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Calculating Maximum Voltage in a circuit.

  1. Jun 5, 2013 #1
    A system is constructed as illustrated. Each resistor is rated at 0.50 watts. What is the maximum Voltage that can pass through the circuit without burning out a resistor.

    http://i.imgur.com/uowKVxU.jpg

    I'm assuming you'd have to narrow it down to which resistor would burn out first, however I'm not certain how you'd do that.

    Thanks in advance :)
     
  2. jcsd
  3. Jun 5, 2013 #2
    First calculate the current passing through each resistor.

    Hint: The Power in each resistor should be less than rated power.
     
  4. Jun 5, 2013 #3

    CWatters

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    What darkxponent said.

    Pick any starting voltage (eg 1V). Work out the current & power in each resistor.

    In this linear circuit the resistor that dissipates the most power will allways do so.

    Write an equation for the power dissipated in that resistor in terms of V.

    Solve for V.
     
  5. Jun 5, 2013 #4
    Using the formula P= v^2/r

    0.5 = V^2 / 300
    V =12.25

    0.5 = V^2 / 600
    V = 17.3

    Since these two are in series, they will both receive the same voltage. So 12.25 is the maximum Voltage that can be present after the first resistor.

    0.5 = V^2 / 250
    V= 11.18

    So the maximum Voltage that can go through the first resistor is 11.18V, which is less than the 12.25V that the smallest parallel resistor can take. Kind of odd, and I don't think it's that easy.



    Also, the total resistance of the system is:

    1/rt = 1/300 + 1/600
    Rt= 200 +250
    Rt=450
     
    Last edited: Jun 5, 2013
  6. Jun 5, 2013 #5

    haruspex

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    I'm not exactly sure what darkxponent was advising. You seem to have interpreted it as finding the limiting current for each resistor, which fits with darkxponent's hint. OTOH, CWatters is suggesting setting the voltage to some unknown V and finding all the currents in terms of V. That does seem a more fruitful approach.
    Another possibility is to work through the same sequence as you have in assessing the net resistance, but at each stage determining which resistor is the limiting element. E.g. with R2 and R3 being in parallel, the one with the lower resistance will draw the greater power.
     
  7. Jun 5, 2013 #6
    Could you have a look at what I've done above? Revamped the post while you were quoting me :).
     
  8. Jun 5, 2013 #7
    Follow these steps

    1) Take the given voltage as V volts.

    2) Find the current in each resistor as a function of V.

    3) Follow the hint i gave in first post.
     
  9. Jun 5, 2013 #8
    So if I relate the two currents, the current of the two parallel points must equal the current along the series portion.

    V= I * R

    If we leave express I in terms of V, we get I = V/250

    And that I must equal the other two Is added together, which is V/600 + V/300 or 3V/600 or V/200 (which is the Rt of the parralel). So if those I are equal, then you get that the Voltage total is both Voltages added, so 450v, where V can be calculated by using the formula p= V^2/r.
    0.5 = V^2 / 450
    V= 15V
    15 * 450 =6750 V

    There's no way that's right. can you tell me where I went wrong?
     
  10. Jun 5, 2013 #9

    haruspex

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    parallel
    250? You want the resultant of R2 and R3, right?
    What is the logical relationship between these two voltages you have calculated? How is it interesting which is larger?
     
  11. Jun 5, 2013 #10
    Ya, parralel sorry. I did my calculation for parralel, so it's all good.

    For the calculating both together, the Voltage will be the same through both lines, calculating the maximum for each line gives you the maximum for both, and you can't exceed the maximum of one of resistances.

    Finally, the significance of calculating all of those gives me my maximum because if I can only run 11v through the first resistor, and that is the lowest voltage possible, then it is my maximum Voltage before I break that resistor.
     
  12. Jun 5, 2013 #11
    This is not correct. you have to do some circuit analysis. Current in each resistor is different as these are not in series. Take three different values of current for each resistor. Now the current passing through R1 will divide into parts, one goes into R2 other in R3. (How is it going to divide?)

    Now the values of current are obtained. you will get it as a function of V. Just calculate the power in each resistor. Now power in each resistor should be less than the rated power.
     
  13. Jun 5, 2013 #12
    So if I do three different one:

    R1= V/250

    R2= V/300

    R3 = V/600

    2/3 of the current will go through R2, where as 1/3 of the current will go through R3.

    Beyond that, I'm unsure what you mean.
     
  14. Jun 5, 2013 #13
    That is wrong. In the Ohms Law V= IR. V is the voltage across the particular resistance taken and I is the current through that particular resistance. In the above problem Voltages across the Resistances is not same as the resistances are not in parallel. R2 and R3 are in parallel but R1 is not. But you don't need to find the voltages. Finding current in each resistor is easier.

    That's right. you are very close. Now you only need to find the current in R1.

    Hint: The current in R1 is the net current in circuit.
     
  15. Jun 5, 2013 #14
    Yea, so I at R1 = I at R2 + I at R3.

    I'll be back in a bit, but I think I've got it. Answer I got is 11.2 + 8.9, where the first us the first voltage at r1 and the 2nd voltage is the common voltage between r2 and r3. I'll write it out fully when I get home. Also, R1 is the limiting, and I used the I in there to find the I, and this the Voltage, in the other two wires.
     
  16. Jun 5, 2013 #15

    haruspex

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    Yes, that looks right. Not sure how you got it, but here's a way working "bottom up":
    Since R2 and R3 are in parallel, consider common voltage across R2||R3. This is 12.25, as you showed.
    Since R2||R3 is in series with R1, consider common current. Max current through R2||R3 = 12.25/200; max current through R1 is sqrt(0.5/250), etc.
     
  17. Jun 6, 2013 #16
    P=I^2 * r
    0.5 = I^2 * 250
    I = 0.048

    V1 = I * R
    = 0.048 * 250
    = 11.18V

    V2= (2/3 * i) × R
    = 2/3 * (0.048) * 300
    = 8.944V

    V3 = (1/3 * i) × R
    = (1/3) * (0.048) *600
    = 8.944V

    V2 = V3, which is true in parallel.

    V = V1 + (V2 or V3)
    V = 11.18 + 8. 94
    V = 20.1 V

    Correct?
     
  18. Jun 6, 2013 #17

    haruspex

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    That's all true, but to work that way you had to know in advance that R1 was the limiting resistor, no?
    Anyway, I was only trying to offer a method in line with your own attempt. I much prefer CWatters' approach:
    let the voltage across the system be V
    resistance of system = 450 Ohm, so total current = V/450.
    power in R1 = (V/450)2250 = V2/810
    power in R2 = (V/450*2/3)2300 = etc.
    power in R2 = (V/450*1/3)2600 = etc.
    then see which is the greatest; set that one equal to 0.5.
     
  19. Jun 6, 2013 #18
    Yea, I used that method to find the limiting resistor (the first one), then did my method to get the actual value :P

    Thanks
     
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