# Question about ohm's law and resistor wattage

Discussion in 'Electronic Basics' started by Rikard Bosnjakovic, Nov 26, 2005.

1. ### Rikard BosnjakovicGuest

A thing struck me right now:

Consider a 10v-circuit with a 1A-load. This gives 10W dissipation (P = U*I
= 10 * 1 = 10).

Suppose I then want to strangle the current down to 0.5A. Ohm's law gives
me R = U/I = 10/0.5 = 20 ohm, so I add that one. With 0.5A load, the
dissipation is 10 * 0.5 = 5W.

For the sake of easy calculation, let's assume the circuit is perfectly
linear. Now for the question: Should the resistor be 5W, i.e. the latter
value above, or 10W, the first value?

I tried to figure it out but couldn't come up with a reasonable answer of
why I should pick the one or other size.

2. ### John PopelishGuest

Add what to what? The original circuit consisted of a 10 volt source
and a 10 ohm resistor (or something else that limited the current to 1
amp.

If you add 10 more ohms in series with the original 10 ohms, the
current will fall to 0.5 amp.
And that 5 watts will be divided equally between each of the two 10
ohm resistors, 2.5 watts each.
As long as tow 10 ohm resistors are in series, each needs to be
capable of getting rid of at least 2.5 watts.

3. ### ehsjrGuest

The power dissipated in a resistor can be computed by
P = I^2R

The *selection* of the resistor wattage is a bit different
than the formula. You use the formula to determine how much
power will be dissipated - and then (generally) you pick a
resistor of about 2X higher wattage. In a high ambient
temperature environment, you may go even higher; in some cases
you may go lower than 2X. 2X is just a rule of thumb,
not an exact design rule. But always go higher is. You
would need a very good reason to go lower than, or equal to,
the exact computed value.

Ed  