[EET 1025C - Fundamentals of AC Circuts] Series-Parallel RLC Circuits

[@Rahel-iz1qr]

do have more about rlc series and parallel circiuts to solve plz

[@profashleyevans]

I don't have any at this time, but I might post some in the future!

[@Rahel-iz1qr]

@@profashleyevans then where i can find more videos about complex series and parallel rlc circiuts and also with power calculations like complex and others?

[@Zapgod]

How would you find Z1 using a calculator?

[@profashleyevans]

It depends on what kind of calculator you have. For example, a TI-89 can automatically convert from rectangular to polar and you can type in the entire calculation in one step. If you don't have an advanced calculator, you would need to do each calculation in multiple steps on your calculator. This might mean first converting from rectangular to polar, or polar to rectangular depending on what value you're trying to find, and then doing additional combination of rectangular values or polar values. I go over the conversion from rectangular to polar, and polar to rectangular in an earlier video within the AC circuits playlist. If you need to see that information, look for the introduction to AC circuits video I posted.

[@Zapgod]

@@profashleyevans ok thank you ill look for the video

[@joeferreira-qr7iq]

when you do current division , you seems to use the denominator as only one branch of the parallel impedances, the denominator should be the sum of the parralel impedances

[@profashleyevans]

Hi Joe, thanks for watching! You might be thinking of a voltage divider here. When you have a voltage divider, the denominator is the sum of all resistances in series. But with the current divider, it is the numerator that is the total resistance of all the parallel paths you are dividing the current among. So in these problems, you'll see that the numerator is the combine resistance of all parallel paths, while the denominator is the resistance of the path in question, where the path we want the current for.

[@leitaylor6042]

Wonderful lecture.when getting Z2 it's easier to compute (1.0k +j-500)/(1.0k*j-500)

[@profashleyevans]

Yes, you can also use product-over-sum to also compute Z2 as well. But, take a closer look at how you typed this, above ;-) I like to show different methods of calculating values, just to show there may be more than one path to a correct answer. Thanks for pointing that out!

[@alexa8309]

Nice work explaining this. The R2 current should be 1.88 not 1.86 ... typo perhaps

[@profashleyevans]

Yes, it was a typo. Thanks for pointing it out!

[@octaviosandoval4633]

How do I calculate the voltage at B?

[@profashleyevans]

The voltage from terminal B to ground is the voltage across R2, which is also the voltage across X_L since they are in parallel. Since we have the current through I_C2, you can use that along with ohm's law to find the voltage across the parallel combination R2 // X_L, which would then be the voltage across each of those individual components.

[@rockyamisola7142]

Is resistor always the real number and conductor, inductor the imaginary? Help.. Tyy :)

[@profashleyevans]

Hi Rocky, yes, the resistance is always the real value of your polar coordinate. Your capacitance and/or inductance will make up the imaginary component.

[@jar01usmc]

Your R2 value is incorrect in the product over sum equation at time 9:53 of the video. You have 9k ohms vice 8k ohms.

[@profashleyevans]

You are correct! It was a typo, my apologies. Even though a 9k is displayed on the screen, 8k was used for the actual calculation. I will work on getting this fixed. Thank you.

[@rockyamisola7142]

Why 0.5k ohms has 90°?

[@profashleyevans]

Hi Rocky, you'll notice the 0.5 k-ohms ( around 4 minutes in) has a j component attached to it. This j means that the value lies on the imaginary axis, at either +90 or -90 degrees. Since the term is "+ j0.5 k-ohms", it lies on the positive imaginary axis and thus has a phase angle of +90 degrees.