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Question

Find the total energy of the capacitors present in the circuit given below

  1. 80.5 μJ
  2. 100 μJ
  3. 120.25 μJ
  4. 90.7 μJ


Solution

The correct option is A 80.5 μJ
When steady-state condition is achieved it means that now capacitor is fully charged, no current will flow and will start acting as an open circuit and in beginning, it acts as a wire without showing any resistance when it is not charged
Energy stored in each capacitor is given as E=12Q2C=12CV2


Now to find the potential difference between E and C
Req=3 Ω+3 Ω+2 Ω+2 Ω=10 Ω
I=10 V10 Ω=1 A

The potential difference between E and C is,
V=1(2+3)=5 V
Energy stored, UEC=12×1×106×52(1)
Potential difference betweenn A and E is same as in between E and B,
VAE=10(2×1)=8 V
Energy stored, UAE=12×2×106×82(2)
Potential difference between E and D is 2 V
UED=12×2×106×22(3)
Total energy stored,
Utotal=12×106[25+128+8]
Utotal=1612×106=80.5 μJ

 

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