A part of circuit in a steady state along with the currents flowing in the branches, the values of resistance etc., is shown in the figure. Calculate the energy stored in the capacitor C (4muF)`

Options

A. `8xx10^(-1)J`
B. `8xx10^(-2)J`
C. `8xx10^(-3)`J
D. `8xx10^(-4)J`

Answer

D

Explanation

When the capacitor plates get fully charged, there will be no current in branch ab, remember capacitance acts as the open circuit since capacitance offers infinite resistance to d.c. The capacitance simply collect the charge. Applying kirchoff's first law to the junctions a and b, we find `i_(1)=3A` and `i_(2)=1A`. Now applying kirchhoff's second law tot he closed mash aefba, we get
`3xx5+3xx1+1xx2=V_(a)-V_(b)`
`V_(a)-V_(b)=20C`
Energy stored in the capacitor
`U=(1)/(2)C(V_(a)-V_(b))^(2)=(1)/(2)xx4xx10^(-6)x(20)^(2)`
`=8xx10^(-4)J`