1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kirchoff's circuits and the Electric Field

  1. Aug 6, 2012 #1
    1. The problem statement, all variables and given/known data


    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
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Aug 7, 2012 #2


    User Avatar

    Staff: Mentor

    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.
  4. Aug 7, 2012 #3
    No I am using $$\vec{E}$$ in my picture as the electric field.

    Also, if you go to this site


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook