As an electrician this is great video to brush up on theory. Great work !
@stepbystepscience4 жыл бұрын
Great to know and thanks for watching and commenting.
@MM-il6xn4 жыл бұрын
1 comprehensive and fun science video by step-by-step science channel = 1 whole term of a boring university course
@stepbystepscience4 жыл бұрын
Thanks for the great comment.
@alejandroalmarza84472 жыл бұрын
Excellent explanation….very detailed and crystal clear ….thanks
@stepbystepscience2 жыл бұрын
You are very welcome!
@aliabdi76304 жыл бұрын
tremendous job, this is something I never see before,
@stepbystepscience4 жыл бұрын
Glad you like it and thanks fo the comment.
@Shannon_Robbie4 ай бұрын
I'm confused. I was told that the inductor is always positive 90 degrees and the capacitor is always -90 degrees. But in your Current calculations it's the opposite. Why? 4:50
@Festus20224 ай бұрын
Nice video: One question that no one ever addresses is WHERE does the transition from one current to another occur? For instance, where exactly does the 1.56amp, 11-degree, Total current transition to the 1.52amp, zero degree current through the resister? Does this occur inside the passive element or in the wire before/after the passive element? Thanks
@andrewfortus26294 жыл бұрын
Thank you for the series I went through all 19 and will go through more. At 11:04 I think it's supposed to be (Bc-Bl)^2 under the square root for total admittance? Thanks again! EDIT: NEVERMIND ! I see now you are adding them up VECTORILY :)
@stepbystepscience4 жыл бұрын
Great that found the videos helpful and thank you very much for commenting.
@eljuancho9138 Жыл бұрын
Thanks
@stepbystepscience Жыл бұрын
Welcome
@LuferXVII9 ай бұрын
Hello, I have a question about knowing the phase angle in the source voltage, can I determine an angle φ according to the impedance values and then do Ohm's Law, please see the equation I made and if it is coherent. Z=147 ohm, φ=tan^-1(((259-193)/(150)))=23.7495. IT= 1.56 A φ= 11 ... V=Z*I V=147*1.56= 229.32 V φ= 23.7495+11=34.7495 V=229.32 ∠ 34.7495....ok?
@stepbystepscience9 ай бұрын
I don't really have time to evaluate the details of this problem, sorry.
@GymChess4 жыл бұрын
Cosine instead of tan: 1,53/1,56 equals almost the same - 11 degrees.
@asandamvusongobese21092 жыл бұрын
would you advase me to to solve using angular form??
@seanarsenault60972 жыл бұрын
What if R is in series and C and L are in parallel? Whats the formula then?
@mahsumoral49604 жыл бұрын
In series RLC circuit, why the scope of 𝑉R vs. 𝑉out graph is displayed as an oval loop?
@marvik6662 жыл бұрын
Hello just started watching you doing numbers if I may ask you to do some rlc circuit in serials also iam just stated to do rlr and RLCR R RESISTOR L INDUCTOR C CAPACITOR if you can please do a Capple of video like this please I thank you very much Kindl regards Anthony
@ronvaliant93373 жыл бұрын
I was taught that if the reactance was inductive it was positive and if it was capacitive it was negative. And it 's the reactance instead of current. Is this wrong?
@rohan_nyc4 жыл бұрын
Again nice video 👍. Can you please solve the problem taking polar angles into consideration. Z=R+jX and Y=G+jB.
@stepbystepscience4 жыл бұрын
I can look into it.
@rohan_nyc4 жыл бұрын
@@stepbystepscience Thank you so much. Im looking forward to watch the video:)
@theherheoroie4 жыл бұрын
Very helpful
@stepbystepscience4 жыл бұрын
Glad to hear that
@sachinrath2193 жыл бұрын
when current lags the voltage,does it mean it flows in the reverse direction to the one taken in the resistor.
@stepbystepscience3 жыл бұрын
No
@khatarinakramer71454 жыл бұрын
What if there are multiple resistors and capacitors and inductors
@stepbystepscience4 жыл бұрын
Then it gets very complicated
@havard80314 жыл бұрын
Can you explain the formula Zt = 1/sqrt(1/R^2 + (1/Xl - 1/Xc)^2)?
@stepbystepscience4 жыл бұрын
I could make a video about it but mean time check out this page: under "Impedance of a Parallel RLC Circuit" www.electronics-tutorials.ws/accircuits/parallel-circuit.html
@kissdarren42332 жыл бұрын
Greeting please I have a problem don't know if you could be of help