Q49 / 142 · NTA Abhyas (JEE Main) 2020 · Physics · Electrostatics

The figure shows a capacitor of capacitance C connected to a battery via a switch, having a total charge Q on it, in steady-state. When the switch S is turned from position A to position B, the energy dissipated in the circuit is

Figure

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

Solution

Q0=Cε

Q1C=Q23C        Q1+Q2=Q0

⇒Q1=Q04; Q2=3Q04

Energy dissipated,
E=12Q0 2C-12Q1 2C-12Q2 23C

=12CQ02-Q0216-9Q023×16

=Q0232C16-1-3=38Q02C

Hence, the correct answer is option: 'B'.

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