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Homework Help: Load resistor in circuit: Find the no-load voltage.

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img62/5467/homeworkprob14.jpg [Broken]

    A. Find the no-load voltage of V0

    B. Find Vo when RL is 450 kΩ

    C. How much power is dissipated in the 30 kΩ resistor if the load terminals are accidentally short circuited

    D. What is the maximum power dissipated in the 50k resistor?

    2. Relevant equations
    V = IR

    Voltage Division:
    (Voltage across series resistor) = [(resistance) / total series resistance)](total input V)

    Current Division (for 2 parallel resistors):
    (current across parallel resistor) = [(other resistor) / (sum of parallel resistors)](total incoming current)]

    3. The attempt at a solution

    I forgot how to deal with load-resistors, and not sure how they apply to a circuit like this.

    I don't need help with part B because, well, that's straight forward, and you find parallel resistance first between the load and other one by V

    because: (1/450 + 1/50)^-1 = 45Ω

    then using voltage division when you have only 30k and 45k in series:

    V knot = [45Ω / (45 + 30)Ω] (120V) = 72V

    Still need help for other parts though
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 18, 2013 #2

    Simon Bridge

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    The load resistor is treated exactly the same as any other resistor.
    I cannot help with the circuit without seeing it though.
  4. Jan 18, 2013 #3


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    Staff: Mentor

    I'm able to see the circuit, Simon. Is ImageShack blocked by your firewall? Here is the URL if it helps:

    http://imageshack.us/a/img62/5467/homeworkprob14.jpg [Broken]

    Last edited by a moderator: May 6, 2017
  5. Jan 18, 2013 #4

    Simon Bridge

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    Ah - I can see it now.
    It didn't show up when I posted.

    The questions A-D can be understood by redrawing the circuit diagram each time according to the description. i.e. for A, just erase the load resistor part. for C - how does "accidentally short circuiting the load terminals" change the circuit diagram?

    I think there is an equation missing too - relating power to voltage and resistance.
  6. Jan 19, 2013 #5
    Yeah I kept editing my post at first to change it to correct info because I got this mixed up with another problem (although similar). Sorry about that. So anyways...

    That would be P = V^2/R added to the equations then.

    For part C, I set the circuit to be the same as in part A (with just 30k and 50k in series) and then did voltage division for the 30k resistor:

    V (30k) = (30k / 30k+50k)(120V) = 45V


    P = V^2 / R = (45)^2 / 30000 which makes it 0.1125 W but this isn't the same as the given correct answer.

    The correct answer would mean that V = 120V when doing P = V^2 / R which is (120)^2 / 30000 = 0.48W

    Assuming the given correct answer for part C really is correct, then why do you have to use 120V for P = V^2 / R?

    I got part D though, just P = 75^2/50000 and then P = 0.1125 W. 75 V because that's the potential across the 50k resistor found earlier in part A.
  7. Jan 19, 2013 #6

    Simon Bridge

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    That's what you get if the load terminals are open circuit.
    Which is part A.

    The question wants:

    C. How much power is dissipated in the 30 kΩ resistor if the load terminals are accidentally short circuited

    ... a short circuit means you put a wire across the load resistance.
  8. Jan 19, 2013 #7
    Alright, I admit I didn't know the difference (now I'm sad, heh). I was never really told about that (enough) in my physics class.

    I need to brush up on this then too, so if you have another wire that bypasses a resistor that means you can disregard potential drop across that resistor then, right (because here the wire across the load resistor would then also bypass the 50k ohm resistor too, right)? Assuming it's the path of least resistance, though.

    Thanks for all your help, by the way.
  9. Jan 20, 2013 #8

    Simon Bridge

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    Many people have heard the term "short circuit" but don't realize what is involved.
    "short circuit" is a wire, "open circuit" is a gap... usually means "no load".

    So, in part C, the circuit only has one resistor... well done.

    You notice that you can draw a better circuit diagram freehand than with a computer?
    Price of progress I guess - LaTeX comes to the rescue there too:
    ... don't think PFLatex supports it though.
  10. Jan 21, 2013 #9
    So that would be a yes then (in the sense that you really can disregard potential drop across a resistor anytime if the current can go around the resistor in another path (short circuited))

    And there's no current through short-circuited resistors either then?

    and I get this error when trying to download the circuitikz files:

    Thanks again anyways though.
  11. Jan 21, 2013 #10

    Simon Bridge

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    Sometime the term "sort circuited" is used incorrectly to refer to a resistor (or other component) that has blown up (or otherwise fatally damaged) as a result of a short circuit - somewhere.

    A blown up resistor is likely to be an open circuit.
    Either way - no current.

    Rats about the circuitiks files ... you are right, link seems to be out of date.
    I get mine with gnu/linux and Mac users get it with MacTex by default, I keep forgetting how annoying Windows can be.

    There is a ctan page about them.

    ... review with links.
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