Consider the circuit shwon in fig. The circuit is in steady state.

The value of `i_1` is

Options

A. `7//9A`
B. `14//13A`
C. `14//3A`
D. `17//23A`

Answer

B

Explanation

In the steady state, there will be no current in capacitor branch
Loop EABE : `6 = i_1 +(1+1)(i_1 +i_2)+2i_1 or 5i_1+2i_2 = 6` (i)

Loop ADCBA : `4 = 2i_2 +(1+1)(i_1+i_2)+2i_2 or 6i_2 +2i_1 = 4`
or `6i_2 + 2i_1 =4`
Solving Eqs. (i) and (ii), `i_1 =14//13 A, i_2 = 4//13A`
Given `V_E = 0, V_E +6 -1i_1 - 1(i_1+i_2) = V_B`
or `V_B =6-2i_1 -i_2 = 6-2xx(14)/(13) - (4)/(13) = (46)/(13)V`
Now along BADC : `V_B +1(i_1+i_2)+2i_2 =V_C`
or `V_C - V_B = i_1+3i_2 = (14)/(13)+3xx(4)/(13) =2V`
Charge on capacitor is `q = C(V_C - V_B) = 4xx2 = 8muC`