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Voltage drop across resistor

  1. May 23, 2013 #1
    I'm curious the amount the voltage would drop across a resisitor.
    If you know the power going into the resistor and the maximum voltage allowed is say, 50V. How would that resistor effect the voltage? I understand the voltage would decrease, and when substituting and solving equations the new voltage would be sqrt(Power*Resistance). Is there anything else someone could add so I can better understand this? Thank you.
  2. jcsd
  3. May 23, 2013 #2


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    Odd wording of a question....but I'll try.

    KVL = Kirchoffs Voltage Law states that the sum of the voltages across the loads in a loop must equal the source.

    If you are saying you have a 50 volt source and one resistor, then all 50 volts drops accross that resistor.
    The current is then simply V=IR.

    Power is based off of V=IR....not the other way around.
    P=IV, so once the I and V are established from V=IR...there ya go. Or you can use P=I^2*R or P= V^2/R

    If there is more than one resistor, say two....then voltage division is used.
    The voltage across the resistor you are interested is R1/(R1 + R2) * the voltage source.
  4. May 23, 2013 #3


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    For a proper answer, you would need to define the circuit (a diagram?). It's not clear what your actual question is.

    But, if you apply 50V from an ideal power supply across a resistor then that defines the volts and the power. If the supply is not ideal then you'd have to give more details (as above)
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