Finding resistor's voltage with current and voltage sources present

  1. JJBladester

    JJBladester 286
    Gold Member

    What is the voltage across R2? You should convert the 4 V source and resistor into a current source as your first step.

    [​IMG]

    The first thing I did is create a second current source out of the 4 V voltage source and the series resistor R2. I called it I_2.

    I_2 = E/R_2 = 4V/100Ω = 40 mA

    This gives us:

    [​IMG]

    We can combine the parallel current sources I_1 and I_2:

    I = I_1 + I_2 = 10 mA + 40 mA = 50 mA

    [​IMG]

    By the current divider rule,

    I_R2 = (R_T*I)/R2 = (20Ω*50mA)/100Ω = 10 mA

    V_R2 = I_R2*R2 = 10mA * 100Ω = 1 V

    However, when I run a simulation of the original circuit using MultiSim, I get 3 V through R2. Where am I going wrong?
     
  2. jcsd
  3. SammyS

    SammyS 8,876
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    If you get 3 V across R2 in the original circuit, that means that there voltage dropped across R1 in the original circuit is 1 V, which makes sense.

    The difficulty appears to be that you are assuming the resistor, R2 in the original circuit has the same voltage across it as the resistor, R2, in the last two circuits.
     
  4. JJBladester

    JJBladester 286
    Gold Member

    I'm stuck. I think I understand what you mean that R2 in the new (simplified) circuit cannot be treated the same as R2 in the original circuit. Am I correct up to the point where I found a combined 50 mA current source acting on two parallel resistors R1 and R2?

    If so, how do I get back to finding the voltage for R2 in the original circuit.
     
  5. SammyS

    SammyS 8,876
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    Find the voltage across R1 in the simplified circuit. It hasn't been changed. Then use that to find the voltage across R2 in the original circuit.
     
  6. NascentOxygen

    Staff: Mentor

    [strike]There are two mistakes in this line, yet you end up with 10 mA and that's the right answer[/strike][​IMG]
    Oops, my mistake. What you are using is not what I think of as current divider rule, but it's okay.
     
    Last edited: Jul 17, 2012
  7. JJBladester

    JJBladester 286
    Gold Member

    So I can't use the "new" R2 even though it is the same R2 as before?

    Anyway, using your advice to find the voltage across R1 in the simplified circuit, I've found the voltage across R2 using Kirchhoff's Voltage Law on the original circuit:

    [​IMG]

    Using the current divider rule:

    [tex]I_{R1}=\frac{(20\Omega )(.05 A)}{(25 \Omega )}=.04A[/tex]

    [tex]V_{R1}=\left (I_{R1} \right )\left (R_1 \right )=\left ( .04A \right )\left (25\Omega \right )=1V[/tex]

    E - VR2 - VR1 = 0
    VR2 = E - VR1 = 4V - 1V = 3V
     
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