# $P=VI$ and $P=V^2/R$ [duplicate]

How come equation a) $P=VI$ says if you double voltage, power is doubled while according to equation b) $P=V^2/R$ says power quadruples? Which one is correct?

• If you double the voltage, what happens to the current?
– jim
Commented Mar 25, 2016 at 19:53

$P=IV$ is always correct.

$P=V^2/R$ is correct for an ohmic device, or a device that obeys Ohms law. Ohm's law states that $V=IR$

Typically problems will involve either a constant resistance and Voltage, then you use $P=V^2/R$, a constant current and resistance, use $P=I^2 R$, or constant voltage and current, then use $P=IV$

To answer your question about power doubling and quadrupling it depends on the device. Increasing voltage for an ohmic device will increase the current as well and thus both formulas will give you the same result. For a device that does not obey Ohm's law, anything is fair game. The only formula that is still correct is $P=IV$

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

If we double $V$ and keep the resistance constant the power changes to:

$$P' = \frac{(2V)^2}{R} = \frac{4V^2}{R} = 4P$$

So the power is quadrupled as you say:

Now take:

$$P = IV$$

We double the voltage so $V \rightarrow 2V$, but if the resistance is constant we also double the current so $I \rightarrow 2I$. So the power changes to:

$$P' = (2I)(2V) = 4IV = 4P$$

Just as before. Both equations tell you that the power quadruples if you double the voltage.

For the case $P=I\cdot V$, power gets doubled when the voltage is doubled only when the current I is constant and resistance can vary.

Similarly power $P$ quadruples when $1/R$ is kept constant => resistance $R$ is constant and current can vary.

How power changes depends on what quantity you keep constant.

Consider this equation $P=I^2\cdot R$, $P$ increases with increase in $R$ when $I$ is constant. Where as considering this equation $P=V^2/R$, $P$ decreases with increase in $R$ keeping $V$ constant.