1
$\begingroup$

I was watching this video from Walter Lewin and while watching these two videos, I noticed there is a "contradiction" in what he is doing. All links direct you exactly to where he begins, so you don't have to search it yourself

LR Circuit

http://www.youtube.com/watch?v=UpO6t00bPb8#t=10m22s

Resistor Circuit (Kirchoff's Law)

http://www.youtube.com/watch?v=59eTiTa9Tvk#t=3m02s

In the LR Circuit video, he sets his current to run CCW and traverses CCW. Then he mentions about the electric field present in the circuit and he sets up the equation:

$$+IR - V = -L\frac{\mathrm{d} I}{\mathrm{d} t} $$

In the Resistor Circuit (Kirchoff's Law) video, he talks about potentials going up and down. He prepared beforehand for us the direction of the current in the circuit. He begins at point P, he traverses the circuit CCW (in the direction of the current). This time, however, he didn't mention anything about the electric field present in the circuit. As he traverses the outer loop, he gets the equation:

$$-6 + 20 - 1 - 2 - \varepsilon_2 - 4 = 0$$

Notice that he gets $-6$ this time when he traverses in the direction of the current? As opposed from the LR Circuit video, he gets $+IR$ when he traverses in the direction of the current.

Could someone please clarify what is going on? Thank you very much. I am unable to progress because of this confounding concept

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Lewin does appear to be using a different sign convention in the two lectures. In the first he takes a voltage drop as positive while in the second he takes a voltage drop as negative.

But it doesn't matter which convention you use as long as you are consistent. The point is that if you go round any closed loop the total voltage change must add up to zero. It doesn't matter whether you take resistors as negative and batteries as positive or the other way round. Either way the voltage changes must total to zero.

In any case in the first lecture Lewin makes the point that you can't always tell which way the current will flow. But if you get the direction of the current wrong all that happens is all your voltage differences change sign and they still add up to zero.

Improve this answer
$\endgroup$
2
  • $\begingroup$ In the LR Circuit video, he is saying the sum of voltage equals the chage in flux. In the convention he adapts LR circuit video, he sums up the voltages equal to the negative rate of change of LI'. $\endgroup$
    – Lemon
    Commented Aug 8, 2012 at 17:59
  • $\begingroup$ Ldi/dt is just the voltage drop/rise across the inductor, so he's just adding voltages. Putting the Ldi/dt on the other sign of the equals sign is just a rearrangement of the equation. I guess he's doing it this way so he can calculate the inductor voltage drop. $\endgroup$
    – John Rennie
    Commented Aug 9, 2012 at 5:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.