# Create an expression for impedance on the form Z = R + jX

I'm supposed to create an expression for impedance on the form: $$Z = R + jX$$ I didn't know how to that so I looked at my teacher solution, but I don't really understand that either. The question information states that we have an electrical engine with these values: $$V_{rms}=230V, I_{rms} = 7.5A, f=50Hz$$ The current (through the electrical engine) is also phase shifted 30°.
My teachers solution looks like this: $$U=ZI <=>Z= \frac{U}{I}$$ $$=\frac{230V}{7.5A \cdot e^{-j \cdot\frac{\pi}{6}}}=(26.5581+15.3333j) \Omega$$
And the alternative solution: $$|Z|=\frac{230V}{7.5A}$$ $$Z=|Z| \cdot (cos(30°)+j \cdot sin(30°))=(26.5581+15.3333j) \Omega$$
The only formula for impedance that I seen is: $$Z = (R^2+(X_L-X_c)^2)^\frac{1}{2}$$ What method is my teacher using to find the answer to the question?

• This formula is irrelevant. What is relevant is that graph you are not showing. But in general, it sounds that you have the graph of the voltage on that specific impedance, from which you can derive the phase. The amplitudes can be derived from the RMS values. Then you write it in a phasor notation, and then you separate it into real and imaginary parts. Commented Nov 2, 2023 at 17:04
• @EugeneSh. What do you mean? There is no graph included in the pdf file. Commented Nov 2, 2023 at 17:07
• You said "The graph is also shifted by 30°." ? Commented Nov 2, 2023 at 17:08
• @RussellH Thanks, I changed it. I hope that it is correct now. Commented Nov 2, 2023 at 17:09
• @EugeneSh. Sorry I probably phrased that incorrectly. It just states that the current (through the engine) is phase shifted 30°. Hopefully I translated it correctly this time. Commented Nov 2, 2023 at 17:13

I think you need to review the basics of complex numbers and phasor analysis. The solutions are just the definition of impedance:

$$\hat V = \hat IZ$$ $$Z = \frac {\hat V} {\hat I}$$

Your voltage is 230 V. Your current is 7.5 A with a phase shift of -30 degrees. -30 degrees is $$\-\pi/6\$$ radians. So your phasors are:

$$\hat V = 230\ \mathrm V$$ $$\hat I = 7.5\angle{-30^\circ}\ \mathrm A = 7.5\angle{-\frac{\pi}{6}} \ \mathrm A = 7.5e^{-j\frac{\pi}{6}}\ \mathrm A$$

And the impedance is the voltage phasor divided by the current phasor:

$$Z = \frac {\hat V}{\hat I} = \frac {230\ \mathrm V} {7.5e^{-j\frac \pi 6}\ \mathrm A} = 30.67e^{j\frac \pi 6}\ \Omega$$

You can also use angle notation and degrees to write the same thing:

$$Z = \frac {\hat V} {\hat I} = \frac {230\ \mathrm V} {7.5 \angle{-30^\circ}\ \mathrm A} = 30.67 \angle{30^\circ}\ \Omega$$

Either way, the $$\R + jX\$$ values are:

\begin{align} Z &= 30.67(\cos 30^\circ + j\sin 30^\circ)\ \Omega \\ &= 30.67(\cos \frac \pi 6 + j\sin \frac \pi 6)\ \Omega \\ &= 26.56 + j15.33\ \Omega \end{align}

The last formula you wrote is the formula for the magnitude of the impedance:

$$|Z| = \sqrt{R^2 + X^2}$$

There's also a formula for the phase of the impedance:

$$\theta = \arctan{\frac X R}$$

These formulas let you go from $$\R + jX\$$ to $$\|Z|\angle{\theta}\$$:

$$|Z| = \sqrt{26.56^2 + 15.33^2} = 30.67$$ $$\theta = \arctan {\frac {15.33} {26.56}} = 30^\circ$$ $$Z = |Z|\angle{\theta} = 30.67\angle{30^\circ}$$