Understanding Voltage Drops and KVL in Circuit Analysis

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The discussion revolves around the application of Kirchhoff's Voltage Law (KVL) and the interpretation of voltage drops across resistors. It clarifies that a resistor does not generate power; rather, a voltage source provides power, with internal resistance consuming some of it. The confusion arises from the treatment of voltage drops, particularly when using the equations V = RI and V = -RI, leading to potential misinterpretations of polarity. It is noted that even if initial assumptions about voltage drops are incorrect, KVL can still yield correct results, as a negative value indicates the actual direction of the voltage drop. Understanding these concepts is crucial for accurate circuit analysis.
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p here stands to power (v in power equation is magnitude), in addition to this : is it also true that in the first one : V = RI (V upper - V lower) and in the second one : V = - RI (V lower - V upper) ?
(current is magnitude in all previous equations)
 

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mohamed el teir said:
second one : V = - RI (V lower - V upper) ?

It's not clear what your question is, but on the quoted part, no. A resistor does not generate power. You can have a voltage source with an internal resistance value, but the voltage source is what is generating the power, and the internal resistance of the power source consumes some of that power as the load current flows through it.
 
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so is this example wrong ? it treated v2 as -RI and v2 appeared to be negative at the end, which stated that resistance + and - ends are reversed i think
 
mohamed el teir said:
View attachment 89670

so is this example wrong ? it treated v2 as -RI and v2 appeared to be negative at the end, which stated that resistance + and - ends are reversed i think

I think they are just trying to illustrate how if you choose the voltage drops incorrectly initially when using KVL, you still get the right answers in the end (because you get a negative number which means the voltage drop is actually in the other direction).
 
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