To determine the amount of energy stored in a 120μF capacitor charged to a potential difference of 240V, we can use the formula for the energy stored in a capacitor:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where:
- [tex]\( E \)[/tex] is the energy stored in the capacitor,
- [tex]\( C \)[/tex] is the capacitance of the capacitor,
- [tex]\( V \)[/tex] is the potential difference across the capacitor.
Let's break down the calculation step by step:
1. Identify the given values:
- Capacitance [tex]\( C \)[/tex]: [tex]\( 120\mu F \)[/tex] or [tex]\( 120 \times 10^{-6} \, F \)[/tex].
- Voltage [tex]\( V \)[/tex]: [tex]\( 240 \, V \)[/tex].
2. Substitute the given values into the formula:
[tex]\[ E = \frac{1}{2} \times (120 \times 10^{-6}) \times (240)^2 \][/tex]
3. Simplify and solve:
[tex]\[ E = \frac{1}{2} \times 120 \times 10^{-6} \times 57600 \][/tex]
4. Calculate the numerical result step-by-step:
- Calculate the square of the voltage:
[tex]\[ (240)^2 = 57600 \][/tex]
- Multiply the capacitance by the squared voltage:
[tex]\[ 120 \times 10^{-6} \times 57600 = 6.912 \][/tex]
- Finally, multiply by [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \frac{1}{2} \times 6.912 = 3.456 \][/tex]
Therefore, the energy stored in the capacitor is [tex]\( 3.456 \)[/tex] joules.
