How Does a Resistor Affect Voltage Drop in a Circuit?

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bhsmith
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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.
 
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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 across 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 you 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.
 
bhsmith said:
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.

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)