Yageo specifies both maximum current and maximum power, see page 5 of the datasheet:

\$P_{max}\$: 100 mW

\$I_{max}\$: 1 A

and you'll also see that for the jumper

\$R_{max}\$: 50 mΩ

That seems inconsistent: 1 A through 50 mΩ is only 50 mW, not 100 mW. In these cases you have to work with the lower value: 50 mW, since 100 mW would mean a 1.4 A current, which exceeds the 1 A limit.

EEs often scoff at the 5 % tolerance specification for the 0 Ω resistor. The engineers at Yageo *know* that that doesn't make sense, and if you take a good look at page 2 you'll see that they don't specify 5 % for the jumper at all:

*F = ± 1 %*

J = ± 5 % (for Jumper ordering, use code of J)

which should be read as "we *use the same code for a jumper* as the tolerance for other values". It does not implied that the 5 % tolerance would apply to the jumper.

Specifying maximum power isn't silly either: the part's weight and specific thermal capacity determine that, regardless of resistance value.

If you know the input power and the output power then the difference is the power lost.... in the feed wires.

Do you know how to calculate the power dissipated in a resistor based on what current is flowing through it? If yes, then you reverse this formula to uncover what current must be flowing to dissipate the power lost in the feed wire. When you have done this you'll find that your prof is right and he also made the assumption that the feed wire between generator and load was purely resistive.

Let's see if you can figure this out by your comments. PF is 0.72 by the way so let's see if you can figure this out too.

## Best Answer

I will not provide the direct answer as this is obviously a homework.

"Power rating" refers to the minimum power that a device can dissipate. When powered, a lamp consume energy and transforms it into light and heat. For this kind a basic problem, we mostly assume that all the energy goes into heat.

The dissipated power is measured in Watts and correspond to the amount of energy per seconds that the lamp will converts to heat. If the lamp is small, or made of plastic for example, it may not be able to dissipate that heat, which mean it will burn (or damaged at least). If it's big enough, or installed on a radiator, it may not burn and the temperature will stabilize before it burns.

An electrical component normally have the capacity to dissipate a specific amount of power. For instance, we can say : this lamp is able to dissipate 90W. More than that, it will burn.

In you problem, finding the power ratings means finding the amount of power (watts) that is consumed by the lamps. Then you can say, this lamps needs to dissipate at least this amount of power.

Electrically speaking, the power dissipated by a element is given by the product of voltage and current across your lamp.

\$P=VI\$

Just find the voltage around each lamps and multiply by the current going across it