A part of a circuit in steady state along with the current flowing in the branches, with value of each resistance is shown in figure. What will be the energy stored in the capacitor ?
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Applying Kirchhoff’s first law at junctions A and B respectively we have, 2 + 1 – i1 = 0
Therefore i1 = 3A
Again, i2 + 1 – 2 – 0 = 0
Therefore i2 = 1A
Now applying Kirchhoff’s second law to the mesh ADCBA treating capacitor as a seat of emf V in open circuit
– 3 × 5 – 3 × 1 – 1 × 2 + V = 0
Therefore V= 20 Volt
So, energy stored in the capacitor U=1/2 CV2= (4 ×10-6 )× (20)2 =8×10-4 J
