/sci/ - Science & Math

>>16126206
if you double the voltage across a resistor, you also double the current through it, hence the voltage squared proportionality. basically you get [math]P=\frac{V^2}{R}[/math] by combining [math]P=VI[/math] and [math]V=IR[/math].
however, this naive application of ohm's law doesnt perfectly apply when we're trying to transport power. if we double the voltage, we actually want the current to *halve* (roughly) in order to maintain constant power supplied. we do this using transformers (picking the appropriate transformer ahead of time, knowing what your input voltage will be). obviously if you have two resistors in series then we wont be able to do this, but for a constant load (a neighborhood, eg), transforming the voltage down to 120V will ensure that the load draws the same power no matter what voltage we throw in the transformer.
let [math]V_S[/math] be the voltage the plant is providing, [math]V_L[/math] be the voltage going into the load transformer, [math]P_L[/math] be the power the load is consuming, and [math]R_C[/math] be the resistance of the transmission lines. we know that [math]P_L=V_LI=(V_S-IR_C)I[/math] must remain constant. what happens if we halve the current [math]I[/math]? since [math]V_S[/math] is the only other non-constant in the equation, it will have to change (increase, namely) in order to compensate.
[math]\displaystyle
P_L=(V_S-IR_C)I \\
P_L=V_SI-I^2R_C \\
V_S=\frac{P_L+I^2R_C}{I} = IR_C + \frac{P_L}{I}
[/math]
the left term is the voltage drop across the transmission lines, and the right term is the voltage drop across the load. notice that halving the current will cause the source voltage to increase if the voltage drop across the load is bigger than the transmission lines (it is), but the voltage wont literally double unless [math]R_C[/math] is zero. if we halve the current but *not quite double* the voltage of the source, the plant must now necessarily be outputting less power while still supplying the same power to the load.