How Does a Resistor Affect Voltage Drop in a Circuit?

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The voltage drop across a resistor in a circuit is determined by Ohm's Law (V=IR) and Kirchhoff's Voltage Law, which states that the total voltage in a loop equals the sum of the voltage drops across each component. When a 50V source is applied to a single resistor, the entire voltage drop occurs across that resistor. If multiple resistors are present, voltage division can be used to calculate the voltage across each resistor based on their resistances. Understanding power in relation to voltage and current is crucial, as power equations (P=IV, P=I^2R, P=V^2/R) help clarify how voltage and current interact. A clear circuit diagram would enhance the understanding of specific scenarios.
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)
 
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