Question

A 2$\mu$F capacitor is charged to a potential 10 V. Another 4$\mu$F capacitor is charged to a potential 20 V. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. Then,

• A. Final energy in 2$\mu$F capacitor is 100 $\mu J$
• B. The net amount heat of dissipated in the circuit is 600 $\mu J$
• C. Final energy in 4$\mu$F capacitor is 200 $\mu J$
• D. The net amount of heat dissipated in the circuit is 200$\mu J$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Final energy in 2$\mu$F capacitor is 100 $\mu J$
B The net amount heat of dissipated in the circuit is 600 $\mu J$
C Final energy in 4$\mu$F capacitor is 200 $\mu J$

Before connection
$Q_1=2\times 10=20\mu C$
$Q_2=4\times 20=80\mu C$

$U_i=\frac{1}{2}2(10{)}^2+\frac{1}{2}4(20{)}^2=900\mu J$

Since connected as shown

After connection,
$Q_{\text{net}}=-20+80=60\mu C$
$V=\frac{60}{2+4}=10Volt$
$U_f=\frac{1}{2}\times 6\times (10{)}^2=300\mu J$
|$\Delta U|=U_f-U_i=600\mu J$

$U_f$ for 2$\mu$F capacitor $=\frac{1}{2}\times 2\times (10{)}^2=100\mu J$

$U_f$ for 4$\mu$F capacitor $=\frac{1}{2}\times 4\times (10{)}^2=200\mu J$