Find V0 in this network using Loop analysis.
Since the network has two loops, or two meshes, we will need two equations to determine all the currents.
Note that since I2 goes directly through the current source, I2 must be 2∠0°A. therefore, one of our two equations is ready solved and…
I2 = 2∠0°…If we now apply KVL to the loop #1, we obtain…
−12 + I1through and times the two impedances (2 − j1) and since for the next two impedances the current will be the sum of (I1 - I2) so we must add the product of (I1 − I2) times (4 + j2) = 0…We now have two equations which will yield the mathematical solutions for the two currents….
Substituting for I2 from the first equation into the second now leaves us with one equation and one unknown, I1…removing the brackets yields…
-12 + 2I1 - jI1 + 4I1 - 8 + j2I1 - j4 = 0… I have coloured the I1 terms for clarity and collecting like terms moving the whole numbers to the right of the equation and keeping the variable on the left we get…I1(6 + j1) = 20 + j4…cross multiplying…which is the same as dividing both sides of the equation by (6 + J1)…