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?


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.

  • $\begingroup$ So by increasing the supple voltage, we are just decreasing the supply current and hence the reduction in Power loss. So increasing Vd rather than Vs would increase the power loss. Correct me if wrong! $\endgroup$ – Polisetty Apr 16 '16 at 11:42
  • 1
    $\begingroup$ @Polisetty, assuming a constant source power, it is true that increasing the source voltage decreases the source current and, as you correctly point out, decreases the power loss. $\endgroup$ – Alfred Centauri Apr 16 '16 at 11:43

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


$$P=VI$$ (since I=$\frac{dQ}{dt}$)

Also note that $P=I^2R$ for resistor circuits only.

  • $\begingroup$ $P = I^2R $ is correct for all circuits in world. $\endgroup$ – Anubhav Goel Apr 16 '16 at 11:41
  • $\begingroup$ No, I had this confusion too, If you remember NCERT AC circuits, check out he formula for power in a circuit for RLC or LC circuits. $\endgroup$ – Prayas Agrawal Apr 16 '16 at 11:42
  • $\begingroup$ I had this in mind . Let me give you a link. $\endgroup$ – Anubhav Goel Apr 16 '16 at 11:45
  • $\begingroup$ physics.stackexchange.com/questions/248039/… $\endgroup$ – Anubhav Goel Apr 16 '16 at 11:47

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.


$V_{s} I = V_{1} I + V_{2} I ...... V_{n} I$

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.


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.

  • $\begingroup$ What's wrong with the answer, may I ask? $\endgroup$ – Procyon Apr 16 '16 at 11:51
  • $\begingroup$ I gave it down vote. Vd is different for different components and Vs ≠ Vd in general. Also, answer seems irrelevant. $\endgroup$ – Anubhav Goel Apr 16 '16 at 12:32
  • $\begingroup$ If you read the question the way it should be read, it clearly says that $V_s$ is the supply voltage and $V_d$ is the voltage drop. By Kirchhoff's voltage law $V_d$=$V_s$ always. I dare say you have no knowledge of Kirchhoff's laws. I am afraid a school-going student knows better than you do. $\endgroup$ – Procyon Apr 16 '16 at 12:51
  • $\begingroup$ Read physics.stackexchange.com/questions/248229/… Vd ≠ Vs . Instead Vd is Vc and Vl seperately. $\endgroup$ – Anubhav Goel Apr 16 '16 at 13:13
  • $\begingroup$ $V_{d}$ may be $V_{l}$ , or $V_{c}$ or $V_{l}$ + $V_{c}$ depending on context. One may be asking voltage drop against across supply wires or across resistor or any other component or across whole apparatus. Only when voltage drop is asked for all circuit Vd = Vs . Its only a special case. $\endgroup$ – Anubhav Goel Apr 16 '16 at 13:18

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