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Finding output voltage given two resistances and voltages

  1. Dec 4, 2017 #1
    • Member advised to use the provided formatting template when starting a new thread in a homework forum.
    Given two sets of load resistances and output voltages:
    Rload = 108 ohms - Vout = 9.6V
    Rload = 10 Ohms - Vout = 2.3V

    Find the source voltage and the output voltage when R1 and R2 are equal.

    When the resistance is very large, the voltage seems to be around 9.6. My assumption was that then the source voltage would be 9.6. However, that seems to be incorrect. I know how to solve the rest of the problem, by using the voltage divider equation ((Vin(R2/(R1+R2))) = Vout), but this initial part is very difficult for me.
     
  2. jcsd
  3. Dec 4, 2017 #2

    scottdave

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    What you have is 2 equations and 2 unknowns. The first equation: Vout is 9.6 Volts and R2 is Rload = 108homs.
    So you have
    $$9.6 ~V = \frac {(Vin)*(108Ω)} {R1 + (108Ω)} $$
    Then with the other situation, you have an equation with Vin and R1. You can now solve for Vin and R1.
    Now that you know Vin and R1, they want you to set R2 equal to R1.
     
  4. Dec 4, 2017 #3
    Is there any other way to go about this process, or is systems of equations the only way to do it.
     
  5. Dec 4, 2017 #4

    gneill

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    Your problem statement is not clear (and you didn't employ the homework help formatting template).
    What are R1 and R2? You haven't clearly defined the circuit.

    In general, when you are given two or more scenarios with given resulting values and you are asked to find the resulting conditions for another situation, then you have a problem that will require solving a system of equations.
     
  6. Dec 4, 2017 #5
    There is an output. Connect to ground with a 108 Ohm resistor and measure 9.6 Volts across the resistor. Connect to ground with a 10 Ohm resistor and see 2.3 Volts. We are going to solve for the thevenin equivalent output open circuit Voltage and the source impedance.

    9.6 V= Vo 108 / 108 + Ro ....per Voltage divider formula
    2.3 V = Vo 10 /10 + Ro ...per Voltage divider formula
    108 + Ro = Vo 108 / 9.6V.... rearranging terms
    10 + Ro = Vo 10 / 2.3 V ... rearranging terms
    108 - 10 = Vo 108/9.6V - Vo 10/2.3 ..... subtracting equations and Ro cancels out
    98 = [108/9.6 - 10/2.3] Vo .... subtraction and distributive property
    Vo = 98/[108/9.6 - 10/2.3] = 98[11.25 - 4.35]= 14.2

    9.6 V /108 Ohms = 0.089 Amps ... per Ohms law
    [Vo - 9.6Volt ] /Io = Ro
    [14.2 Volts - 9.6 Volts ]/ 0.089 Amps = Ro = 51.6 Ohms

    What is R1 and R2 that are equal?

    I am beginning to suspect that Ro is a 50 Ohm source and the real question is that if we load it with a 50 Ohm scope and probe we will see 7.1 Volts
     
  7. Dec 5, 2017 #6

    CWatters

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    Writing and solving two simultaneous equations is the accurate way but you could also use a graphical method....

    1) Calculate the output current for the data points you have (eg I = V/R)
    2) Plot a graph of I vs V.
    3) Extrapolate that to find the open circuit voltage (eg V at I=0).

    The open circuit voltage looks to be about 14V

    You can also do the same method to find the short circuit current.
    Then use the open circuit voltage and short circuit current to calculate the internal resistance.

    load line.jpg

    PS I think I used the right voltages but excel rounded them.
     
  8. Dec 5, 2017 #7

    OmCheeto

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    From scottdave's response, I interpolated the following:

    2017.12.04.pm.pf.hw.problem.png
    Though, now we have at least 4 different annotations for the "internal voltage" of the supply.
    Hence, I feel your pain.

    As far as I can tell, I didn't use "a system of equations".
    Though, in the end, we all came up with the same solution, so we may have a different definition for that phrase.

    hmmm....
    "load" Ω = "load" Ω - Vout = 9.6V ?

    From my meager understanding of electrical stuff, you can't mix ohms and volts like that.
    This is all I know: V=I*Ω

    ps. My method of solving the problem is available, upon request. Though, I will not give the answer.
     
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