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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?

Qmechanic
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Polisetty
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4 Answers4

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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.

Alfred Centauri
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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.

oops
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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.

Anubhav Goel
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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.

Procyon
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