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QJEEM19Ph8500: JEE Main 2019 Physics - Class 12 JEEM19Ph8500 Solution: In the figure shown, after the switch 'S' is turned from…

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Capacitor Question 119 English

Multiple Choice Options

  • A 18Q2C
  • B 58Q2C
  • C 34Q2C
  • D 38Q2C

Step-by-Step Verified Solution

Vi = 12CE2

Vf = (CE)22×4c = 12CE24

ΔE = 12CE2 × 34 = 38 CE2

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