total energy of capacitors $=C{V}^2/2$
Therefore,
work done by battery $=V\delta Q=V(CV-2CV/3)=C{V}^2/3$
Now,
work done by battery $=$ (Final energy - Initial energy) of capacitors + Heat
$C{V}^2/3=C{V}^2/6+HEAT\Rightarrow HEAT=C{V}^2/6=2\times {10}^{-6}\times 900/6=0.3mJ$