Where we use the fact that - j2 = 2∠-90°. As we know the values of V and Vad, we can easily find the value of Vdc, or:
Vdc = V - Vad = 50∠0° - (-20∠0°) = 70∠0°
Now, knowing the value of Vdc we can calculate the value of I3.
I3 = Vdc / (4 - j3) = 14∠36.87°
Note que I3 = I1 + I2. Logo, temos:
I2 = I3 - I1 = 14∠36.87° - 10∠- 90°
Carrying out the calculation, we find:
I2 = 11.2 + j18.4 = 21.54 ∠58.67°
Knowing the value of I2 we can
calculate the value of Vbd, or:
Vbd = 5 ∠-90° . I2 = 5 ∠-90° . 21.54 ∠58.67°
Carrying out the calculation, we find:
Vbd = 107.70∠-31.33°
With the knowledge of Vad and Vbd we can calculate the value of Vab , or:
Vab = Vad - Vbd = = 20∠0° - 107.70∠-31.33°
Carrying out the calculation, we find:
Vab = - 112 + j56 = 125.22∠153.43°
Knowing the value of Vab it is possible to calculate the value of I, or:
I = Vab /4 = 31.305∠153.43° = - 28 + j14
Using the circuit, we see that I = I2 + IZ. Then the value of IZ will be:
IZ = - 28 + j14 - (11.2 + j18.4)
Carrying out the calculation, we find:
IZ = - 39.2 - j4.4 = 39,45∠-173.6°
And from the circuit, we see that - V + Vab + Vbc = 0. Then the value of Vbc will be:
Vbc = V - Vab = 50 + 112 - j56
Carrying out the calculation, we find:
Vbc = 162 - j56 = 171.4∠-19.07°
Therefore the value of Z will be:
Z = Vbc / IZ = 171.4∠-19.07° / 39.45∠-173.6°
Carrying out the calculation, we find:
Z = - 3.92 + j1.87 Ω
Important Note
The impedance Z can be represented by a series circuit between a resistor and an inductor. However,
NOTE that the resistor value is negative. So, for this circuit to be implemented in practice,
it is necessary that the imedance Z contains active circuits, such as transistors, operational amplifiers, etc...