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A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearly
- a)7 × 10_4 J
- b)3.5 × 10_4 J
- c)14 × 10_4 J
- d)7 × 10_3 J
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A conductor of capacitance 0.5mF has been charged to 100volts. It is n...
Given,
C1=0.5 mf
C2=0.2mf
V=100v
So,
U=(1/2) CTV2
CT=C1C2/XC1+C2=(0.5x0.2/0.5+0.2)mf
CT=0.14mf
U=(1/2)(0.14mf)(100V)2
=(10.07x1-6x104)J
=(7x10-2x10-6x104)J
=7x10-4J
C1=0.5 mf
C2=0.2mf
V=100v
So,
U=(1/2) CTV2
CT=C1C2/XC1+C2=(0.5x0.2/0.5+0.2)mf
CT=0.14mf
U=(1/2)(0.14mf)(100V)2
=(10.07x1-6x104)J
=(7x10-2x10-6x104)J
=7x10-4J
Most Upvoted Answer
A conductor of capacitance 0.5mF has been charged to 100volts. It is n...
The initial potential energy of the charged conductor is given by:
U = (1/2)CV^2
where C is the capacitance and V is the voltage.
Substituting the values given:
U = (1/2)(0.5mF)(100V)^2
U = 2500mJ
When the two conductors are connected, charge will flow from the first conductor to the second until the voltage is equalized between them. The final voltage can be calculated using the principle of conservation of charge:
Q1 = Q2
C1V1 = C2V2
Substituting the values given:
(0.5mF)(100V) = (0.2mF)V2
V2 = 250V
The final potential energy can then be calculated using the same formula:
Uf = (1/2)C2V2^2
Uf = (1/2)(0.2mF)(250V)^2
Uf = 7812.5mJ
The loss in potential energy is therefore:
U - Uf = 2500mJ - 7812.5mJ
U - Uf = -5312.5mJ
Rounding to the nearest integer, the loss in potential energy is approximately 5313mJ or 5.3J. Answer: (d) 5.
U = (1/2)CV^2
where C is the capacitance and V is the voltage.
Substituting the values given:
U = (1/2)(0.5mF)(100V)^2
U = 2500mJ
When the two conductors are connected, charge will flow from the first conductor to the second until the voltage is equalized between them. The final voltage can be calculated using the principle of conservation of charge:
Q1 = Q2
C1V1 = C2V2
Substituting the values given:
(0.5mF)(100V) = (0.2mF)V2
V2 = 250V
The final potential energy can then be calculated using the same formula:
Uf = (1/2)C2V2^2
Uf = (1/2)(0.2mF)(250V)^2
Uf = 7812.5mJ
The loss in potential energy is therefore:
U - Uf = 2500mJ - 7812.5mJ
U - Uf = -5312.5mJ
Rounding to the nearest integer, the loss in potential energy is approximately 5313mJ or 5.3J. Answer: (d) 5.
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A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearlya)7 × 10_4 Jb)3.5 × 10_4 Jc)14 × 10_4 Jd)7 × 10_3 JCorrect answer is option 'A'. Can you explain this answer?
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A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearlya)7 × 10_4 Jb)3.5 × 10_4 Jc)14 × 10_4 Jd)7 × 10_3 JCorrect answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearlya)7 × 10_4 Jb)3.5 × 10_4 Jc)14 × 10_4 Jd)7 × 10_3 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearlya)7 × 10_4 Jb)3.5 × 10_4 Jc)14 × 10_4 Jd)7 × 10_3 JCorrect answer is option 'A'. Can you explain this answer?.
A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearlya)7 × 10_4 Jb)3.5 × 10_4 Jc)14 × 10_4 Jd)7 × 10_3 JCorrect answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearlya)7 × 10_4 Jb)3.5 × 10_4 Jc)14 × 10_4 Jd)7 × 10_3 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearlya)7 × 10_4 Jb)3.5 × 10_4 Jc)14 × 10_4 Jd)7 × 10_3 JCorrect answer is option 'A'. Can you explain this answer?.
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