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Homework Help: Voltage drop across resistors

  1. Feb 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine the voltage at each labelled point with respect to ground

    2. Relevant equations

    3. The attempt at a solution

    I'm doing a practice problem and I can't seem to get the right answer.

    I just want direction in the way to solve it, these are my "ideas" so far is to
    Find Req, find source current, use current divider rule and find current in each branch. Then calculate the voltage at the points(I assume its just current*resistance of the resistor in front of the point?) I have no idea if this is the way I'm meant to solve it but I tried it anyways.

    So left side total resistance is 650,
    Right side is 1100,
    top is 1000
    all are in parallel? Or is 1000 in series with two parallel? I assumed in parallel in my calculations.

    650ohm||1100 ohm = 393.93 ohm
    393.93 ohm||1000 ohm=282.6 Req
    50V/282.6 Req=0.1769 is

    0.1796 mA*1000/2393.93 ohm = 0.073907 mA for current through the entire circuit excluding top

    0.073907 mA * 1100ohm/1750 ohm=0.0464 current through left side

    voltage at B = 0.0464*100ohm=4.64 V

    The answers are meant to be
    A=3.85 V
    B=42.31 V
    C=50 V
    D=5 V

    Attached Files:

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  3. Feb 9, 2015 #2


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    I think that's the wrong approach (edit: although it would work).

    The three branches (left, right, top) are indeed all in parallel (between node C and earth/0V) but there is an easier way to work out the current in each branch. Have another look at the circuit. It's possible to work out the voltage at C without doing any calculations.

    Edit 2: Actually you don't need to work out any of the currents. Start by working out the voltage at node C. Then have you heard of the potential divider rule?
    Last edited: Feb 9, 2015
  4. Feb 9, 2015 #3


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    If you want to do it your way...

    = 408.6 not 393.93
  5. Feb 9, 2015 #4
    I tried calculating with the changed value of 408.6 A


    50/290ohm=0.172 is
    0.164 * 1100/1750=0.10335
    0.1033*50=5 V

    I can't seem to make it work calculating currents, I looked into voltage divider rule..

    so 50 V * 50/650 = 3.84 A(correct) for A
    50 V* 550/650= 42.3 A(correct) for B
    50 V* 100/1100 = 4.54 V(not correct) for D
    and C is 50 V

    The answer for D is meant to be 5, I'm not sure if I made an error or if they simply rounded the value up.
  6. Feb 10, 2015 #5


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    That's by far the fastest way to solve the problem.
    I agree with your answer for D. I've no idea why they rounded that to 5V yet they didn't round the voltage for A to 4V.

    Which branch is that for? I have assumed the top one ...

    0.172mA total current is correct. You made a mistake in the next line. It should be...

    ITop = ITotal * (650||1100) / {650||1100 + 1000}
    = ITotal * 408 / {408 + 1000}
    = ITotal * 408 / 1408
    = 0.172mA * 408 / 1408
    = 50uA (edit: well ok it's actually 49.8uA)

    As a check..
    50V/1MOhm = 50uA

    Sorry for all the edits I made to this post.
    Last edited: Feb 10, 2015
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