$ P = \frac {V^2}{R} $ derivation confusion I have always had a confusion of why we use $ P = V I $ or $ P = I^2 R $ and not $ P = \frac {V^2}{R} $  for relating to power loss due to heat in high tension lines. I know there are a lot of questions here but it still doesnt seem clear. I realize that the supply voltage and voltage drop are different things. But then, how did the equation come up in the first place? Lets say Supply voltage is $ V_s$ and the voltage drop is a $ V_d $
So this would give, $ P = V_s I $ (Joule's Law) and $ V_d = I R$ (Ohm's Law)
Wikipedia says we get $P = \frac {V^2}{R} $ by combining these both. How can you do that when $ V_s$ and $ V_d $ are two different parameters? Can we just combine them both in this case too- the high voltage power lines case? I we do, what is the $ V $ that is to be used to calculate the power loss using the $P = \frac {V^2}{R} $ formula?
Could anyone explain with hypothetical values?
 A: 
How can you do that when Vs and Vd are two different parameters?

One must keep track of the variables.  The power delivered to a resistor is
$$P_R = V_R \cdot I_R = V_R \left( \frac{V_R}{R} \right) = \frac{V^2_R}{R}$$
where I have subscripted the variables so it is clear that the voltage and current variables are the voltage across and current through the resistor.
The power delivered by a source is
$$P_s = V_s \cdot I_s $$
Since the transmission lines have a non-zero resistance $R$, there is a voltage across due to the source current through
$$V_d =  I_s R$$
and an associated power loss $P_\mathrm{loss}$
$$P_\mathrm{loss} = V_d \cdot I_s = V_d \left( \frac{V_d}{R} \right) =  \frac{V^2_d}{R}$$
we could have also written
$$P_\mathrm{loss} = V_d \cdot I_s = (I_s R) I_s =  I^2_sR$$
Now, the voltage across the load is
$$V_L = V_s - V_d$$
and so the power delivered to the load is
$$P_L = V_L \cdot I_s =  (V_s - V_d)I_s = P_s - P_\mathrm{loss}$$
as expected.
A: Yeah, I had this confusion too. But know that$$P=\frac{v^2}{R}$$ for RESISTOR circuits only. Actually Power for any circuit is (Instantaneous power more precisely) $$P=VI$$ Here's how:
We know that $P=\frac{dW}{dt}$
Lets first calculate dW. dW is the elemental amount amount of work done on elemental charge dQ in moving it through a potential difference of V across the battery.Therefore
$$dW=VdQ$$
$$P=VI$$ (since I=$\frac{dQ}{dt}$)Also note that $P=I^2R$ for resistor circuits only.
A: P = Vs I and P = Vd I
, both are correct.
Former means Power consumed by circuit and latter power consumed by different circuit components like wires.
Or
$V_{s} I = V_{1} I + V_{2} I ...... V_{n} I$ 
Where V1 and Vn are voltage across different circuit components.
If you use $P_{s} = V_{s} I$ and $V_{s} = IR$
You get , $P_{s} = \frac{V_{s}^2}{R}$ which is power consumed in circuit.
If you use $P_{d} = V_{d} I$ and $V_{d} = IR$
You get , $P_{d} = \frac{V_{d}^2}{R}$
which is power consumed by circuit component.
A: I think that by Kirchhoff's voltage law the voltage drop $V_d$ must equal the supply voltage $V_s$. So $V_d=IR=V_s$ and hence the result.
