I still wonder how he is able to specialize in many fields like maths, physics, mechanics,electricity,chemistry,astrology and teach them with passion. A legend. Huge respect for him.I wish I had teachers like him.Pls like if you agree with me🤗👍✌
First your youtube videos are great - I enjoy them a lot - greetings from Denmark :-) Second I belive you missed an minus sign when converting Z ( 4 -j2 to 4.47 / - 26.565 deg.). Third video (9 of 38) is 2 times in the playlist.
yeah I noticed this aswell, but its good because it shows that we are attentive and are actually understanding the work, also GREAT VIDEOS, Michel you are the GUY, respect sir.
The voltage across a capacitor is always 90 degrees behind the voltage across the resistor, (and the voltage across the inductor is always 90 degrees ahead of the voltage across the resistor). Hence the - 90 degrees.
Thank you so much for your videos! I have a question: If the supply voltage is in polar form, I was told that -by convention- means it is already an RMS-voltage. Therefore, why are we still dividing by 1/2 and using Vmax & Imax?
Both the inductor and the capacitor are devices that essentially do not consume power. The inductor does have a coil that has some resistance, therefore in the real component, there is some power consumption. In the capacitor there is no coil and very little wire, therefore there is very little to no power consumption. (Only resistance consumes power).
Thank you so much for this wonderful resource Sir. I have a question though. Why is it that I can not use the P=VI to calculate power across the components, can you please explain to me when to use the P=VI and when to use P=(1/2)(Imax)(Vmax)cos(ϕi-ϕv)? It would be a huge favor, thank you!
If there is a phase difference between the voltage and the current in the circuit you must use the second equation. If there is no phase difference, the second equation collapses into the first equation.
@@jorgeg9754 it is a good question!simply put,cos(a)cos(b)=1/2[cos(a+b)+cos(a-b)],and when there is a phase angle, the second part of the right side of the equation will become 0.
while calculating the voltage across the capacitance how the phase angle will be 90 deg and while calculating the voltage across the resister , phase angle 0 deg.. can any one tell me please ??
That is all explained in great detail in this playlist: PHYSICS 49 RCL CIRCUITS. The phase angle indicates the difference when the current in the circuit reaches its maximum value, and when the voltage across the component reaches its maximum value.
Since the 2 components are in series, you can just add them up: Z = ristance of the resistor + impedance of the capacitor. (The impedance of the capacitor lags the resistance of the resistor by 90 degrees)
A person who spent a lifetime working as an engineer and as a college professor at various institutions teaching a number of different subjects. I learned a few things along the way.
I still wonder how he is able to specialize in many fields like maths, physics, mechanics,electricity,chemistry,astrology and teach them with passion. A legend. Huge respect for him.I wish I had teachers like him.Pls like if you agree with me🤗👍✌
the angle for the impedance is -26.565
I think he forgot the negative when handling the arctan
First your youtube videos are great - I enjoy them a lot - greetings from Denmark :-)
Second I belive you missed an minus sign when converting Z ( 4 -j2 to 4.47 / - 26.565 deg.).
Third video (9 of 38) is 2 times in the playlist.
First, welcome to the channel! And thank you for your input. You are indeed correct that it should have been - 26.565 degrees
yeah I noticed this aswell, but its good because it shows that we are attentive and are actually understanding the work, also GREAT VIDEOS, Michel you are the GUY, respect sir.
These videos are the only reason I'm passing my electrical course at university
tan^-1(-2/4)= - 26.565
So again,
our Hero is back on stage...
Love you professor.... From EAST AFRICA!!
Welcome to the channel!
You are a wonderful teacher and I thank you from the bottom of my heart I wish you all success (I see you from Iraq 😘😍)
Thank you and welcome to the channel!
This man is a Hero .Love u man.
Sir I like to thank you for your help you did your best for us
It's our pleasure
I had a question. Why is X sub c 2 phase angle of -90 degrees? I know where the two comes but where does the phase angle come from?
The voltage across a capacitor is always 90 degrees behind the voltage across the resistor, (and the voltage across the inductor is always 90 degrees ahead of the voltage across the resistor). Hence the - 90 degrees.
@@MichelvanBiezen ooooooooh thank you sooooo much. I really appreciate your videos. Helping me get through Circuits Analysis 2.
Glad you find our videos helpful. 🙂
kind sir , u saved my life !!!
Thank you so much for your videos! I have a question: If the supply voltage is in polar form, I was told that -by convention- means it is already an RMS-voltage. Therefore, why are we still dividing by 1/2 and using Vmax & Imax?
In the video, it is made clear that the I and V in the power equation are in terms of the maximum values, not the RMS values.
@@MichelvanBiezen thank you. And thanks again for the video!
Shouldn't the angle of the impedance be negative?
That is a good catch. The angle for the impedance should have been - 26.565 degrees. Note that the title includes the correction.
Why was the angle negative for X_c when finding the voltage across the capacitor?
The voltage lags the current across a capacitor by 90 degrees.
Is it always in a RC circuit that P absorbed by the capacitor turns out to be 0?
I mean what's the physical reason of this
Both the inductor and the capacitor are devices that essentially do not consume power. The inductor does have a coil that has some resistance, therefore in the real component, there is some power consumption. In the capacitor there is no coil and very little wire, therefore there is very little to no power consumption. (Only resistance consumes power).
@@MichelvanBiezen Thank you so much
Thank you so much for this wonderful resource Sir. I have a question though. Why is it that I can not use the P=VI to calculate power across the components, can you please explain to me when to use the P=VI and when to use P=(1/2)(Imax)(Vmax)cos(ϕi-ϕv)? It would be a huge favor, thank you!
If there is a phase difference between the voltage and the current in the circuit you must use the second equation. If there is no phase difference, the second equation collapses into the first equation.
@@MichelvanBiezen Thank you so much for the answer, its all very clear now! You are an excellent teacher Sir!
@@MichelvanBiezen how does cos(0) make the (1/2) disappear into the first equation P=IV?
@@jorgeg9754 it is a good question!simply put,cos(a)cos(b)=1/2[cos(a+b)+cos(a-b)],and when there is a phase angle, the second part of the right side of the equation will become 0.
the angle should be -26.56 for impedance i guess
You are correct.
Sir Send more lectures😊😊😍
while calculating the voltage across the capacitance how the phase angle will be 90 deg and while calculating the voltage across the resister , phase angle 0 deg.. can any one tell me please ??
That is all explained in great detail in this playlist: PHYSICS 49 RCL CIRCUITS. The phase angle indicates the difference when the current in the circuit reaches its maximum value, and when the voltage across the component reaches its maximum value.
@@MichelvanBiezen thankyou so much professor.
Thanks sir jii aap bhut achha samjhate ho again thanks
An angel sent to save my a*s from failing this course.
Glad you found the videos helpful.
why the angle is +26.56 and not -26.56? can anyone advise please
It is indeed -26.56 degrees. Note the correction at the end of the title.
Please start the astrophysics series
The phase angle that u hv calculated is wrong
Yes, see the correction.
Thank you
How to find Z
Since the 2 components are in series, you can just add them up: Z = ristance of the resistor + impedance of the capacitor. (The impedance of the capacitor lags the resistance of the resistor by 90 degrees)
It should be I=1.118 /_ 56.57°
You are correct. (Note the correction in the title - left off the negative sign).
forgive me sir, the angle of total impedance should be minus 26.56
You are correct. It should have been -26.56 degrees. Good catch.
What type person you are? I am unable to understand you,
You teach electricity, astronomy, mechanics of materials, physics etc
Amazing
A person who spent a lifetime working as an engineer and as a college professor at various institutions teaching a number of different subjects. I learned a few things along the way.