Question

Find heat produced in the capacitors on closing the switch $S$.

Answer 1

Before the switch is closed, current flows in the branch having $4\mu F$ capacitor, charge q= $20\times 4\mu F=80\mu C$

When the switch is closed, a decaying current flows in all the three loops. That is (left, right and bottom)

So the potential at the top most junction ie; between the two capacitors and $2\Omega$ resistors is $x$ volt (let's assume)

Assuming the node connected on the left that connected $20V$ and resistance R and $4\Omega$ is 0 volt.

So, from the nodal equation at $x$, is

=>$(x-20)4+(x-0)5=0$

x= $\frac{80}{9}V$

Work done by cell= q.V

= Final charge on left plate of $4\mu F$ - initial potential on left plate of $4\mu F$

= 4(20-$\frac{80}{9}$)

= 4$\times 11.11$= $44.44\mu J$ (negative)

Energy absorbed by the capacitor= $\frac{1}{2}\times 4\times {(20-\frac{80}{9})}^2+\frac{1}{2}\times 5\times {\left(\frac{80}{9}\right)}^2=246.91+197.53=444.44\mu J\left(negative\right)$

Heat produced= Work done by batteries- energy absorbed by capacitors

= $-44.44-(-444.44)$

= $-44.44+444.44$

= $400.00\mu J$