IDNLearn.com offers a unique blend of expert answers and community insights. Discover prompt and accurate answers from our experts, ensuring you get the information you need quickly.
Sagot :
Sure, let's solve the given problem step-by-step.
Question:
The energy stored in a parallel plate capacitor is 3 Joules. What is the capacitance of the capacitor if the potential difference between the plates is 220 Volts?
Given data:
1. Energy stored (E) = 3 Joules
2. Potential difference (V) = 220 Volts
To find: Capacitance (C) in microfarads ([tex]\( \mu F \)[/tex])
Step-by-step solution:
1. Write down the formula for the energy stored in a capacitor:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
2. Rearrange the formula to solve for capacitance (C):
[tex]\[ C = \frac{2E}{V^2} \][/tex]
3. Substitute the given values into the formula:
[tex]\[ C = \frac{2 \times 3}{220^2} \][/tex]
4. Calculate the capacitance in Farads:
[tex]\[ C = \frac{6}{48400} \approx 0.00012396694214876034 \, \text{Farads} \][/tex]
5. Convert the capacitance from Farads to microfarads:
[tex]\[ 1 \, \text{Farad} = 10^6 \, \mu \text{Farads} \][/tex]
[tex]\[ C \approx 0.00012396694214876034 \, \text{Farads} \times 10^6 \approx 123.96694214876034 \, \mu \text{Farads} \][/tex]
Result:
The capacitance of the capacitor is approximately [tex]\( 124 \, \mu F \)[/tex].
So, the correct answer is:
A. [tex]\( 124 \, \mu \text{F} \)[/tex]
Question:
The energy stored in a parallel plate capacitor is 3 Joules. What is the capacitance of the capacitor if the potential difference between the plates is 220 Volts?
Given data:
1. Energy stored (E) = 3 Joules
2. Potential difference (V) = 220 Volts
To find: Capacitance (C) in microfarads ([tex]\( \mu F \)[/tex])
Step-by-step solution:
1. Write down the formula for the energy stored in a capacitor:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
2. Rearrange the formula to solve for capacitance (C):
[tex]\[ C = \frac{2E}{V^2} \][/tex]
3. Substitute the given values into the formula:
[tex]\[ C = \frac{2 \times 3}{220^2} \][/tex]
4. Calculate the capacitance in Farads:
[tex]\[ C = \frac{6}{48400} \approx 0.00012396694214876034 \, \text{Farads} \][/tex]
5. Convert the capacitance from Farads to microfarads:
[tex]\[ 1 \, \text{Farad} = 10^6 \, \mu \text{Farads} \][/tex]
[tex]\[ C \approx 0.00012396694214876034 \, \text{Farads} \times 10^6 \approx 123.96694214876034 \, \mu \text{Farads} \][/tex]
Result:
The capacitance of the capacitor is approximately [tex]\( 124 \, \mu F \)[/tex].
So, the correct answer is:
A. [tex]\( 124 \, \mu \text{F} \)[/tex]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.