A capacitor is used to store 24 watt hour of enegy at 1200 volt. What should be the capacitance of the capacitor?

Options

A. `120mF`
B. `120muF`
C. `24muF`
D. `24mF`

Explanation

To find the capacitance of a capacitor that stores 24 watt-hours of energy at 1200 volts, we can follow these steps:

### Step 1: Convert watt-hours to joules
The energy stored in the capacitor is given in watt-hours. We need to convert this to joules since the standard unit of energy in physics is joules.

\[ \text{Energy (in joules)} = \text{Energy (in watt-hours)} \times 3600 \, \text{seconds} \]

Given that the energy is 24 watt-hours:

\[ \text{Energy} = 24 \, \text{Watt-hours} \times 3600 \, \text{seconds} = 86400 \, \text{Joules} \]

### Step 2: Use the formula for energy stored in a capacitor
The energy \( E \) stored in a capacitor is given by the formula:

\[ E = \frac{1}{2} C V^2 \]

Where:
- \( E \) is the energy in joules,
- \( C \) is the capacitance in farads,
- \( V \) is the voltage in volts.

### Step 3: Rearrange the formula to solve for capacitance
We need to rearrange the formula to solve for \( C \):

\[ C = \frac{2E}{V^2} \]

### Step 4: Substitute the known values
Now we can substitute the values we have into the equation. We have \( E = 86400 \, \text{J} \) and \( V = 1200 \, \text{V} \):

\[ C = \frac{2 \times 86400}{1200^2} \]

### Step 5: Calculate the capacitance
Now we can calculate \( C \):

\[ C = \frac{172800}{1440000} \]

\[ C = 0.12 \, \text{F} = 120 \, \text{mF} \]

### Conclusion
The capacitance of the capacitor should be \( 120 \, \text{mF} \) (millifarads).

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