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: Simple Combination Circuit, Find Current

  1. Feb 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Combinations of Resistors

    Im not sure if the image will show up. If the image doesn't show up; it is similar to this picture http://www.ceb.cam.ac.uk/data/images/groups/CREST/Teaching/impedence/sparal6.gif
    In the link's picture R1 is in the correct spot, R2 is R6-5-4 and R3 is R3-2. Sorry if there is any confusion.

    Given the circuit shown below, the power supply is set to a voltage V=18.3 V, R2 = 33.2 Ω, and R3 = 46.5 Ω. Suppose you place an ammeter into the circuit and find that the current through R2 is 0.263 A. Answer the following questions:

    A)What is the voltage across resistor R2? 8.74 V This is correct
    B)What is the current through resistor R3? 0.188 A This is correct
    C)What is the current flowing through R1? 0.451 A This is correct

    D)What is the resistance of resistor R1? I NEED HELP ON THIS ONE

    2. Relevant equations
    V2 = R2 * I2
    I3 = V3 / R3
    I1(series)= I2 + I3 (Parallel)
    R1 = V1 / I1

    3. The attempt at a solution
    A) V2 = R2 * I2 = 33.2 *.263 = 8.74
    Voltage going through resistor 2 is 8.74 V

    B) I3 = V3 / R3 = 8.74/46.5 = .188 A
    Current going through resistor 3 is .188 A

    C) I1(series)=I2 + I3 (parallel) = I1 = .188 + .263= .451 A
    Current going though resistor 1 is .451 A

    R1 = V1 / I1 = 18.3/.451 = 40.57 ohms

    This answer is not correct and I am not sure why. I thought that the way to find resistance was R= V/I. I solved for the correct current in the previous problem and we are given that V= 18.3 V in the wording of the problem.

    Any help would be greatly appreciated!
  2. jcsd
  3. Feb 14, 2014 #2


    User Avatar

    Staff: Mentor

    Your image is a bit hard to see, given that it's blue lines on a black background (for my viewer at least). So here I've rendered it as I understand it (note that you can upload gif files if you use the Advanced editing panel):


    Note that your voltage V1 = 18.3 V is not across R1 alone. Try writing KVL around a loop with R1 in it.

    Attached Files:

  4. Feb 14, 2014 #3
    So, the voltage that they gave is for the total circuit then?
    If that's the case then I can subtract the V1(8.74) and V2(8.74) from it.
    Vtot = V1 + V2 + V3
    18.3 = V1 +8.74 + 8.76
    V = .82 volts going though R1

    Then would you take R1 = V1 / I1 = .82/.451 = 1.818 ohms ?
  5. Feb 14, 2014 #4


    User Avatar

    Staff: Mentor

    No, that's not correct.

    Okay, first, voltage does not go "through" a component. Voltage is across components. It's the potential difference from one end to the other. Second, the voltages across parallel components are the same voltage. You don't sum them! They are not in series.

    So when I suggested that you write KVL for a loop including R1, you cannot have both R2 and R3 in that loop, you can only have one or the other. Try tracing a loop through the voltage source and R1...
  6. Feb 14, 2014 #5
    I think I get it. The current would only go down one of the parallel resistors so I shouldn't have subtracted the voltage lost across the resistors for both R2 and R3.
    Vtot= V1 + V2
    18.3 = V1 + 8.74
    V1 = 9.56 Volts lost across R1

    R1 = V1 / I1 = 9.56 / .451 =21.2 ohms
  7. Feb 14, 2014 #6


    User Avatar

    Staff: Mentor

    Huzzah! Yes, that looks good! :smile:
  8. Feb 15, 2014 #7
    All right, thanks so much for your help! I really appreciate it!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted