Question

please help me......​

Answer 1

Answer:

by checking it i get to know that the answer of this question is

D)60%..

Answer 2

ANSWER ::

Initial capacitance = 1 μF

U = 1/2×1×v² = v²/2

Now

Final voltage after switch 2 is :

V =(C1V1) /(C1+C2) = 1×V/(1+4)=V/5

Final energy in both the capacitors

= U

= 1/2(1+4)(V/5)²

= v²/10

Therefore, energy dissipated

= ( v²/2 - v²/10) /(v²/2)

= 4v²/10 / v²/2

= 8/10

= 4/5

Hence the percentage of energy dissipated

= 4/5 ×100%

= 80%

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