A condenser of capacity 2mF charges to a potential 200V is connected in parallel with a condenser of same capacity but charged to a potential 100V. The percentage loss of energy of system is

Options

A. `50 %`
B. `35 %`
C. `20 %`
D. `10 %`

Answer

To solve the problem, we need to calculate the initial energy stored in each capacitor, the total energy before they are connected, the energy after they are connected, and finally the percentage loss of energy.

### Step-by-Step Solution:

1. **Identify the given values**:
- Capacitor 1 (C1) = 2 mF = 2 x 10^-3 F
- Voltage across C1 (V1) = 200 V
- Capacitor 2 (C2) = 2 mF = 2 x 10^-3 F
- Voltage across C2 (V2) = 100 V

2. **Calculate the energy stored in C1 (U1)**:
\[ U1 = \frac{1}{2} C1 V1^2 \]
\[ U1 = \frac{1}{2} \times 2 \times 10^{-3} \times (200)^2 \]
\[ U1 = \frac{1}{2} \times 2 \times 10^{-3} \times 40000 \]
\[ U1 = 40 \text{ joules} \]

3. **Calculate the energy stored in C2 (U2)**:
\[ U2 = \frac{1}{2} C2 V2^2 \]
\[ U2 = \frac{1}{2} \times 2 \times 10^{-3} \times (100)^2 \]
\[ U2 = \frac{1}{2} \times 2 \times 10^{-3} \times 10000 \]
\[ U2 = 10 \text{ joules} \]

4. **Calculate the total initial energy (U)**:
\[ U = U1 + U2 = 40 + 10 = 50 \text{ joules} \]

5. **Calculate the total charge before connecting the capacitors**:
- Charge on C1 (Q1):
\[ Q1 = C1 \times V1 = 2 \times 10^{-3} \times 200 = 0.4 \text{ coulombs} \]
- Charge on C2 (Q2):
\[ Q2 = C2 \times V2 = 2 \times 10^{-3} \times 100 = 0.2 \text{ coulombs} \]
- Total charge (Q):
\[ Q = Q1 + Q2 = 0.4 + 0.2 = 0.6 \text{ coulombs} \]

6. **Calculate the equivalent capacitance (Ceq)**:
\[ Ceq = C1 + C2 = 2 \times 10^{-3} + 2 \times 10^{-3} = 4 \times 10^{-3} \text{ F} \]

7. **Calculate the energy stored in the equivalent capacitor (U')**:
\[ U' = \frac{1}{2} \frac{Q^2}{Ceq} \]
\[ U' = \frac{1}{2} \frac{(0.6)^2}{4 \times 10^{-3}} \]
\[ U' = \frac{1}{2} \frac{0.36}{4 \times 10^{-3}} = \frac{0.36}{8 \times 10^{-3}} = 45 \text{ joules} \]

8. **Calculate the loss of energy**:
\[ \text{Loss of energy} = U - U' = 50 - 45 = 5 \text{ joules} \]

9. **Calculate the percentage loss of energy**:
\[ \text{Percentage loss} = \left(\frac{\text{Loss of energy}}{U}\right) \times 100 \]
\[ \text{Percentage loss} = \left(\frac{5}{50}\right) \times 100 = 10\% \]

### Final Answer:
The percentage loss of energy of the system is **10%**.