Question

Hear produced in the capacitor on closing the switch s is

Answer 1

Answer:

Before the switch is closed, current flows in the branch having 4\mu F 4μF capacitor, charge q= 20\times 4\mu F\\ =\quad 80\mu C20× 4μF

=80μC

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 2Ω resistors is xx volt (let's assume)

Assuming the node connected on the left that connected 20 V20V and resistance R and 4\Omega 4Ω is 0 volt.

So, from the nodal equation at xx, is

=>(x-20)4\quad +\quad (x-0)5=0(x−20)4+(x−0)5=0

x= \dfrac { 80 }{ 9 } V 980 V

Work done by cell= q.V

= Final charge on left plate of 4\mu F4μF - initial potential on left plate of 4\mu F4μF

= 4(20-\dfrac { 80 }{ 9 } 980 )

= 4$\times 11.11×11.11= 44.44\mu J44.44μJ (negative)</p><p>Energy absorbed by the capacitor= \dfrac { 1 }{ 2 } \times 4\times { (20-\dfrac { 80 }{ 9 } ) }^{ 2 }+\quad \dfrac { 1 }{ 2 } \times 5\times { (\dfrac { 80 }{ 9 } ) }^{ 2 }\\ =246.91+197.53\\ =444.44\quad \mu J\quad (negative)$21×4×(20−980 )2 + 21 ×5×(980 ) 2

=246.91+197.53

=444.44μJ(negative)

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

= -44.44- (-444.44)−44.44−(−444.44)

= -44.44+ 444.44−44.44+444.44

= 400.00 \mu J400.00μJ