A part of a circuit in steady state along with the currents flowing in the branches,the values of resistances,etc, is shown in figure.Calculate the energy stored in the capacitor.

Options

A. `6xx10^(-4)J`
B. `8xx10^(-4)J`
C. `16xx10^(-4)J`
D. Zero

Answer

B

Explanation

Applying Kirchhoff.s first law at junctions A and B respectively we have `2 +1 -i_(1)` = 0
i.e, `i_(1)` = 3A and `i_(2) +1 -2-0` i.e., `i_(2)`= 1A
Now applying Kirchhoff.s second law to the mesh ADCBA treating capacitor as a seat of emf V in open circuit  `-3xx5–3xx1–1xx2+V`=0 ie. V=`(V_(A) - V_(B))` = 20 V
So, energy stored in the capacit U=`(1)/(2)CV^(2)`=`(1)/(2)(4xx10^(-6))xx(20)^(2)`=`8xx10^(-4)J`