R1 114 R2 202 R3 304 R4 408 R5 620 I1 A I2 A I3

Solution

We have two separate sources of emf in the two sub-circuits as part of the main circuit. We have, as indicated, currents I1, I2 and I3 flowing through the parts as depicted.

We will employ Kichhoff\'s loop law to form equations involving the currents and then solve them to determine the required the values.

For the upper circuit: -12 + 6.2I1 + 3.04 I1 + 2.02 I3 = 0

For the lower cicuit: -9 - 2.02 I3 + 4.08 I2 + 1.14 I2 = 0

Also by Kirchhoff\'s junction rule, we get:

I1 = I3 + I2, We put this in the above equation to get:

9.24 I2 + 11.26 I3 = 12

5.22 I2 - 2.02I3 = 9

we multiply the second equation with 5.57426 to get:

29.0976I2 - 11.26I3 = 50.16834

Adding the first equation, we get:

38.3376 I2 = 62.16834

Therefore, I2 = 1.622 Amperes

I3 = [5.22 I2 - 9] / 2.02 = -0.264 amperes [The minus sign here denotes that the current is opposite in direction to what has been shown in the figure.]

I1 = 1.622 - 0.264 = 1.358 Amperes