# why ohmic losses increase with resistance?

Electrical newbie here. I'm trying to get understanding how is resistance involved with power dissipation (ohmic losses, eg. heating). Primarily I'm looking at Electromagnet coil, it says the losses are $P=I^2R$, so reducing resistance R reduces heating power loss P, which sounds reasonable as it comes from (I assume):

$P=UI$ and Ohm's law $I=\frac{U}{R}$ (and thus $U=IR$), so by substituting U we get $P=UI=(IR)I=I^2R$

But, if one substitutes I instead, he'd get $P=UI=U(\frac{U}{R})=\frac{U^2}{R}$. Which seems to tell me, that with stable voltage source U one would get lower power loss (heating) with increased resistance R!

Which unfortunately also makes sense to me from empiric point of view: If I connect very high value resistor (or its equivalent - a veeery long wire) across 230V line, it would only heat a little, and I put very low value resistor across 230V line, it would heat so much it would burn (which I guess is what fuses do for living). (replace 230V AC with 9V DC battery if AC/DC distinction matters here)

So I guess I'm missing something basic - would increasing resistance reduce or increase power losses? Or is wire in "put it in 230V socket" example behaving completely different that wire in electromagnet example (and if so, why?)

• A simple reason (barring coil etc stuff ; from electrical POV) : Increasing resistance decreases current through it and vice versa ; So using both the equations, power dissipated decreases. Jan 12 '15 at 3:04
• Thanks @Plutoniumsmugglerwithhat: yes you're right, $I^2$ from $I^2R$ actually contains $\frac{1}{R^2}$ inside, so it actually makes is inversely proportional to R too, as in the $\frac{U^2}{R}$. So higher resistance would actually mean lower power losses in transformer (opposite to what wikipedia on first sight says), but it would also mean lower current and thus weaker magnetic field (which one usually doesn't want). Makes sense now! Jan 12 '15 at 3:22

• In one case you're holding voltage constant, in the other you're holding the magnetic field constant and reducing the voltage proportionally . Since the power is proportional to $V^2$ (P = V^2/R), the power drops with dropping resistance. Jan 12 '15 at 2:59