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A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is
- a)6 nJ
- b)4 nJ
- c)8 nJ
- d)2 nJ
Correct answer is option 'A'. Can you explain this answer?
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A 60 pF capacitor is fully charged by a 20 V supply. It is then disco...
To solve this question, we need to understand the concept of energy stored in a capacitor and how it changes when capacitors are connected in parallel.
1. Energy Stored in a Capacitor:
The energy stored in a capacitor can be calculated using the formula:
E = 0.5 * C * V^2
where E is the energy stored, C is the capacitance, and V is the voltage across the capacitor.
2. Initial Energy of the Capacitor:
Given that the capacitance of the fully charged capacitor is 60 pF and the voltage is 20 V, we can calculate the initial energy stored in the capacitor:
E1 = 0.5 * (60 * 10^-12) * (20^2)
E1 = 0.5 * 72 * 10^-9
E1 = 36 * 10^-9
E1 = 36 nJ
3. Connecting Capacitors in Parallel:
When the fully charged capacitor is disconnected from the supply and connected to an uncharged capacitor of the same capacitance in parallel, the total capacitance becomes:
C_total = C1 + C2
C_total = 60 pF + 60 pF
C_total = 120 pF
4. Final Energy of the Capacitor:
Since the capacitors are connected in parallel, the voltage across both capacitors will be the same. Therefore, the voltage across each capacitor will be half of the initial voltage, which is 20 V / 2 = 10 V.
Using the formula for energy stored in a capacitor, we can calculate the final energy stored in each capacitor:
E2 = 0.5 * (60 * 10^-12) * (10^2)
E2 = 0.5 * 6 * 10^-9
E2 = 3 * 10^-9
E2 = 3 nJ
5. Energy Lost in the Process:
To find the energy lost in the process, we subtract the final energy from the initial energy:
Energy lost = E1 - E2
Energy lost = 36 nJ - 3 nJ
Energy lost = 33 nJ
However, the correct answer given is option A, which states that the energy lost is 6 nJ. This suggests that there might be an error in either the given answer or the question itself. Based on the calculations above, the energy lost should be 33 nJ, not 6 nJ.
1. Energy Stored in a Capacitor:
The energy stored in a capacitor can be calculated using the formula:
E = 0.5 * C * V^2
where E is the energy stored, C is the capacitance, and V is the voltage across the capacitor.
2. Initial Energy of the Capacitor:
Given that the capacitance of the fully charged capacitor is 60 pF and the voltage is 20 V, we can calculate the initial energy stored in the capacitor:
E1 = 0.5 * (60 * 10^-12) * (20^2)
E1 = 0.5 * 72 * 10^-9
E1 = 36 * 10^-9
E1 = 36 nJ
3. Connecting Capacitors in Parallel:
When the fully charged capacitor is disconnected from the supply and connected to an uncharged capacitor of the same capacitance in parallel, the total capacitance becomes:
C_total = C1 + C2
C_total = 60 pF + 60 pF
C_total = 120 pF
4. Final Energy of the Capacitor:
Since the capacitors are connected in parallel, the voltage across both capacitors will be the same. Therefore, the voltage across each capacitor will be half of the initial voltage, which is 20 V / 2 = 10 V.
Using the formula for energy stored in a capacitor, we can calculate the final energy stored in each capacitor:
E2 = 0.5 * (60 * 10^-12) * (10^2)
E2 = 0.5 * 6 * 10^-9
E2 = 3 * 10^-9
E2 = 3 nJ
5. Energy Lost in the Process:
To find the energy lost in the process, we subtract the final energy from the initial energy:
Energy lost = E1 - E2
Energy lost = 36 nJ - 3 nJ
Energy lost = 33 nJ
However, the correct answer given is option A, which states that the energy lost is 6 nJ. This suggests that there might be an error in either the given answer or the question itself. Based on the calculations above, the energy lost should be 33 nJ, not 6 nJ.
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A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them isa)6 nJb)4 nJc)8 nJd)2 nJCorrect answer is option 'A'. Can you explain this answer?
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A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them isa)6 nJb)4 nJc)8 nJd)2 nJCorrect answer is option 'A'. 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 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them isa)6 nJb)4 nJc)8 nJd)2 nJCorrect answer is option 'A'. 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 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them isa)6 nJb)4 nJc)8 nJd)2 nJCorrect answer is option 'A'. Can you explain this answer?.
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