Answer :
To determine how much energy can be stored in a capacitor, we use the formula:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where:
- [tex]\( E \)[/tex] is the energy stored in the capacitor (in joules),
- [tex]\( C \)[/tex] is the capacitance (in farads),
- [tex]\( V \)[/tex] is the voltage across the capacitor (in volts).
Let's solve the problem step-by-step:
1. Convert the Capacitance:
- The capacitance given is 25 microfarads ([tex]\( \mu F \)[/tex]).
- To convert microfarads to farads, use the conversion factor [tex]\( 1 F = 10^6 \mu F \)[/tex].
- So, [tex]\( 25 \mu F = 25 \times 10^{-6} F = 0.000025 F \)[/tex].
2. Input the Voltage:
- The voltage applied to the capacitor is 12 volts.
3. Calculate the Energy:
- Substitute the values into the energy formula:
[tex]\[
E = \frac{1}{2} \times 0.000025 \, F \times (12 \, V)^2
\][/tex]
- Simplifying the equation:
[tex]\[
E = \frac{1}{2} \times 0.000025 \times 144
\][/tex]
- Calculate the product:
[tex]\[
E = 0.0000125 \times 144 = 0.0018 \, \text{J}
\][/tex]
The energy stored in the capacitor is [tex]\( 0.0018 \, \text{J} \)[/tex]. Therefore, the correct answer is [tex]\( 0.0018 \, \text{J} \)[/tex].
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where:
- [tex]\( E \)[/tex] is the energy stored in the capacitor (in joules),
- [tex]\( C \)[/tex] is the capacitance (in farads),
- [tex]\( V \)[/tex] is the voltage across the capacitor (in volts).
Let's solve the problem step-by-step:
1. Convert the Capacitance:
- The capacitance given is 25 microfarads ([tex]\( \mu F \)[/tex]).
- To convert microfarads to farads, use the conversion factor [tex]\( 1 F = 10^6 \mu F \)[/tex].
- So, [tex]\( 25 \mu F = 25 \times 10^{-6} F = 0.000025 F \)[/tex].
2. Input the Voltage:
- The voltage applied to the capacitor is 12 volts.
3. Calculate the Energy:
- Substitute the values into the energy formula:
[tex]\[
E = \frac{1}{2} \times 0.000025 \, F \times (12 \, V)^2
\][/tex]
- Simplifying the equation:
[tex]\[
E = \frac{1}{2} \times 0.000025 \times 144
\][/tex]
- Calculate the product:
[tex]\[
E = 0.0000125 \times 144 = 0.0018 \, \text{J}
\][/tex]
The energy stored in the capacitor is [tex]\( 0.0018 \, \text{J} \)[/tex]. Therefore, the correct answer is [tex]\( 0.0018 \, \text{J} \)[/tex].
