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Question

Shown in the figure is network of capacitors and resistors. The potentials of some of the nodes are given, Find the
(a) potential of P
(b) energy stored in the capacitor connected between R and S
(c) energy stored in the capacitor connected between P and Q
1016398_786b7016ff9c4db2ba0480467662e70e.png

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Solution

At steady state, capacitor offers infinite resistance to DC, therefore, no current passes through 1μF capacitor and current passes through 1Ω resistor across PQ. Since three identical resistors of resistance 13Ω each are connected in series. This series combination is connected in parallel with another resistor of 1Ω across PQ
The effective resistance between P and Q will be 12Ω
Applying KCL at P we obtain
i1+i2+i3=0Vp101+Vp30.3+Vp(20)1=0Vp[1+10.3+1]=10+1020
(b) Vp=0;i1=10amp;i2=10amp,i3=20amp
The potential difference across PQ
VPQ=RPQ.i3VPQ=(1/2)20=10V
Since Vp=0,VQ=010V
ie., VPQ=+10V
The potential difference between R and S =VRS
Since iPR=IPQ/2=i3/2=2010=10;VS=0;VP=0
VR=VP(IPR)(RPR)=[10×1/3]=103V
(b) the energy stored in the capacitor (0.4μF)
=16(0.4×106)(103)2=2.2×106J
(c) the Energy stored in the capacitor (1μF)
=12(1×106)102=50μJ
1039050_1016398_ans_bb9375342c684ab18b7864456db45f36.png

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