A capacitor of 100 µF is charged to 200 V and then it is discharged through a 2 ohm resistance. The total heat generated in the resistance is

Question

Accepted Answer

To determine the total heat generated in the resistance when a capacitor is discharged, we can use the formula for the energy stored in a capacitor and the principle of energy conservation. The energy stored in a capacitor is given by:

$$ E = \frac{1}{2} C V^2 $$

where $ C $ is the capacitance and $ V $ is the voltage across the capacitor.

Given:
- Capacitance, $ C = 100 \, \mu F = 100 \times 10^{-6} \, F $
- Voltage, $ V = 200 \, V $

Substituting these values into the formula, we get:

$$ E = \frac{1}{2} \times 100 \times 10^{-6} \times (200)^2 $$

$$ E = \frac{1}{2} \times 100 \times 10^{-6} \times 40000 $$

$$ E = \frac{1}{2} \times 4 $$

$$ E = 2 \, J $$

This energy is completely dissipated as heat in the resistance when the capacitor is discharged. Therefore, the total heat generated in the resistance is 2 J.

Now, let's explain why each of the other options is incorrect:

1. **Option 1 (20 J):** This value is incorrect because it overestimates the energy stored in the capacitor. The calculation clearly shows that the energy stored is only 2 J, not 20 J.

2. **Option 2 (200 J):** This is incorrect as it significantly overestimates the energy. The energy stored in the capacitor is calculated to be 2 J, which is far less than 200 J.

3. **Option 3 (10 J):** This option is incorrect because it also overestimates the energy stored in the capacitor. The correct calculation shows that the energy is 2 J, not 10 J.

In conclusion, the correct answer is Option 4 (2 J), as it accurately reflects the energy stored in the capacitor and the total heat generated in the resistance when the capacitor is discharged.