Kirchoff's circuits and the Electric Field

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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.