# P=V*I applied on DC resistor circuits.

According to Ohm's law, when a V DC voltage is applied to a R Ohms resistor, a I current will flow through the resistor.
On the other hand, I've learnt that the power that the resistor will have to dissipate could be calculated according to P=V*I. However, if I try with the following example, it seems not making sense:
V = 24 VDC, R = 12 Ohm.
So,
I = 2 A, P = 48 W.
An almost 50 W resistor... ¿¿¿??? It's too much!!! what's wrong here?

It’s too much!!! what’s wrong here?

Why do you think that?
That’s a lot of current.

P=I2R ( 22x12 = 48), so whichever way you do the arithmetic, 48W is the answer.

albuino:
V = 24 VDC, R = 12 Ohm.
So,
I = 2 A, P = 48 W.
An almost 50 W resistor... ¿¿¿??? It's too much!!!

No, it's correct. You need one of these: http://www.ebay.com/itm/250911320379

(Although that's going to get very hot at full power so you should really get 100W. Maybe 150W to be safe...)

@ fungus.. YES GREAT PART...
that's what the little "ears' are for.. Bolting to a heatsink.. That's the only way it can handle the 'rated' power
and heatsink grease is required for best results.

Bob

R = 12 Ohm

An almost 50 W resistor... ¿¿¿??? It's too much!!

Yeah but 12Ohms is little more than a dead short....

Yeah but 12 Ohms is little more than a dead short....

Yep and could look like a glowing heater if you used the right wire, or put it in a bulb and make some light.

be80be:

Yeah but 12 Ohms is little more than a dead short....

Yep and could look like a glowing heater if you used the right wire, or put it in a bulb and make some light.

I've seen those big resistors bolted to metal plates and to heat them up (eg. in 3D printers where they want the base plate to be hot).

Pick too small of a power rating and it will be a LER for a short time.

(Light Emitting Resistor)

(Light Emitting Resistor)

XD ... must remember that one

Albuino

All other responders have basically stated that the dissipated power is indeed 48watts.

However, what you need to ask yourself is "Why do I not trust or believe calculations derived from very basic equations that I claim to understand"

This is not intended as a criticism but rather a suggestion that you need to enhance your experience in the application of theory before you replace it by intuition.

jackrae:
Albuino

All other responders have basically stated that the dissipated power is indeed 48watts.

However, what you need to ask yourself is "Why do I not trust or believe calculations derived from very basic equations that I claim to understand"

This is not intended as a criticism but rather a suggestion that you need to enhance your experience in the application of theory before you replace it by intuition.

Very good way to look at this. If the math is right perhaps there may be something wrong with the circuit design that needs to be looked at.

I use a couple of 10 ohm power resistors (25W each) bolted to a large piece of 6mm thick aluminium plate
as a dummy-load for testing things (like motor drivers, solar panels) - in parallel they give me 5 ohms, in series 20,
or just one of them gives 10.

If I put my 30V 3A power supply across one of them and crank up to full voltage the resistor gets hot
very quickly as its dissipating 30 x 3 = 90W (I don't do this for long!).

The equations make perfect sense once you have hands-on experience like this. Another important
realisation is that P = V x I combines with V = I x R to yield P = I x I x R

Power is proportional to the square of current - this means that high currents can be a real problem
to handle - 10A is a hundred times more "heat generating" than 1A for instance (which is why fuses are
very handy - they stop the wiring catching fire).

Albuino why are you dropping the full 24v with the resistor? Is the resistor in series with another component in your circuit? Could you give a little detail about your project?

This might be redundant for many, but I want to point out that Power is also proportionate to the square of the Voltage. P=V2/R This is something many people learn the hard way after raising the input voltage to a linear regulator. This results in an exponential power dissipation increase just the same as increasing the current flow. What worked coolly with a 9V battery as input, now puts the regulator in shutdown after switching to a 12V supply.

afremont:
This might be redundant for many, but I want to point out that Power is also proportionate to the square of the Voltage.

ie. Volts is directly proportional to amps.

It's the law!

fungus:

afremont:
This might be redundant for many, but I want to point out that Power is also proportionate to the square of the Voltage.

ie. Volts is directly proportional to amps.

It's the law!

Only as long as Xc = Xl
Reminds me of 300,000km/S, it's not just a good idea. It's the law!

Reminds me of 300,000km/S, it's not just a good idea. It's the law!

Well, in a vacuum anyway ...

Power dissipation often comes in handy in transistor selection also.
P=IV=I^2*R = V^2/R
For MOSFETs, current flow & Rds is usually known, so P=I^2 * R is good
(and if you current & Rds, V=IR so you can determine Vds also)

For BJTs, current flow & Vce-sat is usually known, so P=IV is good
Vce-sat of ~0.7V vs Rds of 0.035ohm shows why MOSFETs are preferred for high current loads:
example: 1A & 0.7V = 700mW, while 1A & 0.035ohm = 35mW
typically ignored is the power from the base current - say for an NPN 15mA was used, and Vbe is 0.7V -that's another 10.5mW that is dissipated. 1.5% of the total from the example above, so ignoring it is fairly safe. At Arduino limits, say 35mA, max would be 0.035A * 0.7V = 24.5mW, still a fairly small amount.

For resistors, voltage & resistor value is known, so P=V^2/R is good.
For an LED for example: (Vs - Vf-led - Vtransistor)/current = resistor, or (Vs - Vf-led - Vtransistor)/ resistor = current, depending on what is being solved for (what resistor do I need for 20mA? if I use this 220 ohm resistor, what current will I get?)
Vtransistor is Vce-sat for BJTs, and current * Rds for MOSFETs
How is this applicable? Say you're wondering how much can you do with a 1/8w resistor,as you notice a warm smell coming from your circuit? 0.125W = I^2 * R, so with 20mA (max continuous for most LEDs), smallest R you can get by with is 0.125/(0.02^2) = 312 ohm, so a standard 330 would be safe.
That same 312 ohm resistor with a 1/4W rating could handle more current: P=I^2R, or Sqrt(P/R) = I, so Sqrt(0.25/312) = 28mA.

All things to consider as you select components.

Thanks everyone. The help received has been excellent by far.
Actually it was a general electronic question. There is no project around this. So I have really learnt that I must believe electronic laws since I didn't.

afremont:
This might be redundant for many, but I want to point out that Power is also proportionate to the square of the Voltage. P=V2/R This is something many people learn the hard way after raising the input voltage to a linear regulator. This results in an exponential power dissipation increase just the same as increasing the current flow. What worked coolly with a 9V battery as input, now puts the regulator in shutdown after switching to a 12V supply.

Actually the input to a voltage regulator is a poor example here as linear regulators look like constant current loads (the regulator
keeps the output voltage constant, typically the load then draws a fixed current, thus the input to the regulator is a constant current.)

Increasing the voltage to the linear regulator does not increase the current (unlike the resistor case). Dissipation in regulator is (Vin - Vout) * current.

Also a square law is not exponential - exponential increase is what happens as you increase the voltage across a diode, and makes
a square law look tame!