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QJEEM20Ph69c6: JEE Main 2020 Physics - Class 12 JEEM20Ph69c6 Solution: Two capacitors of capacitances C and 2C are charged to…

Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :

Multiple Choice Options

  • A Zero
  • B 32CV2{3 \over 2}C{V^2}
  • C 92CV2{9 \over 2}C{V^2}
  • D 256CV2{{25} \over 6}C{V^2}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 5th September Morning Slot Physics - Capacitor Question 96 English Explanation

By conservation of charge,

q1+q2=q1′+q2′{q_1} + {q_2} = {q_1}' + {q_2}'

⇒\Rightarrow −CV+(2C)(2V)=(C+2C)V′- CV + (2C)(2V) = (C + 2C)V'

V′=3CV3C=VV' = {{3CV} \over {3C}} = V

Uf=12Cv2+12(2C)V2{U_f} = {1 \over 2}C{v^2} + {1 \over 2}(2C){V^2}

Uf=32CV2{U_f} = {3 \over 2}C{V^2}

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