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. `8 xx 10^(-1) J`
B. `8 xx 10^(-2) J`
C. `8 xx 10^(-3) J`
D. `8 xx 10^(-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 collects the charge

Applying Kirchhoff.s first law to the junctions a and b, we find `i_1 = 3A and i_2 = 1A `
Now applying Kirchhoff.s second law to the closed mash aefba, we get
`3 xx 5 + 3 xx 1 + 1 xx 2 = V_a – V_b`
or `V_a – V_b = 20 V`
` U = 1/2 C(V_a - V_b)^2 = 1/2 xx 4 xx 10^(-6) xx (20)^2`
`= 8 xx 10^(-4) J`