Two identical capacitors each of capacitance 5 mF are charged potentials 2 kV and 1 kV

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:

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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.

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