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A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system
- a)Increases by a factor of 2
- b)Increases by a factor of 4
- c)Decreases by a factor of 2
- d)Remains the same
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A capacitor is charged by a battery. The battery is removed and anothe...
First, let's understand the concept of capacitors in parallel.
Capacitors in Parallel:
When capacitors are connected in parallel, the total capacitance of the system is the sum of the individual capacitances. In this case, when the second identical uncharged capacitor is connected in parallel to the charged capacitor, the total capacitance of the system doubles.
Effect on Electrostatic Energy:
The electrostatic energy stored in a capacitor is given by the formula: U = 1/2 * CV^2, where U is the energy, C is the capacitance, and V is the voltage.
When the second capacitor is connected in parallel, the total capacitance becomes 2C. Since the voltage across both capacitors is the same (as they are connected in parallel), the energy stored in each capacitor remains the same.
Calculation:
Initially, the energy stored in the single charged capacitor is U1 = 1/2 * C * V^2.
When the second capacitor is connected in parallel, the total energy stored in the system becomes U2 = 1/2 * 2C * V^2 = 2 * (1/2 * C * V^2) = 2U1.
Conclusion:
Therefore, the total electrostatic energy of the resulting system decreases by a factor of 2 when the second identical uncharged capacitor is connected in parallel.
Capacitors in Parallel:
When capacitors are connected in parallel, the total capacitance of the system is the sum of the individual capacitances. In this case, when the second identical uncharged capacitor is connected in parallel to the charged capacitor, the total capacitance of the system doubles.
Effect on Electrostatic Energy:
The electrostatic energy stored in a capacitor is given by the formula: U = 1/2 * CV^2, where U is the energy, C is the capacitance, and V is the voltage.
When the second capacitor is connected in parallel, the total capacitance becomes 2C. Since the voltage across both capacitors is the same (as they are connected in parallel), the energy stored in each capacitor remains the same.
Calculation:
Initially, the energy stored in the single charged capacitor is U1 = 1/2 * C * V^2.
When the second capacitor is connected in parallel, the total energy stored in the system becomes U2 = 1/2 * 2C * V^2 = 2 * (1/2 * C * V^2) = 2U1.
Conclusion:
Therefore, the total electrostatic energy of the resulting system decreases by a factor of 2 when the second identical uncharged capacitor is connected in parallel.
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Community Answer
A capacitor is charged by a battery. The battery is removed and anothe...
Concept:
Capacitor is defined as a component in an electrical circuit which has the ability to store energy in the form of an electrical charge producing a potential difference across its plates. It is like rechargeable battery.
Charge on capacitor is given by:
q = CV
where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.
Energy stored in capacitor is calculated using the formula, U:
where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.
Calculation:
Charge on capacitor
where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.
Calculation:
Charge on capacitor
q = CV
When it is connected with another uncharged capacitor
Initial energy
Loss of energy = Ui - Uf
i.e. decreases by a factor (2)
Initial energy
Loss of energy = Ui - Uf
i.e. decreases by a factor (2)Attention JEE Students!
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A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer?
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A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer?.
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer?.
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A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting systema)Increases by a factor of 2b)Increases by a factor of 4c)Decreases by a factor of 2d)Remains the sameCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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