CBSE Class 12-science Answered

when 3μF capacitor is charged to 12 V , it contains 36 μC  . When 6 μF capacitor is charged to 12 V , it contains 72 μC .

when positive plate of 3μF capacitor that contains +36  μC is connected to negative plate of 6 μF capacitor that conatins -72 μC ,

net charge on the coneected plates will become  (+36 - 72 ) μC = -36 μC . Similarly other connected plates will contain +36 μC .

Now 3 μF capacitor and 6 μF capacitor are connected in parallel to give ( 3 + 6 ) μF = 9 μF capacitor and this capacitor

combination has charge magnitude 36 μC on commonly connected plates .

Hence Potential difference V = Q / C = 36 μC / 9μF = 4 V

Initial stored energy on capacitors, U_i = (1/2) C₁ V₁² + (1/2) C₂ V₂² 

U_i = [  (1/2) × 3 × 10^-6 × 12×12 ]  + [ (1/2) × 6 × 10^-6 × 12 × 12] = 648 × 10^-6 J

Energy stored on capacitors after connecting them parallel , U_f = (1/2) C V² 

U_f = (1/2) × 9 × 10^-6 × 4 × 4 = 72 × 10^-6 J 

Energy loss = U_i - U_f = ( 648 - 72 ) × 10^-6 J = 576 × 10^-6 J

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Answer 

(A) - Common potential is 12 V - False , It is calculated above as 4 V

(B) - Loss in energy = 576 μJ , True 

(C) - Common potential = 4 V , True

(D) - Loss in enrgy = 144 μJ , False