When you stack resistors, do their ratings just add up?

[jivetrky]

Say I connect two 30K ohm, 1/2 watt resistors in series...

Do I now have 60k ohm 1 watt ? or is it 60k ohm 1/2 watt? Or does the math work strangely somehow and it's not that at all?

[Deadbot1_1973]

in series they ADD to 60 Ohms at 1/2 watt
in parallel they get funky. I forget the formula off the top of my head.

edit Linky to page with the equations (http://technology.niagarac.on.ca/students/t/ptaylor/calculations.htm)

[jivetrky]

OK Thanks!

[FireMogle]

parrallel is 1/R + 1/r1 + 1/r2 if I remember right.

[Super Nade]

Say I connect two 30K ohm, 1/2 watt resistors in series...

Do I now have 60k ohm 1 watt ? or is it 60k ohm 1/2 watt? Or does the math work strangely somehow and it's not that at all?
Two different things really. Rated wattage has nothing to do with how you arrange the resistors. It describes the individual power handling capabilities of the resistor.

In Series: Requiv = R1+R2

In Parallel: Requiv = R1*R2/R1+R2

The power dissipated by a resistor P = V*I or I2*R
So in your config you neeed to be looking at the power dissipated by each individual resistor.

[kayson]

In Parallel: Requiv = R1*R2/R1+R2
Um that simplifies to R2+R2...and thats not right.

In parallel, its 1/Requiv = 1/R1+1/R2

so Requiv = 1/(1/R1+1/R2)

Edit: I think you meant (R1*R2)/(R1+R2). That works - order of operations ftw.

[Roofles]

ACK! That formula sucks and my professor would yell at us for using it!

I think nade has it right, its just that formatting with one line of text is aweful.

1/Rp = 1/R1 + 1/R2 + .... + 1/Rn is the correct ;) equation.

[Deadbot1_1973]

Two different things really. Rated wattage has nothing to do with how you arrange the resistors. It describes the individual power handling capabilities of the resistor.

In Series: Requiv = R1+R2

In Parallel: Requiv = R1*R2/R1+R2

The power dissipated by a resistor P = V*I or I2*R
So in your config you neeed to be looking at the power dissipated by each individual resistor.

OOps you are right on the wattage Super Nade...It's been 16 years since I took Electronics in high school, and I haven't really put it to use for most of that time..

[Super Nade]

You are not wrong per se, but when working with resistors, it is always a good idea not to exceed their rated wattage. The reason being, with most resistors, resistance is a non-linear fuction of temperature beyound the rated value. Also, thermal noise increases tremendously. This is a big problem if you are working on a sensitive ckt design.

[Overload]

since: power = voltage squared / resistance. if you add them in series the TOTAL power consumbed will be half. each can disipate 1/2 watt so you effectively have room to supply more power to the system (resistor pair). what in hell are you connecting a 30k resistor to that needs more than 1/2 watt! on its own it would take 15k volts to exceed its power rating. high power resistors are usually used where low resistance is needed, ie lots of current and heat.

[jivetrky]

since: power = voltage squared / resistance. if you add them in series the TOTAL power consumbed will be half. each can disipate 1/2 watt so you effectively have room to supply more power to the system (resistor pair). what in hell are you connecting a 30k resistor to that needs more than 1/2 watt! on its own it would take 15k volts to exceed its power rating. high power resistors are usually used where low resistance is needed, ie lots of current and heat.


Well I'm not actually worried about it needing more than 1/2 watt. Was just curious how it worked when you stacked them like that.

[Dragonprince]

since: power = voltage squared / resistance

That's an unusual formula for figuring Watts. Where did you come across that? I'm curious because I've worked with various electronics, ac and dc for years and everyone always uses P=I*E for Watts. Your formula works fine, I've just never seen that way of working the math...:shrug:

[Knacko]

If I forget, I use this site to find the formulas for power, voltage, etc.

http://www.sengpielaudio.com/calculator-ohm.htm

[T'wolves]

That's an unusual formula for figuring Watts. Where did you come across that? I'm curious because I've worked with various electronics, ac and dc for years and everyone always uses P=I*E for Watts. Your formula works fine, I've just never seen that way of working the math...:shrug:

It's the "PIE" wheel, P=I*E and R=I/E therefore E=I/R, therefore P=I*I/R. or I=E*R therefore P=E*E*R = P=E(squared)*R....it keeps going round and round.

[Maviryk]

I believe the pie wheel is used more by electricians, and the formulas by EEs.

[T'wolves]

I believe the pie wheel is used more by electricians, and the formulas by EEs.

But all the wheel is, is formulas...so to use one is to use the other. But I admit electricians probably use the power equation more than electronic tech's (my 2yr degree).

[Super Nade]

It is interesting to see how each profession has its own niche. As a physcist, the only thing I see is that energy needs to be conserved and the equation should be dimensionally consistent. ;)

[dogbert_2001]

It's the "PIE" wheel, P=I*E and R=I/E therefore E=I/R, therefore P=I*I/R. or I=E*R therefore P=E*E*R = P=E(squared)*R....it keeps going round and round.

If E is voltage, you have the 2nd equation (and all subsequent equations) wrong. It should read "R = E/I".

Sounds to me like this "PIE" wheel has its faults.


With regards to power ratings, I think a good rule of thumb is it's only as good as the lowest rated component.

Let's say you have two 30k resistors in series, one rated 1/2W, and the other 1W. The current through each will always be the same (because they're in series) and so the power dissapated by each will always be the same. Therefore, you can only give it as much current until the 1/2W is maxed.

Same for in parallel. The voltage across both must be the same, and so the power dissapated will be the same. So you can only give it as much voltage until the lesser resistor is maxed.

The total "power rating" becomes 1W, and not 1+1/2W for both examples. But this is very misleading, because you don't drive circuits with power, you drive them with either voltage or current.

So in conclusion, when you have multiple components connected the same way (all parallel or all series), to calculate "power rating", take the lowest and multiply by the number of components.

<--EE/CompE

[Quailane]

That's an unusual formula for figuring Watts. Where did you come across that? I'm curious because I've worked with various electronics, ac and dc for years and everyone always uses P=I*E for Watts. Your formula works fine, I've just never seen that way of working the math...:shrug:

I learned that formula in a physics class.

[Gautam]

That's an unusual formula for figuring Watts. Where did you come across that? I'm curious because I've worked with various electronics, ac and dc for years and everyone always uses P=I*E for Watts. Your formula works fine, I've just never seen that way of working the math...:shrug:
I=E/R, no? ;)

energy needs to be conserved and the equation should be dimensionally consistent. ;)

I wonder what sort of profession gets to dodge this?! :eek:

[Super Nade]

I wonder what sort of profession gets to dodge this?! :eek:

The same profession that has to build a Magneto-optic atom trap without being able to buy the necessary electronic components off the shelf. :bang head

[T'wolves]

If E is voltage, you have the 2nd equation (and all subsequent equations) wrong. It should read "R = E/I".

Sounds to me like this "PIE" wheel has its faults.


<--EE/CompE

Your right that I fubar'ed the equation, it has been a while since I last saw it. But the PIE Wheel is perfect. The second link given in post #13 has the PIE wheel shown, and from it you can derive all the formulas correctly.

Never doubt the wheel...

[Deadbot1_1973]

The same profession that has to build a Magneto-optic atom trap without being able to buy the necessary electronic components off the shelf. :bang head

Or mine. I install telephone systems from May until December(hey it's cold in Minnesota in the winter). I guess you could say we conserve energy in the winter, but our expenditure easily outweighs our conservation.:santa:

[johan851]

In your series explanation, you're off - although the ampere's would stay the same in both resistors, the voltage would drop from one resistor to the next - therefore the power dissipated by each resistor would NOT be equal.
No sir. The voltage drop across each resistor is equal, and voltage drop across the resistor is what you're using in p = VI. You wouldn't use the potential difference between the resistor and ground for that equation - it simply wouldn't make sense. You said it yourself - the current is the same through each, and power is also (I^2)*R. If the current through each is equal, and the resistances are both 30Kohm, then the power must be the same.

In the parallel example, he's talking about the case where both resistors share a common node at each of their ends. In that case, voltage across, current through, and power dissipated for each remains equal.

Back to school for you, Adak. :D

[Adak]

No sir. The voltage drop across each resistor is equal, and voltage drop across the resistor is what you're using in p = VI. You wouldn't use the potential difference between the resistor and ground for that equation - it simply wouldn't make sense. You said it yourself - the current is the same through each, and power is also (I^2)*R. If the current through each is equal, and the resistances are both 30Kohm, then the power must be the same.

In the parallel example, he's talking about the case where both resistors share a common node at each of their ends. In that case, voltage across, current through, and power dissipated for each remains equal.

Back to school for you, Adak. :D

I intended it to be a bit of a satire, to dogbert's tome, but it just didn't come out right - I've deleted it.

Adak