Answer :
To find out how much energy can be stored in the capacitor, 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 joules,
- [tex]\( C \)[/tex] is the capacitance in farads,
- [tex]\( V \)[/tex] is the voltage in volts.
Let's go through the steps:
1. Convert Capacitance to Farads:
The capacitance is given as [tex]\( 25 \, \mu F \)[/tex]. To convert microfarads ([tex]\( \mu F \)[/tex]) to farads (F), use the conversion factor [tex]\( 1 \, F = 10^6 \, \mu F \)[/tex].
[tex]\[
25 \, \mu F = 25 \times 10^{-6} \, F = 2.5 \times 10^{-5} \, F
\][/tex]
2. Calculate Energy Stored:
Now plug the capacitance in farads and the voltage (12 volts) into the energy formula:
[tex]\[
E = \frac{1}{2} \times 2.5 \times 10^{-5} \, F \times (12 \, V)^2
\][/tex]
Calculate the voltage squared:
[tex]\[
12^2 = 144
\][/tex]
Then calculate the energy stored:
[tex]\[
E = \frac{1}{2} \times 2.5 \times 10^{-5} \times 144
\][/tex]
[tex]\[
E = 0.5 \times 2.5 \times 10^{-5} \times 144
\][/tex]
[tex]\[
E = 0.0018 \, \text{joules}
\][/tex]
So, the amount of energy that can be stored in the capacitor is approximately 0.0018 joules.
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
Where:
- [tex]\( E \)[/tex] is the energy stored in joules,
- [tex]\( C \)[/tex] is the capacitance in farads,
- [tex]\( V \)[/tex] is the voltage in volts.
Let's go through the steps:
1. Convert Capacitance to Farads:
The capacitance is given as [tex]\( 25 \, \mu F \)[/tex]. To convert microfarads ([tex]\( \mu F \)[/tex]) to farads (F), use the conversion factor [tex]\( 1 \, F = 10^6 \, \mu F \)[/tex].
[tex]\[
25 \, \mu F = 25 \times 10^{-6} \, F = 2.5 \times 10^{-5} \, F
\][/tex]
2. Calculate Energy Stored:
Now plug the capacitance in farads and the voltage (12 volts) into the energy formula:
[tex]\[
E = \frac{1}{2} \times 2.5 \times 10^{-5} \, F \times (12 \, V)^2
\][/tex]
Calculate the voltage squared:
[tex]\[
12^2 = 144
\][/tex]
Then calculate the energy stored:
[tex]\[
E = \frac{1}{2} \times 2.5 \times 10^{-5} \times 144
\][/tex]
[tex]\[
E = 0.5 \times 2.5 \times 10^{-5} \times 144
\][/tex]
[tex]\[
E = 0.0018 \, \text{joules}
\][/tex]
So, the amount of energy that can be stored in the capacitor is approximately 0.0018 joules.
