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I want your feedback on my approach to solving the question. I also want clarification on the phase angle for the inductive reactance.

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Well, the impedance of your circuit is given by:

$$\underline{\text{Z}}_{\space\text{in}}=\frac{\text{R}\cdot\text{j}\omega\text{L}}{\text{R}+\text{j}\omega\text{L}}=\frac{\text{R}\cdot\text{j}\omega\text{L}}{\text{R}+\text{j}\omega\text{L}}\cdot\frac{\text{R}-\text{j}\omega\text{L}}{\text{R}-\text{j}\omega\text{L}}=\frac{\text{R}\cdot\text{j}\omega\text{L}\left(\text{R}-\text{j}\omega\text{L}\right)}{\text{R}^2+\left(\omega\text{L}\right)^2}=$$ $$\frac{\text{R}\cdot\text{j}\omega\text{L}\text{R}-\text{R}\cdot\text{j}\omega\text{L}\text{j}\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}=\frac{\text{R}^2\omega\text{L}\text{j}+\text{R}\left(\omega\text{L}\right)^2}{\text{R}^2+\left(\omega\text{L}\right)^2}\tag1$$

The magnitude can be found using:

$$\left|\underline{\text{Z}}_{\space\text{in}}\right|=\sqrt{\left(\frac{\text{R}\left(\omega\text{L}\right)^2}{\text{R}^2+\left(\omega\text{L}\right)^2}\right)^2+\left(\frac{\text{R}^2\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}\right)^2}=\frac{\text{R}\omega\text{L}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}=$$ $$\frac{5\cdot2\pi\cdot60\cdot10\cdot10^{-3}}{\sqrt{5^2+\left(2\pi\cdot60\cdot10\cdot10^{-3}\right)^2}}\tag2$$

And the angle is given by:

$$\arg\left(\underline{\text{Z}}_{\space\text{in}}\right)=\arg\left(\frac{\text{R}^2\omega\text{L}\text{j}+\text{R}\left(\omega\text{L}\right)^2}{\text{R}^2+\left(\omega\text{L}\right)^2}\right)=$$ $$\arg\left(\text{R}^2\omega\text{L}\text{j}+\text{R}\left(\omega\text{L}\right)^2\right)-\arg\left(\text{R}^2+\left(\omega\text{L}\right)^2\right)=$$ $$\arctan\left(\frac{\text{R}^2\omega\text{L}}{\text{R}\left(\omega\text{L}\right)^2}\right)-0=\arctan\left(\frac{\text{R}}{\omega\text{L}}\right)=$$ $$\arctan\left(\frac{5}{2\pi\cdot60\cdot10\cdot10^{-3}}\right)\tag3$$

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  • \$\begingroup\$ Yeah, I found this formula on the web somewhere earlier, thanks. What's confusing me is that, if the phase angle of the inductive reactance is -90 will that not imply -j in place of j in the above formula? \$\endgroup\$ – octopus Apr 11 at 18:40
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    \$\begingroup\$ you shouldn't be "finding formulas on the web", honestly, you need an introduction to complex circuit theory, which should just be a couple pages down in whatever material you're reading that gives you these exercises. This is really basics of the basics – any introduction to electronic circuits of the last 100 years will introduce you to complex impedances. \$\endgroup\$ – Marcus Müller Apr 11 at 18:55
  • \$\begingroup\$ @Jan Try to show us how we can find the impedance magnitude as well. \$\endgroup\$ – G36 Apr 11 at 19:00
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    \$\begingroup\$ Yeah, seen it. That was very helpful. But as @MarcusMüller said I need to get a solid foundation. We are just being introduced to the topic so, I think with time I will get deeper understanding. Thank you so much for your feedback. It was very helpful. \$\endgroup\$ – octopus Apr 11 at 21:08
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    \$\begingroup\$ @octopus of course you will if you study hard and willingly ;) good luck and I am glad that I could help you. \$\endgroup\$ – Jan Apr 11 at 21:12
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As this is homework, and you asked for feedback:

This is exactly the point where you need to realize that writing things as complex impedances can significantly simplify things. (The results are the same as if were to calculate it in your angle / amplitude notation, it's just easier and more general.)

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  • \$\begingroup\$ Okay, thanks for the feedback. The face angle for the inductive reactance, is it right? and to what extent does it include the formula stated by @Jan above? \$\endgroup\$ – octopus Apr 11 at 18:38
  • \$\begingroup\$ It is what Jan does. you just need to learn how to do it. It's easy. \$\endgroup\$ – Marcus Müller Apr 11 at 18:40
  • \$\begingroup\$ What's confusing me is that, if the phase angle of the inductive reactance is -90 will that not imply -j in place of j in the formula stated by Jan? \$\endgroup\$ – octopus Apr 11 at 18:42
  • \$\begingroup\$ that just means the -90° for your \$X_L\$ is incorrect; an inductor does have an impedance of \$j\omega L\$, which has positive imaginary part. \$\endgroup\$ – Marcus Müller Apr 11 at 18:54
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    \$\begingroup\$ That's a direct result of how things are defined, it's what directly follows from the model behind that model. Really, read an introductory text on complex circuit theory. Again, this is ultra-basic stuff, covered by every Network Theory textbook for EE students – this isn't obscure knowledge. \$\endgroup\$ – Marcus Müller Apr 11 at 19:56
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A more common approach is the following:

ZR = 5

ZL = j·ω·10^(-8)

Zeq = (ZR·ZL) / (ZR + ZL)

Simplify Zeq and you get something like:

Zeq = a + j·ω·b


The phase angle of L comes in when you think about its equation in the time domain:

v(t) = L·(di(t) / dt)

Now transform it in the phasors domain:

V(ω) = j·ω·L·I

V and I, in the inductor L, are out of phase by j, that is by 90 degrees.

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