For the loop ABCDEFA what is the corresponding loop equation

For the loop ABCDEFA, what is the corresponding loop equation? 1. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 2. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 3. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 4. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 5. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 6. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 7. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 8. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 9. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 10. epsilon_1 - i_1 R_1 - i_3 R_3 + epsilon_2 - i_4 R_4 - i_2 R_2 = 0 Choose the correct relationships among the currents. A1: i_1 + i_3 = i_5; A2: i_1 = i_3 + i_5; B1: i_1 > i_2 and i_3 > i_4; B2: i_1 i_4; B3: i_1 > i_2 and i_3

Solution

loop equation of ABCDEFA    - (4)

current flows from +ve to -ve and E1 and E2 are opposite in polarity

015:

total current entering the juncion = ttal current leaving the junction

i1 + i3 = i5

current in a closed loop is same across all elements

i1 = i2 and i3 = i4

hence (5) A1 and B4

016:

i1 = i3 = 0.7/(1.3*3) = 0.179

i5 = i1 +i2 = 0.358 A

017:

i1 =0 as there is no closed loop and no current flows.

018:

time constant = RC = 240*17.2e-6 = 4.128 msec.