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, we need to find the heat generated in the resistance until the capacitor becomes fully charged. Here’s a step-by-step solution:

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

### Step 2: Calculate the charge stored in the capacitor when fully charged
The charge \( Q \) stored in a capacitor is given by the formula:
\[ Q = C \times V \]
Substituting the values:
\[ Q = (2 \times 10^{-6} \, F) \times (100 \, V) = 200 \times 10^{-6} \, C \]
So, the charge \( Q = 200 \, \mu C \).

### Step 3: Calculate the heat generated in the resistance
The heat \( H \) generated in the resistance when the capacitor is fully charged can be calculated using the formula:
\[ H = \frac{1}{2} C V^2 \]
Substituting the values:
\[ H = \frac{1}{2} \times (2 \times 10^{-6} \, F) \times (100 \, V)^2 \]
Calculating \( V^2 \):
\[ (100 \, V)^2 = 10000 \, V^2 \]
Now substituting back:
\[ H = \frac{1}{2} \times (2 \times 10^{-6}) \times 10000 \]
\[ H = \frac{1}{2} \times 2 \times 10^{-2} \, J = 10^{-2} \, J \]

### Step 4: Convert the heat from Joules to millijoules
Since \( 1 \, J = 1000 \, mJ \):
\[ H = 10^{-2} \, J = 10 \, mJ \]

### Final Answer
Thus, the heat generated in the resistance until the capacitor becomes fully charged is:
\[ \boxed{10 \, mJ} \]

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