Question

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

• A. $0.0002J$
• B. $0.0005J$
• C. $0.00075J$
• D. Zero

Answer 1

The correct option is A $0.0002J$

Heat produced = Work done by battery - Energy absorbed by capacitors

Before S is closed, $q$ on capacitor during steady state = 20 \times 4 = 80 \mu C

Finally when S is closed;

$\frac{U_x-20}{4}+\frac{x-0}{5}=0$

$\Rightarrow V_x=\frac{80}{9}V$

Work done by all = $q\times V=4(20-\frac{80}{9})=-35.56\mu J(-ve)$

Energy absorbed by capacitors $=\frac{1}{2}\times 4(20-\frac{80}{9}{)}^2+\frac{1}{2}\times 5\times (\frac{80}{9}{)}^2-\frac{1}{2}\times 4\times {20}^2=-355.56\mu J$

Heat produced = Work done by battery - Energy absorbed by capacitors $=-35.56+355.56=320\mu J$