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ELI the ICE manby leright
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#1
Feb1906, 11:43 PM

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This is probably a stupid question, and I realized the answer before, and I know my professor explained this in class, but why does voltage lead current for inductors and current lead voltage for capacitors. I don't want a physical explaination but a mathematical one. I am looking at my circuits notes and my professor states that it is obvious from the following equations that this is true.
V = j(omega)LI and I = j(omega)CV, where omega is frequency and I and V are the current and voltage phasors. I know this is stupid because I was able to answer this question myself fairly easily just by looking at the equations, but now I look at them again and cannot answer this question for the life of me. SOMEONE HELP ME!! :( If I can't figure this out on my own then I will ask my professor tomorrow, but I'd rather not ask such a question since he states that it is obvious in his lecture notes. Thanks a lot. edit: I suppose if you plug the phasors into the above equations and then convert them from exponential form to trigonometric form using euler's identity the sine expression becomes the real part (formerly the imaginary part) and then the cosine expression becomes the imaginary part (formerly the real part). Making the sine expression the real part would make the wave 90 degrees out of phase. Is this correct? I remember realizing the answer to this question before and my reasoning seemed much simpler. 


#2
Feb2106, 10:30 AM

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P: 41,098

Just write the differential equations for voltage and current in the two situations.
V = L di/dt I = C dv/dt And remember that when you differentiate a sine wave, you get the cosine function. So current lags the voltage in an inductor, and voltage lags the current in a capacitor. Make sense? 


#3
Feb2106, 12:12 PM

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#4
Feb2106, 01:08 PM

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ELI the ICE man
Do you know where the phasor notation comes from? For the inductor case, write the current i(t) = exp(jwt) and differentiate. So di(t)/dt = jwi(t). You can also think of a multiplication by j as a counterclockwise rotation by 90 degrees on the phasor diagram....



#5
Feb2106, 03:47 PM

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Realizing that [tex] j = e^{j\frac{\pi}{2}} [/tex], multiplying by [tex]j[/tex] amounts to adding 90 degrees to the exponential. In elaboration of what berkeman has described:
[tex] V = L \frac{di}{dt}=L j \omega e^{j \omega t} = L \omega e^{j( \omega t + \frac {\pi}{2}) [/tex]. The output voltage (V) is ahead (leads) the input current (I) by 90 degrees through an inductor. (Plot it and you can see). By analogous reasoning, you can show current leading voltage by 90 degrees across a capacitor. 


#6
Feb2106, 07:27 PM

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