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For computing the power dissipated over a resistor the correct formula is $P = R \times I^2$. Why, in other purposes equivalent, $P = \frac{V^2}{R}$ is not good?

I can imagine that the actual current flowing through the resistor has its impact, but $V = R \times I$, so I'm looking for a better explanation.

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Your statement

For computing the power dissipated over a resistor the correct formula is P = R x I^2. Why, in other purposes equivalent, P = V^2 / R is not good?

is news to me.

The instantaneous electrical power $P$ in a circuit element is defined as $P = VI$ where $v$ is the potential difference across the circuit element and $I$ is the current passing through it.
For a resistor for which $V=IR$ where $R$ is the resistance of the resistor the power equation can be written as $P = I^2R$ and $P=\frac {V^2}{R}$.

Which of the three variations of the formula you use depends on the situation.
For a resistor of constant resistance all the three formulae are equivalent.

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  • $\begingroup$ It's about following exercise in The Art of Electronics 3rd edition practice: New York City requires about 10^10 watts of electrical power, at 115 volts(this is plausible:10 million people averaging 1 kilowatt each). A heavy power cable might be an inch in diameter. Let's calculate what will happen if we try to supply the power through a cable 1 foot in diameter made of pure copper. Its resistance is 0.05 microhms(5 x 10^-8 ohms) per foot. (A) Calculate the power lost per foot from "I^2R losses" >>>>>>>>> If I use the P=V^2/R I get a different answer than using P=I^2 x R $\endgroup$ – alex Oct 9 '16 at 10:04
  • $\begingroup$ I now understand. If you want to use the voltage it must be the voltage drop across the cables which is equal to the voltage at the power station minus the voltage at the consumer end. Please have a look at my answer to this question: physics.stackexchange.com/questions/248229/… $\endgroup$ – Farcher Oct 9 '16 at 11:46
  • $\begingroup$ Thank you very much! It was so simple but needed to be stated. $\endgroup$ – alex Oct 9 '16 at 13:44

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