Solve the following two equations for I2:
(2+j2)I1 - 2I2 = 8
-2 I1 + (4-j2)I2 = 6
Life is short! I put this into MathCAD and found
I2 = 3.4+j0.2 A = 3.41 A \/ -3.4 deg
Solve the following two equations for V1 and V2:
Eq. 1: jV1 + V2 = 8
Eq. 2: V1 + (-1+j)V2 = -j6
Multiplying Eq. 1 times j and then adding to Eq. 2 yields:
(-1+j2)V2 = j2,
V2 = 0.8-j0.4
(after conversion of the two complex numbers to polar form, division, then conversion back).
Back substitution into Eq. 1 yields:
V1 = 0.4-j7.2
V1-V2 = -0.4-j6.8
I2 = (V1-V2)/(-j2) = (-0.4-j6.8)/(-j2) = 3.4-j0.2
Life is too short for that too!
Answer for P4.16
The current source pushes 3 A of current through the lower 2 ohm resistor, thus there is a 6 V difference across it with polarity as shown,
In the top portion, the 8 V source across the 2+j2 ohm impedance would create a current of 2-j2 A through the top 2 ohm resistor; thus it has a voltage of 4-j4 V across it.
The sum of the two, 10-j4 V is the Thevenin voltage.
|Killing the Sources:|
Shorting the voltage source and opening the current source yields this combination of passive elements. The j2 ohm and 2 ohm elements at the top are in parallel and have the equivalent impedance of 1+j ohm. The Thevenin impedance is thus 3+j ohm.
|Using the Generator:|
The total impedance of the equivalent Thevenin circuit is 3-j ohm,
so the current through the resistor is
I = (10-j4)/(3-j) = (10.8\/-21.8) / (3.16\/-18.4) = 3.41\/-3.4a