AP EAPCET 08-Jul-2022 Shift 2 Paper

Initial stored energy =Energy‌of‌capacitor‌1
=

1

2

C1V12
=

1

2

×500×10−12×(100)2...[∵C1=500pFV1=100V]
‌=250×10−12×104=250×10−2×10−6
‌=2.5µJ
When both capacitors are joined, common potential of combination is
V‌common ‌=V=‌

Q1+Q2

C1+C2

‌ or ‌V=‌

C1V1+C2V2

C1+C2

or
V=

500×10−12×100

2×500×10−12

=50V
Final stored energy =

1

2

(C1+C2)V2
=

1

2

×2×500×10−12×(50)2
=500×10−12×2500
=125×10−8
=1.25×10−6J
=1.25µJ
Loss of energy, ∆U=Final‌energy−Initial‌energy
=25−1.25µj=1.25µj
(No option is matching)