Difference between $I^2R$ and $V^2/R$ and $VI$ for measuring power $P$

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Solution 1

$P = IV$ 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

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

Solution 2

All the formulae are valid. But sometimes people confuse what variables to to plug to the equations when using them.

For example consider a DC circuit with a battery $10V$ and with two resistances $R_1$ and $R_2$ in series . Suppose the question asks to find the power dissipated by the $R_1$.

We have 3 formulae :

$P=IV$

$P=I^2R$

$P=\frac{V^2}{R}$

Ofcourse all of them are valid but you cannot apply every one of them directly.

Consider using $P=\frac{V^2}{R}$. Some people will directly use it on $R_1$ to get the answer as

$P=\frac{10^2}{R_1}$

But this is wrong. Because the $V$ in the equation actually means potential across the resistor $R$ (and not potential of the circuit).

Therefore you have to first find the potential across $R_1$ by using Ohms Law. ($V=IR$) . For series $I=\frac{E}{R_1 + R_2}$

Put the variables in $P=\frac{V^2}{R}$.

You get $P=\frac{V^2R_1}{(R_1+R_2)^2}$.

Oh wait! Thats the same thing you would get if you would have used $P=I^2R$ . Its not a coincidence. In series, the total current flowing thought the circuit is equal to the current flowing through any resistor. Therefore you donot need to calculate individual currents for the resistors.

The same is for $P=VI$ . You need potential across $R_1$ and not of the whole circuit. Plug in the correct variablesyou would get the answer from all of them.

Side note :

As you just saw from the example, it is always easy to find Power from a Resistor in series by using $P=I^2R$. Similarly for parallel circuits, it is easy to get the power by $P=\frac{V^2}{R}$. But in then end of the day, both of them will yield the correct results.

Solution 3

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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Dilshad Hossain
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Updated on August 01, 2022

Comments

  • Dilshad Hossain
    Dilshad Hossain over 1 year

    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?

  • The Photon
    The Photon about 9 years
    @BMS, please don't gratuitously add MathJax where it's not needed. It takes much longer to load for many users.
  • Dilshad Hossain
    Dilshad Hossain about 9 years
    but sometimes P=I2R and P = V2/R are not equal . when exactly when we use I2R and V2/R ?
  • BMS
    BMS about 9 years
    @ThePhoton Thanks for the info about MathJax, didn't know. You can roll-back my edit.
  • The Photon
    The Photon about 9 years
    @DilshadHossain, can you give an example? Was the device in question an ideal resistor?
  • Dilshad Hossain
    Dilshad Hossain about 9 years
    like when there is capacitor or inductor
  • Martin Petrei
    Martin Petrei about 9 years
    @DilshadHossain For a capacitor or inductor, the formula gives the instantaneous power, i.e. the power dissipated in a particular time t.
  • Qmechanic
    Qmechanic about 9 years
    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.
  • Martin Petrei
    Martin Petrei about 9 years
    @DilshadHossain The general law is p(t) = i(t) v(t).