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We use $I^2R$ or $V^2/R$ or $VI$ for measuring power $P$. Are all of these applicable for all circuits? I have seen in some circuit $V^2/R$ is not equal to $I^2R$. Why is that?

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$P = IV$ applies to all circuit branches.

$P = I^2R$ or $P = V^2/R$ are restatements of the general rule that apply when we are considering power delivered to an ideal resistor that behaves according to Ohm's law $V = IR.$

I have seen in some circuit $V^2/R$ is not equal to $I^2R$ (like when there is capacitor or inductor). Why is that?

Those components are not ideal resistors. The forms with R are a special case for when we are considering an ideal resistor.

For other components in static (DC) circuits, you should use the general form $P=IV$.

As Tinchito says, when dealing with a time-varying circuit, you should use the instantaneous form

$p(t) = i(t) v(t)$.

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  • $\begingroup$ @BMS, please don't gratuitously add MathJax where it's not needed. It takes much longer to load for many users. $\endgroup$ – The Photon Aug 11 '14 at 19:17
  • $\begingroup$ but sometimes P=I2R and P = V2/R are not equal . when exactly when we use I2R and V2/R ? $\endgroup$ – Dilshad Hossain Aug 11 '14 at 19:20
  • $\begingroup$ @ThePhoton Thanks for the info about MathJax, didn't know. You can roll-back my edit. $\endgroup$ – BMS Aug 11 '14 at 19:21
  • $\begingroup$ @DilshadHossain, can you give an example? Was the device in question an ideal resistor? $\endgroup$ – The Photon Aug 11 '14 at 19:22
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    $\begingroup$ In general, Phys.SE encourages to implement MathJax, even in such trivial case as here, in order to get correct mathmode rendering, cf. @BMS's edit. $\endgroup$ – Qmechanic Aug 11 '14 at 19:32
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After solving problems on circuits and power dissipation in them, I observed that $V^2/R$ is used when the voltage is constant across the elements in the circuit and $I^2R$ is used when current is constant through the elements in the circuit.

They yield the same result when a purely resistive load is used. Even the formula $P=VI$ will give the answer. This is because when only resistors are used, the real power comes into action. With capacitors and inductors, reactive power come into action. That is why the answers don't match. When solving problems with inductors and capacitors, the impedance (which is a complex quantity) is used.

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The measure $ R $ of resistance is an invented one. It was deduced long ago by experiment that many materials had a constant ratio $ \frac {V}{I}$ between the voltage applied and the current flowing. Thus the quantity 'resistance' was defined to be precisely this ratio. Later, when inductive and capacitive effects were observed, 'reactance' $ Z $ was defined to be this quantity instead, while resistance now refers to the value of this quantity when no capacitive or inductive effects are at play.

Voltage and current are measurable, physical, well defined quantities. Resistance and reactance are quantities which are defined in the process of modelling electrical circuits. The equations you give are correct only in the context of modelling circuits without capacitive or inductive effects. If you replace $ R $ with $ Z $, then the equations would be correct in the context of a more general model-- one accounting for capacitors and inductors too.

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The formulas are of course all true if used and interpreted correctly. But human error is the wild card. For practical reasons, I-squared-R is the most reliable formula because it's almost impossible to apply it incorrectly.

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