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A capacitor of capacitance C1 is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, is
- a)1
- b)1/2
- c)1/√2
- d)1/3
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A capacitor of capacitance C1is charged by connecting it to a battery....
Explanation:
Given: Capacitance of first capacitor, C1
Capacitance of second capacitor, C2
Charge is distributed equally between the two capacitors after connection
Initial energy stored in the capacitors:
The initial energy stored in the first capacitor is given by:
\[ U_{1} = \frac{1}{2} C_{1} V^{2} \]
where V is the voltage across the first capacitor.
The initial energy stored in the second capacitor is zero as it is uncharged.
Therefore, the total initial energy stored in the capacitors is:
\[ U_{initial} = \frac{1}{2} C_{1} V^{2} \]
Final energy stored in the capacitors:
After connecting the two capacitors in parallel, the total capacitance becomes:
\[ C_{eq} = C_{1} + C_{2} \]
Since the charge is distributed equally between the two capacitors, the voltage across both capacitors will be the same. Therefore, the total final energy stored in the capacitors is:
\[ U_{final} = \frac{1}{2} C_{eq} V^{2} = \frac{1}{2} (C_{1} + C_{2}) V^{2} \]
Ratio of total final energy to total initial energy:
The ratio of total final energy to total initial energy is given by:
\[ \frac{U_{final}}{U_{initial}} = \frac{\frac{1}{2} (C_{1} + C_{2}) V^{2}}{\frac{1}{2} C_{1} V^{2}} = \frac{C_{1} + C_{2}}{C_{1}} \]
Given that the charge is distributed equally between the two capacitors, we have:
\[ \frac{C_{1} + C_{2}}{C_{1}} = 1 + \frac{C_{2}}{C_{1}} \]
Since the charge is distributed equally, \(\frac{C_{2}}{C_{1}} = 1\), so the ratio simplifies to:
\[ 1 + 1 = 2 \]
Therefore, the ratio of the total final energy to total initial energy is 2, which is equivalent to 1/2.
Hence, the correct answer is option B - 1/2.
Given: Capacitance of first capacitor, C1
Capacitance of second capacitor, C2
Charge is distributed equally between the two capacitors after connection
Initial energy stored in the capacitors:
The initial energy stored in the first capacitor is given by:
\[ U_{1} = \frac{1}{2} C_{1} V^{2} \]
where V is the voltage across the first capacitor.
The initial energy stored in the second capacitor is zero as it is uncharged.
Therefore, the total initial energy stored in the capacitors is:
\[ U_{initial} = \frac{1}{2} C_{1} V^{2} \]
Final energy stored in the capacitors:
After connecting the two capacitors in parallel, the total capacitance becomes:
\[ C_{eq} = C_{1} + C_{2} \]
Since the charge is distributed equally between the two capacitors, the voltage across both capacitors will be the same. Therefore, the total final energy stored in the capacitors is:
\[ U_{final} = \frac{1}{2} C_{eq} V^{2} = \frac{1}{2} (C_{1} + C_{2}) V^{2} \]
Ratio of total final energy to total initial energy:
The ratio of total final energy to total initial energy is given by:
\[ \frac{U_{final}}{U_{initial}} = \frac{\frac{1}{2} (C_{1} + C_{2}) V^{2}}{\frac{1}{2} C_{1} V^{2}} = \frac{C_{1} + C_{2}}{C_{1}} \]
Given that the charge is distributed equally between the two capacitors, we have:
\[ \frac{C_{1} + C_{2}}{C_{1}} = 1 + \frac{C_{2}}{C_{1}} \]
Since the charge is distributed equally, \(\frac{C_{2}}{C_{1}} = 1\), so the ratio simplifies to:
\[ 1 + 1 = 2 \]
Therefore, the ratio of the total final energy to total initial energy is 2, which is equivalent to 1/2.
Hence, the correct answer is option B - 1/2.
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Community Answer
A capacitor of capacitance C1is charged by connecting it to a battery....
If Q is the initial charge on capacitor C1, then the initial energy is given by Ui = Q2/2C1.
As the two capacitors are connected together and the charge is distributed equally, the charge on each capacitor is Q/2.
Since the potential difference (in a parallel connection) across the two capacitors is also the same, it follows that their capacitance are equal (since C = Q/V).
Thus, C1 = C2 = C (say)
Also, Q1 = Q2 = Q/2
Therefore, the final energy stored in the two capacitors is
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A capacitor of capacitance C1is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, isa)1b)1/2c)1/√2d)1/3Correct answer is option 'B'. Can you explain this answer?
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A capacitor of capacitance C1is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, isa)1b)1/2c)1/√2d)1/3Correct answer is option 'B'. 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 of capacitance C1is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, isa)1b)1/2c)1/√2d)1/3Correct answer is option 'B'. 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 of capacitance C1is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, isa)1b)1/2c)1/√2d)1/3Correct answer is option 'B'. Can you explain this answer?.
A capacitor of capacitance C1is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, isa)1b)1/2c)1/√2d)1/3Correct answer is option 'B'. 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 of capacitance C1is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, isa)1b)1/2c)1/√2d)1/3Correct answer is option 'B'. 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 of capacitance C1is charged by connecting it to a battery. The battery is then removed and this capacitor is connected to a second uncharged capacitor of capacitance C2. If the charge gets distributed equally on the two capacitors, then the ratio of the total energy stored in the capacitors, after connection to the total energy stored in them before connection, isa)1b)1/2c)1/√2d)1/3Correct answer is option 'B'. Can you explain this answer?.
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