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Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential is
- a)300 volt
- b)500 volt
- c)425 volt
- d)400 volt
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to...
First find the charge in 1st capacitor
Q1= C1xV1
=0.003*300
=0.9 C
Similarly, find Q2
i.e. Q2=2.5 C
After connecting the capacitors in parallel you get a net capacitance C=8 mF (c1+c2).
Thus, final potential is V=Q/C
Q=Q1+Q2=3.4 C
V=3.4/0.008
= 425 V
Q1= C1xV1
=0.003*300
=0.9 C
Similarly, find Q2
i.e. Q2=2.5 C
After connecting the capacitors in parallel you get a net capacitance C=8 mF (c1+c2).
Thus, final potential is V=Q/C
Q=Q1+Q2=3.4 C
V=3.4/0.008
= 425 V
Most Upvoted Answer
Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to...
Solution:
Given, capacitance of first conductor, C1 = 3.0mF
Capacitance of second conductor, C2 = 5.0mF
Potential of first conductor, V1 = 300V
Potential of second conductor, V2 = 500V
When the two conductors are connected, the charges will redistribute until they reach a common potential.
Let the final potential be V.
Charge on the first conductor, Q1 = C1V1 = 3.0mF x 300V = 900mC
Charge on the second conductor, Q2 = C2V2 = 5.0mF x 500V = 2500mC
Total charge before connection, Q = Q1 + Q2 = 900mC + 2500mC = 3400mC
Total capacitance after connection,
1/C = 1/C1 + 1/C2 = (C1 + C2) / (C1C2) = 8 / 15mF
C = 15/8mF
Using the formula, Q = CV
V = Q/C = (3400 x 10^-6) / (15/8 x 10^-3) = 680V
Therefore, the final potential is 680V.
But this is not one of the options given in the question.
To find the correct answer, we need to use the principle of conservation of charge.
The total charge after connection should be the same as the total charge before connection.
Let Qf be the charge on the first conductor after redistribution.
Then, the charge on the second conductor after redistribution is Q - Qf.
Using the formula for capacitance, Q = CV, we can write:
Qf = C1V and Q - Qf = C2V
Substituting the given values, we get:
Qf = 3.0mF x V and Q - Qf = 5.0mF x V
Adding these two equations, we get:
Q = 8.0mF x V
Substituting the total charge, Q = 3400mC, we get:
3400mC = 8.0mF x V
V = 425V
Therefore, the final potential is 425V, which is option (c).
Given, capacitance of first conductor, C1 = 3.0mF
Capacitance of second conductor, C2 = 5.0mF
Potential of first conductor, V1 = 300V
Potential of second conductor, V2 = 500V
When the two conductors are connected, the charges will redistribute until they reach a common potential.
Let the final potential be V.
Charge on the first conductor, Q1 = C1V1 = 3.0mF x 300V = 900mC
Charge on the second conductor, Q2 = C2V2 = 5.0mF x 500V = 2500mC
Total charge before connection, Q = Q1 + Q2 = 900mC + 2500mC = 3400mC
Total capacitance after connection,
1/C = 1/C1 + 1/C2 = (C1 + C2) / (C1C2) = 8 / 15mF
C = 15/8mF
Using the formula, Q = CV
V = Q/C = (3400 x 10^-6) / (15/8 x 10^-3) = 680V
Therefore, the final potential is 680V.
But this is not one of the options given in the question.
To find the correct answer, we need to use the principle of conservation of charge.
The total charge after connection should be the same as the total charge before connection.
Let Qf be the charge on the first conductor after redistribution.
Then, the charge on the second conductor after redistribution is Q - Qf.
Using the formula for capacitance, Q = CV, we can write:
Qf = C1V and Q - Qf = C2V
Substituting the given values, we get:
Qf = 3.0mF x V and Q - Qf = 5.0mF x V
Adding these two equations, we get:
Q = 8.0mF x V
Substituting the total charge, Q = 3400mC, we get:
3400mC = 8.0mF x V
V = 425V
Therefore, the final potential is 425V, which is option (c).
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Community Answer
Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to...
First find the charge in 1st capacitor
Q1= C1*V1
=0.003*300
=0.9 C
Similarly, find Q2
i.e. Q2=2.5 C
After connecting the capacitors in parallel you get a net capacitance C=8 mF (c1+c2).
Thus, final potential is V=Q/C
Q=Q1+Q2=3.4 C
V=3.4/0.008
= 425 V
Hope you understood. Thank you
Q1= C1*V1
=0.003*300
=0.9 C
Similarly, find Q2
i.e. Q2=2.5 C
After connecting the capacitors in parallel you get a net capacitance C=8 mF (c1+c2).
Thus, final potential is V=Q/C
Q=Q1+Q2=3.4 C
V=3.4/0.008
= 425 V
Hope you understood. Thank you
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Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. Can you explain this answer?
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Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. 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 Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. 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 Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. Can you explain this answer?.
Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. 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 Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. 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 Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Two spherical conductors of capacitance 3.0mF and 5.0mF are charged to potentials of 300volt and 500volt. The two are connected resulting in redistribution of charges. Then the final potential isa)300 voltb)500 voltc)425 voltd)400 voltCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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