Why current lags voltage in an inductive circuit (explanation
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- Опубликовано: 29 сен 2024
- In a purely resistive circuit, current and voltage are in phase with one another. In a purely inductive circuit, voltage and current are 90 degrees out of phase with each other. This video explains why that is.
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The best explanation I've found so far. Now stuck in my head without having to learn anything by heart or through complex math.
This is just explaining the waveform with the mathematical equations.Actual fact is that when current flows through an inductor an emf is induced which will oppose the cause(Lenze s law) ,so there will be an opposing current in the inductor which opposes the main current.Overall effect is the lagging of the current.
Exactly this is why current lags voltage in an inductive load. You explain in two lines what that guys couldn't in 9 mins
is the induced back emf goes to maximum point instantly, i mean when the switch is on? and then decrease with increasing flux?
My question is why there is no back voltage associated with back current...back current opposes current , back voltage will also be associated with back current
@@SamiUllah-xs3tm good question and observation
In a purely inductive circuit, if voltage applied equals voltage induced ; how is there a current at any moment in time ?
0 marks for not labelling your graph. Current? Voltage? Voltage at the source? Voltage over the inductor?
How there is no voltage when current is max ? When current at its max,back emf vanishes batteries all potential so voltage seems like become zero but can it really be ? How particles gonna move without potential.Sorry im a mechanical engineer and dont really have much electrical background..
I was hoping for more of a physical explanation as opposed to a purely mathematical one. A sort of image of what is actually happening..
Did you find a video for it?
Yeah I can understand Mathematically not physically,to understand I must go through formula of inductive and capacitive voltage and it's origin.
I believe it has something to do with the formation and disintegration of the magnetic field around the coil, because of the changing direction of the current.
there is a mechanical analogy to inductance. it is called the flywheel effect. like trying to change the speed of a heavy spinning flywheel.
Pro Tip: Use a legend, you have different color traces and only have a mouse pointer. It is hard to keep track when you're talking to a just a plot with no labels...
If self-inductance (induced back emf) of an ideal inductor is always equal and opposite to the applied voltage then how does the current flow in that inductor, as the net voltage will always be zero?
Lenz’s law states that induced voltage, created by the applied voltage/current flow, opposes the the applied voltage/current flow. The intensity of the opposition to the applied voltage/current flow is directly related to the “inductance of the inductor, represented by the letter L”. Practically speaking the intensity of the applied current flow is always stronger than the induced current flow. The applied current flow pushes through the induced current. This opposition to the applied current is measured in Ohms, other wise know as reactance, X sub L. Simply, usually never a net 0. The sine wave plotting and vector representations do not try to show the intensity of each, but normally show that the applied is 180 degrees opposite the induced with no lagging or leading angle, and that’s q whole other conversation.
@@jmbravo-stugatti5867 thank you for more clarification.
@@jmbravo-stugatti5867 is the induced back emf goes to maximum point instantly, i mean when the switch is on? and then decrease with increasing flux?
Generator's energy > inductor's EMF
EMF is the result of current.
applied voltage case CURRENT, then, CURRENT cause EMF, so current first then EMF
You did everything except explain why current lags voltage in an inductive circuit. Some of the things made sense. However, when you have a graph depicting voltage and you are talking about current, and there is no current depicted on the graph, then this doesn't explain the current lagging voltage in an inductive circuit. It's a description of where the voltages are, but not the current . You brought up current at the end of the video, but it didn't explain why current lags voltage. Give it a rest ! Most people who want to teach and claim to have some good way of explaining things are actually very bad teachers and can't explain anything that is very technical and can't use physics or electrical theory, mathematics, or an in-depth understanding of the subject matter to make their point. This is simply another example of somebody trying to teach electronics, but doesn't fully understand the subject matter themselves, and can not fully explain why their equation is important to why current lags voltage.
Try again. You will need your equation. Explain where that came from. Next, your graph needs the applied voltage, the induced voltage, and the current superimposed on the graph. Show where 0, 90, 180, 270, and back to 360 degrees. Now, point out the current, voltage, induced voltage, electromagnetic charge on the inductor, and show the relationships that fall into play to produce the induced voltage and the out of phase current. I've listened to your video several times and It's very obvious that you did not achieve your goal.
One of the few acronyms I recall from engineering school: ICE and ELI ICE: in a Capacitive circuit (C), I (current) leads E (voltage)ELI: in an Inductive circuit (L), E (voltage) leads I (current)
@@briant7652 CIVIL Civ viL
In simple terms, inductive circuits have a delay because of the magnetic field, and that's why current doesn't instantly follow voltage when you turn them on.
Why does the initial equation have a negative sign for inductance? It makes it seem like the equation is for the CEMF voltage value.
Simple answer : Because current thru an inductor cannot change instantaneously.. The voltage can..
"Then it drops like its hot" 3:35, I guess that is a technical term in circuit analysis. Student, "When in rate of change max, again?", Teacher "When it drops like its hot!".
Spent a whole day looking for a decent explanation. Thank you!
Yes understqnd but why lag or lead in different circuits
ok this is math explanation why the hell is there current when voltage is zero
when voltage is at minimum change current is at maximum change.
but to get any current at all you need to have an applied voltage right? so when the applied voltage is 0 (the red graph) how is it that you have any value of current at all?
If I am right, it's bleeding off the collapsing mag field in the inducter.
rather than saying applied voltage you can say voltage across inductor. and if its pure inductor it has no resistance and there wont be a voltage across it and a lot of current would flow.
Thank you so much for this video. This was what was missing for me to understand capacitance.
I didn't got it. I have to watch it again and again.
Great explanation. Thank You
that is in AC only? what happens in DC?
You audio levels are way too low
Brilliant
at 5:31 when you said that the current is supposed to be at maximum, why did you mark it at its negative peak value and not its positive?
Ayms he was marking the graph of the current with respect to the back EMF graph(the green graph)
so the maximum back EMF is produced when you have maximum rate of change of current which you get when current is at 0
i also want to know that..he hasnt got any explanation for that..anyone can derive that mathematically, but hardly anyone knows what happens physically
Current or voltage at their lowest peak value are actually at their peak value in terms of effectiveness in a circuit. Simply changing directions in an alternating circuit
Thank You So much !! 🙌🏻👌🏻🙏🏻👍🏻
Around time 4:00 when you're discussing the rate of change of current, dI / dt, I've found that over half of students also get it wrong and claim that the positive and negative peaks are the max rate of change, and at the zero reference line is the minimum. The reason is, the (incorrect) association 'max current must mean maximum change' and 'zero current must mean zero change' and completely ignoring the slope of a tangent line along different points of the sine wave.
Easiest way to get folks past this intellectual hurdle is:
1) dI/dt is the slope of the tangent line at a point on the sine wave of current
2) find on the graph where the change in the current line changes the most for the smallest increment of time (the visual intuition approach)
3) calculate the slope of the tangent line (by using two values of (I, t)) at the positive and negative peaks of the sine wave, then do the same at the zero reference line, and you find that the plunge through the zero reference line is the steepest slope and thus the greatest rate of change. The slope of the line is zero at the positive and negative peaks, ie. rate of change of current is zero.
This works because all STEM students learn how to calculate the slope of a line in Jr. high or high school.
.
In a purely inductive circuit, if voltage applied equals voltage induced ; how is there a current ?
Same augment can be held accountable for capacitors also, then why current graph is 90 degree lead to source voltage in the beginning. Are you discussing mutual or self inductor??
I'm so confused
why applied voltage and self induced emf in inductor cancel each other? please explain ...
r!shi kesh Because of Lenzes law. the induced emf will be created in the direction which creates sucha current which produces a magnetic field that opposes the original field producing the induced emf in the first place.
The existence of a sinusoidal current resulting from the application of a sinusoidal voltage to an inductor is a characteristic of the non-Coulomb electric field that is proportional to the rate of change in current causing a changing magnetic field.
The current is a result of an opposing Coulomb electric field, which is a result of polarization by the non-Coulomb electric field associated with the changing magnetic field, and the current is a consequence of the resultant field of the applied field and the Coulomb electric field in the inductance coil.
In an inductor, the opposition to the applied voltage which is changing the current is instantaneous and so, the current can only follow after the applied field has overcome the opposing emf. In an inductor for sinusoidal currents, the current lags the voltage by 90 degrees if the inductor is pure, and less if a resistance is in series with it; the inductor fights before current flows. If an inductor weren’t to fight, you will get energy for free!
Electrostatics and circuits belong to one science not two. To learn the operation of inductive circuits it is instructive to understand Current, the conduction process, resistors and Voltage at the fundamental level as in the following two videos:
i. ruclips.net/video/TTtt28b1dYo/видео.html and
ii. ruclips.net/video/8BQM_xw2Rfo/видео.html
Inductors find applications as filters in power supplies and in resonant circuits in tuned amplifiers.
If we increase the “frequency” of the input voltage to an inductor, the “rate of change” of the input voltage and the applied field is “greater than” the rates obtained with applied voltages at lower frequencies. At low frequencies, this causes a smaller induced opposing electric field and emf, therefore, large currents will flow within small intervals of time in the coils of the inductor.
In the limit, if the input is a dc voltage, the current will become so large that the inductor will burn out.
Low pass filters are built using L-R circuits in simple applications. Therefore, we say the inductor “blocks high frequencies” and acts as a “short circuit” at low frequencies as if the inductor was not there but a wire.
It is not possible in this post to discuss in more detail current in inductor circuits and inductive reactance with sequential diagrams (textbook 4, see below).
The last frame References in video #1 lists textbook 4 which discusses these topics in more detail using a unified approach and provides an intuitive understanding of reactance.
why doesn't the applied voltage have an effect on the induced voltage? or why doesn't the current flowing in the circuit is the resultant for applied voltage and the induced voltage?
You explained why they are out of phase. You didn't explain why the current lags
is the induced back emf goes to maximum point instantly, i mean when the switch is on? and then decrease with increasing flux?
man you're awesome ! perfect explanation.
"perfect" (my ass)
U r better than my high school teacher... Good grief how didn't I think about derivatives and slopes in the way u explained before I watched ur video... U r the best no caps
wouldn't voltage be zero regardless if delta time or delta current was zero based on the equation you placed?
You're a penny richer because mine has finally dropped
Wow !
How is it that the induced voltage is exactly opposite to the applied voltage across the inductor in AC circuit?
If u say ∆i\∆t as slope it would be easy to understand 😉
Thank you so much for the beat explanation i have been struggling with this
OMG you explained it so smoothly. I always wanted to know why this happens
The best explaination which erases my doubt
I honestly dun know why some people dislike this kind of videos...
At 1:42 the equation E = -L di/dt is your back EMF, the force the current pushes against, this is NOT the voltage across the inductor. In this equation E represents the induced voltage also called back EMF which opposes the magnetic field created by current flow. But this same symbol E also represents the voltage across the resistor in the previous equation E=IR, and this is confusing. The back EMF equation should use the lower case Epsilon symbol for back EMF instead of E. The symbol E should properly represent the real voltage and the relationship between voltage across and current through the circuit is E = L di/dt , or V(t) = L di/dt. Note positive L value.
I was just about to comment on that.. Thanks for pointing that out, in all my texbooks that I've read, E or V(t)= L(di/dt)
Thanks, my asking how to make 180 degree phase shft voltge and current?
You would need a power source, instead of a load, to get a 180 degree phase shift. Inductors and capacitors can at most, get you a 90 degree phase shift. You would need a system that can add power to the circuit, instead of draw power from it, for voltage and current to have a 180 degree phase shift.
Nice Work
Please subtitle your videos
+Sultan Naeem RUclips have an option to show subtitles, just click on cc on the bottom of the video screen
Thanks For Reply
Thanks your explanation is awesome
please upload about reactive power
This made 0 sense to me. I don't understand how there can be any current in a circuit with 0 voltage. I still have absolutely no idea why current and voltage are not in phase.
@Hearing.Chanting Remembering.Krsna That seems like an answer to a different question. I didn't pick up any of that.
Change of North Pole and
South Pole? Lagging
Perfect explanation!!! Thanks!
Thank you for making this and the capacitor video so easy to understand.
Thanks. The explanation in my eddy current testing (NDT) Level II textbook might as well be in Greek.
+jockellis I hope my explanation sounds a little more like english ;)
Nice
This explanation is so clear that i actually had a 'eureka' moment because of your explanation i realised that (and please do correct me if i'm wrong) when the voltage is pushing hard through the rise and fall sections of the sine wave (which happens to cross the zero point), the current is at maximum flow but when it hits the peaks of the sine wave the voltage is almost stable for a brief moment and therefore the current flow is at its minimum because the electrons almost stop moving before heading very quickly back towards the zero point and on to the opposite polarity peak. i hope that makes sense, is that very basically correct or am i still missing something? thank you for your time and great videos :)
Current starts late. He should draw blue back to beginning. The voltage leaves the zero line first and current doesnt hit zero line for first time until voltage is at 90 degrees so voltage has a 90 degree head start. Current needs time to start hes reving his engine getting the power built up
@@nephilimshammer9567 getting the revs going that's a good analogy. ive since realised that I was wrong in my initial understanding after doing more research and studying. It's all about the journey though 👍🏼 Thanks for the reply.
Thanks brodie you the man
Excellent explanation
Canadian...... A?
excellent video.
Thank you 😊 sir
Thank you !
Thanks
What causes loss of power in an inductive circuit?
There is no loss of power in a purely inductive circuit. An ideal inductor is a 100% reversible energy storage device, that stores energy during half the cycle, and releases it 1 quarter cycle later.
What causes the loss of power in a REAL circuit with an inducitve power factor, is the non-zero resistance of the wires. Suppose we desire to transmit 1 kW of real power to a load with an inductive nature, from a 120V AC source (the RMS voltage). If this were a pure resistive load, it would draw 8.3 Amps of current (the RMS current). However, suppose it really draws 10 Amps, due to its inductive nature. The reason it takes 10A instead of 8.3A, is that there is a 33.6 degree phase shift between the current and voltage waveforms, caused by the inductive load. If you integrate the shifted waveforms to find the average power, you will get 1 kW, instead of 1.2 kW as you'd get from multiplying 120V * 10A. We call this, 1.2 kVA, when you multiply RMS values of current and voltage, with non-synchronized waveforms.
This means the power distribution network that supplies power to it, now has a loss of power associated with carrying 10A instead of 8.3A. If the wires totaled 1 ohm of resistance, the power loss in the wires line would increase from 68.9W (corresponding to the 8.3A resistive load) to 100W (corresponding to the 10A that actually supply this load).
Rate of change is at it's minimum when current flow is at it's maximum, that doesn't make any sense.
when you are going to the top of the curve, near the top, the slope is smaller(more gentle) than the starting point at 0.
slope = rate of change = delta i / delta t
Draw a tangent line near the top of the curve AND also a tangent line near 0 and check the slopes.
Near 0 the slope is steeper. That means the rate of change is faster.
ruclips.net/video/6LV63WtuvJg/видео.html
That makes sense review your math and the meaning of the graph.
That makes sense, review your and the meaning of the graph.
What is i here ?
The i stands for current in this context. Not to be confused with the imaginary unit "i", which is commonly called "j" in electrical engineering, due to "i" already standing for current.
Why do we use "i", when there isn't even an "i" in the word current? The answer is historical reasons. At one point, what we call current today, was called current intensity.
But why physically??
I feel like engineering would be more welcoming to women if men would stop calling abstract concepts, electrical components and even software components "guys". Every engineer I know does that.
3:29 counter-intuitive? How about obvious? By visual inspection it is obvious current is "flattening" at the peak.
David James. Man I wish it was obvious. I have seen many a student struggle with that concept. I am glad you see it as obvious though.
The Electric Academy It IS obvious. Ask them where the curve looks the flattest. Obviously at the peak it is. When it crosses the x axis it is also obvious it is steep. An easy illustration of this would be if you had a small car "ride" the wave and ask people where is the car most level. Obviously at the peak ( and in the valley) it would be level. Even a child could easily understand this so I find it hard to believe the average person, and especially a student, cannot see this easily.
David James Maybe if you are too smart you should be making videos for us.
I have been a spark since 1970, had trouble following all this then still vague on it all.I have not needed any of this mathematics in the 50 years that passed in the trade.I worked in generation, maintenance and oil rigs.I am looking for practical examples of how all this works in relation to measurement of real power in household situations /W inductive motors and heating elements.
I would love a simple explanation to pass on.
+bullsnutsoz I'll see if I can come up with an explanation for you.
Yes thanks I need to explain KVA and KVAR to people who ask me why having fluoro lighting and induction motors do not fully show up as KW s on their power bills.
I worked on rigs where KVA was measured alongside KVAR. The KVAR from huge induction motors dragged the generators down even when they were lightly loaded with KWs.
Does that make sense?