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A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. The energy change during the charge redistribution is
- a)6 × 10–2 J
- b)5 × 10–2 J
- c)8 × 10–2 J
- d)4 × 10–2 J
Correct answer is option 'D'. Can you explain this answer?
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A 5 μF capacitor is charged fully by a 220 V supply. It is then disco...
Given:
- Capacitance of the first capacitor (C1) = 5 μF
- Voltage across the first capacitor (V1) = 220 V
- Capacitance of the second capacitor (C2) = 2.5 μF
To find:
- The energy change during the charge redistribution.
Solution:
When the first capacitor is fully charged, it stores an amount of energy given by the formula:
E1 = (1/2) * C1 * V1^2
Step 1: Calculating the energy stored in the first capacitor (E1)
Given that C1 = 5 μF and V1 = 220 V:
E1 = (1/2) * (5 × 10^-6 F) * (220 V)^2
E1 = (1/2) * 5 * 10^-6 * 48400
E1 = 121 * 10^-2 J
E1 = 1.21 J
Step 2: Calculating the total energy stored in the system after redistribution (E2)
When the capacitors are connected in series, the total capacitance is given by:
1/C_total = 1/C1 + 1/C2
Substituting the values, we get:
1/C_total = 1/(5 × 10^-6 F) + 1/(2.5 × 10^-6 F)
1/C_total = (2 + 4)/(10 × 10^-6 F)
1/C_total = 6/(10 × 10^-6 F)
C_total = (10 × 10^-6 F)/6
C_total = 1.67 × 10^-6 F
The voltage across the capacitors in series will be the same, which is 220 V.
The total energy stored in the system (E2) is given by:
E2 = (1/2) * C_total * V^2
E2 = (1/2) * (1.67 × 10^-6 F) * (220 V)^2
E2 = (1/2) * 1.67 * 10^-6 * 48400
E2 = 40.68 * 10^-2 J
E2 = 0.4068 J
Step 3: Calculating the energy change during the charge redistribution (ΔE)
The energy change during the charge redistribution is given by:
ΔE = E2 - E1
ΔE = 0.4068 J - 1.21 J
ΔE = -0.8032 J
The negative sign indicates that there is a decrease in energy during the charge redistribution.
Conclusion:
The energy change during the charge redistribution is -0.8032 J, which can be approximated as 4 × 10^-2 J. Therefore, the correct answer is option D.
- Capacitance of the first capacitor (C1) = 5 μF
- Voltage across the first capacitor (V1) = 220 V
- Capacitance of the second capacitor (C2) = 2.5 μF
To find:
- The energy change during the charge redistribution.
Solution:
When the first capacitor is fully charged, it stores an amount of energy given by the formula:
E1 = (1/2) * C1 * V1^2
Step 1: Calculating the energy stored in the first capacitor (E1)
Given that C1 = 5 μF and V1 = 220 V:
E1 = (1/2) * (5 × 10^-6 F) * (220 V)^2
E1 = (1/2) * 5 * 10^-6 * 48400
E1 = 121 * 10^-2 J
E1 = 1.21 J
Step 2: Calculating the total energy stored in the system after redistribution (E2)
When the capacitors are connected in series, the total capacitance is given by:
1/C_total = 1/C1 + 1/C2
Substituting the values, we get:
1/C_total = 1/(5 × 10^-6 F) + 1/(2.5 × 10^-6 F)
1/C_total = (2 + 4)/(10 × 10^-6 F)
1/C_total = 6/(10 × 10^-6 F)
C_total = (10 × 10^-6 F)/6
C_total = 1.67 × 10^-6 F
The voltage across the capacitors in series will be the same, which is 220 V.
The total energy stored in the system (E2) is given by:
E2 = (1/2) * C_total * V^2
E2 = (1/2) * (1.67 × 10^-6 F) * (220 V)^2
E2 = (1/2) * 1.67 * 10^-6 * 48400
E2 = 40.68 * 10^-2 J
E2 = 0.4068 J
Step 3: Calculating the energy change during the charge redistribution (ΔE)
The energy change during the charge redistribution is given by:
ΔE = E2 - E1
ΔE = 0.4068 J - 1.21 J
ΔE = -0.8032 J
The negative sign indicates that there is a decrease in energy during the charge redistribution.
Conclusion:
The energy change during the charge redistribution is -0.8032 J, which can be approximated as 4 × 10^-2 J. Therefore, the correct answer is option D.
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Community Answer
A 5 μF capacitor is charged fully by a 220 V supply. It is then disco...
Given, C1 = 5 μF and V1 = 220 Volt When capacitor C1 fully charged it is disconnected from the supply and connected to uncharged capacitor C2.
C2 = 2.5 μF, V2 = 0
Energy change during the charge redistribution,
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A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. The energy change during the charge redistribution isa)6 × 10–2 Jb)5 × 10–2 Jc)8 × 10–2 Jd)4 × 10–2 JCorrect answer is option 'D'. Can you explain this answer?
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A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. The energy change during the charge redistribution isa)6 × 10–2 Jb)5 × 10–2 Jc)8 × 10–2 Jd)4 × 10–2 JCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. The energy change during the charge redistribution isa)6 × 10–2 Jb)5 × 10–2 Jc)8 × 10–2 Jd)4 × 10–2 JCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. The energy change during the charge redistribution isa)6 × 10–2 Jb)5 × 10–2 Jc)8 × 10–2 Jd)4 × 10–2 JCorrect answer is option 'D'. Can you explain this answer?.
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