AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Consider the below arrangement of the electrical circuit.
The energy in IF capacitor is calculated as,
E1‌‌=‌

1

2

CV2
‌‌=‌

1

2

C(‌

Q

C

)2
‌‌=‌

1

2

‌

Q2

C

‌‌=‌

1

2

×‌

42

1

‌‌=8J
Similarly,
E2‌‌=‌

1

2

‌

Q2

C

‌‌=‌

22

2×2

‌‌=1J
The total energy is calculated as,
E0=E1+E2
=8+1
=9J
The common potential of the capacitors is calculated as,
V‌common‌‌‌=‌

C1V1+C2V2

C1+C2

‌‌=‌

1×4+2×1

1+2

‌‌=‌

6

3

‌‌=2V
For the new arrangement of the circuit the energy is calculated as
E1‌‌=‌

1

2

C1Vcommon2
‌‌=‌

1

2

×1×4
‌‌=2J
Similarly,
E2‌‌=‌

1

2

C2V‌common ‌2
‌‌=‌

1

2

×2×4
‌‌=4J
From the conservation of the energy, the energy stored in the inductor is calculated as,
EL‌‌=E0−(E1+E2)
‌‌=9−(2+4)
‌‌=3J