A `2muF` capacitor that was initially uncharged is connected to a battery of EMF 100V and a resistance and the switch is closed. The heat generated in the resistance until the capacitor becomes fully charged is (in mJ):

Options

A. 10
B. 20
C. 100
D. 200

Explanation

To solve the problem of finding the heat generated in the resistance until the capacitor becomes fully charged, we can follow these steps:

### Step 1: Identify the given values
- Capacitance \( C = 2 \mu F = 2 \times 10^{-6} F \)
- EMF of the battery \( E = 100 V \)

### Step 2: Calculate the charge stored in the capacitor
The charge \( Q \) stored in a capacitor is given by the formula:
\[ Q = C \times E \]
Substituting the values:
\[ Q = 2 \times 10^{-6} F \times 100 V = 2 \times 10^{-4} C \]

### Step 3: Calculate the work done by the battery
The work done \( W \) by the battery to move this charge across the potential difference is given by:
\[ W = Q \times E \]
Substituting the values:
\[ W = 2 \times 10^{-4} C \times 100 V = 2 \times 10^{-2} J = 0.02 J \]

### Step 4: Calculate the potential energy stored in the capacitor
The potential energy \( U \) stored in the capacitor is given by:
\[ U = \frac{1}{2} C E^2 \]
Substituting the values:
\[ U = \frac{1}{2} \times 2 \times 10^{-6} F \times (100 V)^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 10000 = 0.01 J \]

### Step 5: Calculate the heat generated in the resistance
The heat \( H \) generated in the resistance is the difference between the work done by the battery and the potential energy stored in the capacitor:
\[ H = W - U \]
Substituting the values:
\[ H = 0.02 J - 0.01 J = 0.01 J \]

### Step 6: Convert the heat generated to millijoules
Since \( 1 J = 1000 mJ \):
\[ H = 0.01 J = 10 mJ \]

### Final Answer
The heat generated in the resistance until the capacitor becomes fully charged is **10 mJ**.

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