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Rikard Bosnjakovic
Guest
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 Posted:
Sat Nov 26, 2005 7:57 am Post subject:
Question about ohm's law and resistor wattage |
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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.
--
Sincerely, | http://bos.hack.org/cv/
Rikard Bosnjakovic | Code chef - will cook for food
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John Popelish
Guest
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| Posted:
Sat Nov 26, 2005 8:14 am Post subject:
Re: Question about ohm's law and resistor wattage |
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Rikard Bosnjakovic wrote:
| Quote: | 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.
|
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.
| Quote: | With 0.5A load,
the dissipation is 10 * 0.5 = 5W.
|
And that 5 watts will be divided equally between each of the two 10
ohm resistors, 2.5 watts each.
| Quote: | 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?
|
As long as tow 10 ohm resistors are in series, each needs to be
capable of getting rid of at least 2.5 watts.
| Quote: | 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. |
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ehsjr
Guest
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| Posted:
Sat Nov 26, 2005 9:35 am Post subject:
Re: Question about ohm's law and resistor wattage |
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Rikard Bosnjakovic wrote:
| Quote: | 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.
|
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
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