Kirchoff's circuits and the Electric Field

AI Thread Summary
The discussion revolves around the relationship between electric fields and circuit analysis, specifically using Kirchhoff's laws. A video suggests solving circuit problems by following the direction of the electric field and current, leading to questions about the validity of this approach compared to traditional methods. The integral equation for electric fields and potential differences is debated, particularly in non-closed loops, raising concerns about its applicability in resistor circuits. The conversation also touches on the interpretation of voltage across components in relation to current direction. Overall, the thread explores whether the electric field approach aligns with established circuit theory or presents discrepancies.
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Homework Statement



circuitlry.jpg


I have drawn the direction of the electric field in the picture.

I saw this on a video on youtube where this guy solves circuit problems solely on looking at the direction of the electric field. Basically he follows the current and the electric field

$$-\varepsilon_1 + IR_1 + \varepsilon_2 + IR_2 = 0$$

What is the different theory behind the approach? Is it a coincidence that they both will give the same answer or is one of them wrong? For instance between $$b$$ and $$c$$, the electric field and the current is in the same direction so we have

$$\int_{b}^{c} \mathbf{E}\cdot d\mathbf{s} =\int_{b}^{c} Eds = -\Delta V$$


Which means the potential should be minus, but "according to Ohm's Law, the electric field and the current are in the same direction, so we get +IR" and in the battery we go "against the electric field, so we get $$-\varepsilon$$.


The circuit the guy on youtube () does involves an inductor, but I thought I could apply the same principle to regular resistor circuits. Does the equation $$\int_{b}^{c} \mathbf{E}\cdot d\mathbf{s} = -\Delta V$$ no longer hold? Note that the integral isn't a closed loop.

Thank you for reading
 
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In your Kirchoff's analysis, you are using symbol E for the voltage across each component, so E has units of volts. Once you mark in a loop's current arrow, you can determine the voltage E across each component because to cause current to flow in a resistor in the agreed direction, one particular end of that component must be positive relative to the other.

In ∫E.ds the term E is the voltage gradient, in volts/metre. You don't know E in the circuit, nor s, so this equation is of no use here.
 
No I am using $$\vec{E}$$ in my picture as the electric field.

Also, if you go to this site

http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/

and open the file "Non-conservative Fields - Do Not Trust Your Intuition". On page 2/3 in the pdf, you see him does the same thing again. For the left loop he has $$+I_1 R_i$$ even though he assumed the direction of $$I_1$$ is clockwise and he traverses clockwise
 
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