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Elctric circuits

  1. Nov 13, 2004 #1
    please help i'm am having lots of trouble undersanding electric circuits. my question says " use kirchhoff's loop rule and ohm's law to derive the equation for the equivalent resistance of four resistors connected in a series. concisely explain each step. " so i know that the eqaution for equivalent resistance is R=R1+R2+R3+R4 and that kirchhoff's loop rule is EV=0 and that ohm's law is I^2R. but how do i use that to get the equivalent i'm so confused. please help.
  2. jcsd
  3. Nov 13, 2004 #2


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    Let's consider this case for a really simple circuit with one loop consisting of an ideal voltage source and your four resistors, all in series.

    First lets look at Kirchoff's Loop Rule, what we usually call Kirchoff's Voltage Law (KVL) in my class. It states that the sum of the potential differences around the loop must be equal to zero. In this case, that means that the voltage across the source is equal to the sum of the potential drops across the four resistors:

    [tex] V_s - (V_1 + V_2 + V_3 + V_4) = 0 [/tex]

    Let's talk about the current in the circuit. This being a loop (all elements in series), the current must be the same everywhere in the circuit. Detectors at any two different points along the loop should measure the same rate of charge per second flowing past, otherwise charge is piling up somewhere, or charge is not conserved. Since we have a steady current, and charge is conserved, the current is constant and equal to [itex] I [/itex] everywhere.

    Ohm's law tells us two things:

    1. This total current [itex] I [/itex] in the circuit is equal to the voltage of the source divided by the total (or equivalent) resistance of the circuit:

    [tex] I = \frac{V_s}{R_{eq}} [/tex]

    [tex] \therefore \ \ V_s = R_{eq}I [/tex]

    2. The potential difference (ie voltage drop) across each resistor is given by the current through it times its resistance:

    [tex] V_1 = R_{1}I [/tex]
    [tex] V_2 = R_{2}I [/tex]
    [tex] V_3 = R_{3}I [/tex]
    [tex] V_4 = R_{4}I [/tex]

    Substitute the results of 1 and 2 (from Ohm's law) into the the very first equation (from KVL):

    [tex] V_s - (V_1 + V_2 + V_3 + V_4) = 0 [/tex]

    [tex] R_{eq}I - (R_{1}I + R_{2}I + R_{3}I + R_{4}I) = 0 [/tex]

    Cancel out I and rearrange terms:

    [tex] R_{eq} = R_{1} + R_{2} + R_{3} + R_{4} [/tex]

    That's the result we were trying to prove. :smile:
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