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Two identical capacitors each of capacitance 5 mF are charged potentials 2 kV and 1 kV respectively. The -ve ends are connected together. When +ve ends are also connected together, the loss of energy of the system is
Options
(a) 160 J
(b) 0 J
(c) 5 J
(d) 1.25 J
Correct Answer:
1.25 J
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Electrons used in an electron microscope are accelerated by a voltage of 25 kV
- The drift velocity of the electrons in a copper wire of length 2 m under
- The distance of the closest approach of an alpha particle fired at a nucleus
- Two waves are represented by the equations y₁ = a sin (?t + kx + 0.57) m
- When 10¹⁹elecrons are removed from a neutral metal plate through some process,
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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Initially, the total electrostatic energy stored in two capacitors is Ui =1/2CV²1 + 1/2CV²2.
Initial charges on two capacitors are q1=CV1 and q2=CV2.
Let q′1 and q′2 be the charges on capacitors when the same terminals of two capacitors are connected to each other. By charge conservation
= q′1+q′2=q1+q2
The potential V′ across two capacitors is equal = q′1/C =q′2/C
Thus, q′1 = q′2 =(q1+q2)/2, and V′ = (V1+V2)/2
Thus, final electrostatic energy stored in the two capacitors is –
Uf = 1/2CV² + 1/2CV² = 1/4C ( V²1 + V²2 + 2V1V2)
and loss in energy will be = Ui − Uf = 1/4C(V1−V2)².