A `2muF` capacitor is charged to a potential = 10 V. Another `4muF` capacitor is charged to a potential = 20V. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. What heat is evolved in the circuit?

Options

A. ` 300 mu J `
B. ` 600 mu J`
C. ` 900 mu J`
D. ` 450 mu J`

Answer

To solve the problem of heat evolved in the circuit when two capacitors are connected, we will follow these steps:

### Step 1: Calculate the charge on each capacitor
1. **For the 2 µF capacitor charged to 10 V:**
\[ Q_1 = C_1 \times V_1 = 2 \, \mu F \times 10 \, V = 20 \, \mu C \]

2. **For the 4 µF capacitor charged to 20 V:**
\[ Q_2 = C_2 \times V_2 = 4 \, \mu F \times 20 \, V = 80 \, \mu C \]

### Step 2: Determine the equivalent charge when connected in a loop
When the capacitors are connected in a loop with the positive plate of one connected to the negative plate of the other, the effective charge \( Q_f \) can be calculated as:
\[ Q_f = Q_2 - Q_1 = 80 \, \mu C - 20 \, \mu C = 60 \, \mu C \]

### Step 3: Calculate the equivalent capacitance
Since the capacitors are connected in series, the equivalent capacitance \( C_f \) is given by:
\[ \frac{1}{C_f} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_f} = \frac{1}{2 \, \mu F} + \frac{1}{4 \, \mu F} = \frac{2 + 1}{4} = \frac{3}{4} \Rightarrow C_f = \frac{4}{3} \, \mu F \approx 1.33 \, \mu F \]

### Step 4: Calculate the initial energy stored in the capacitors
1. **Energy in the 2 µF capacitor:**
\[ E_1 = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 2 \, \mu F \times (10 \, V)^2 = 100 \, \mu J \]

2. **Energy in the 4 µF capacitor:**
\[ E_2 = \frac{1}{2} C_2 V_2^2 = \frac{1}{2} \times 4 \, \mu F \times (20 \, V)^2 = 800 \, \mu J \]

### Step 5: Calculate the total initial energy
\[ E_{initial} = E_1 + E_2 = 100 \, \mu J + 800 \, \mu J = 900 \, \mu J \]

### Step 6: Calculate the final energy stored in the equivalent capacitor
\[ E_{final} = \frac{1}{2} C_f Q_f^2 = \frac{1}{2} \times \frac{4}{3} \, \mu F \times (60 \, \mu C)^2 \]
Calculating this:
\[ E_{final} = \frac{1}{2} \times \frac{4}{3} \times 10^{-6} \, F \times 3600 \times 10^{-12} \, C^2 = \frac{4 \times 3600}{6} \times 10^{-18} = 2400 \times 10^{-18} = 2.4 \, \mu J \]

### Step 7: Calculate the heat evolved in the circuit
The heat evolved \( Q \) is given by the difference in energy:
\[ Q = E_{initial} - E_{final} = 900 \, \mu J - 2.4 \, \mu J = 897.6 \, \mu J \]

### Final Answer
The heat evolved in the circuit is approximately:
\[ \boxed{897.6 \, \mu J} \]