Question

Initially, the switch is open for a long time and capacitors are uncharged. If it is closed at $t=0$ ,then

• A. The battery supplies power after switch is closed
• B. When the capacitors are fully charged, work done by the battery is $9\mu J$
• C. A long time after switch is closed, the heat generated in the circuit is $9\mu J$
• D. The ratio of energy stored in capacitor 'A' before and after the switch is closed is 1:1.

Answer 1

Answer: (A), (B), (D)

The correct options are
A The battery supplies power after switch is closed
B When the capacitors are fully charged, work done by the battery is $9\mu J$
D The ratio of energy stored in capacitor 'A' before and after the switch is closed is 1:1.

1) When 'S' is open, charge in the circuit = $6\mu C$
When 'S' is closed, total capacitance = $3\mu F$
Total charge = $9\mu C$
Charge supplied by battery is $\Delta q=3\mu C$ $\text{Work done by battery,}{W}_B=3\times 3=9\mu V$
Initially energy stored in capacitor $U_i=\frac{1}{2}\times 2\times 3^2=9\mu J$
After a long time, energy stored in system
$U_f=\frac{1}{2}\left(3\right)3^2=\frac{27}{2}\mu J$
Using first law of thermodynamics, heat generated = $W_B+U_i-U_f=9+9-\frac{27}{2}=\frac{36-27}{2}=\frac{9}{2}\mu J$
The potential difference across the capacitor 'A' does not change.
Hence, the energy stored remains the same.