Equivalent Power In series and parallel combination

[Shivansh Mathur]

How can we derive the formula for finding equivalent power in a series and parallel combination of 'n' resistors (with fixed resistance)?

 

[Qwertywerty]

How can we derive the formula for finding equivalent power in a series and parallel combination of 'n' resistors (with fixed resistance)?

Use that formula for the particular combination , in which a particular variable is common to all .
Example - Use P = ∑i²R_n . What is the advantage of using this ?

Thus , what would you use for parallel ?

Hope this helps .

 

[suranna]

How can we derive the formula for finding equivalent power in a series and parallel combination of 'n' resistors (with fixed resistance)?

nR(series)=2xnR(parallel)
n=no. of resistors.
R=value of resistor.
power=(voltage)square/resistance

 

[Shivansh Mathur]

I think i should put forward the question as under:

we have 2 identical resistors R(1) and R(2),
P(1) = Power of R(1)
P(2) = Power of R(2)
P(s) = Equivalent power of resistors R(1) and R(2) when connected in series.

P(s) = 1/P(1) + 1/P(2) HOW?

and if

P(p) = Equivalent power of resistors R(1) and R(2) when connected in parallel.

then,

P(p) = P(1) + P(2) HOW?

 

[berkeman]

nR(series)=2xnR(parallel)

Welcome to the PF.

Can you give more details about that equation that you wrote?

 

[Shivansh Mathur]

The same question is put up on this link -
https://answers.yahoo.com/question/index?qid=20080607083645AAqHHPw

 

[berkeman]

The same question is put up on this link -
https://answers.yahoo.com/question/index?qid=20080607083645AAqHHPw

The question still makes no sense. Are any things being held constant? Like the source voltage or the source current? You need to show the circuit diagrams involved, along with the source voltages in each configuration. Then show your math based on the circuit diagrams.

 

[sophiecentaur]

P(p) = P(1) + P(2) HOW?

Energy conservation principle. Where else would the input power come from or where would any surplus power go?

 

[Shivansh Mathur]

Sir, i myself have not done any math. I just seek to find a short way to calculate equivalent power in a circuit where devices of equal power rating are connected.
The source VOLTAGE is kept constant as you said.

for eg-

When three bulbs of 60W-200V rating are connected in series to a 200V supply, the power drawn by them will be : (Ans= 20W)

Now, i am asking for the proof of the formula that i have highlighted.

Have provided the circuit diagram also this time :)

 

[Jony130]

Well the answer is very simple;
If bulbs have 60W at 200V this means that the bulbs resistance is equal to:
P = V^2/R ---> R = V^2/P = 200V^2/60W = 667Ω and because we have three light bulbs connected in series the total resistance is equal to 3*667 = 2kΩ so the total power is equal to P = 200V^2/2kΩ = 20W This means that the book give you the correct equation.
But you should remember that this formula is only true for a device with a constant resistance. But this equation is not true for the light bulb. Becomes the light bulb resistance is not constant but will change with the apply voltage. See the graph of resistance of a 100W/230V light bulb in function of a supply voltages.

The resistance for 230V is equal 530Ω so current is equal 0.434A
But for 120V the current will not be equal to I = 120V/530Ω = 0.226A
As we can read form the graph the current will be equal I = 120V/ 380Ω = 0.315A

 

[sophiecentaur]

Well the answer is very simple;
If bulbs have 60W at 200V this means that the bulbs resistance is equal to:
P = V^2/R ---> R = V^2/P = 200V^2/60W = 667Ω and because we have three light bulbs connected in series the total resistance is equal to 3*667 = 2kΩ so the total power is equal to P = 200V^2/2kΩ = 20W This means that the book give you the correct equation.
But you should remember that this formula is only true for a device with a constant resistance. But this equation is not true for the light bulb. Becomes the light bulb resistance is not constant but will change with the apply voltage. See the graph of resistance of a 100W/230V light bulb in function of a supply voltages.

View attachment 87405

The resistance for 230V is equal 530Ω so current is equal 0.434A
But for 120V the current will not be equal to I = 120V/530Ω = 0.226A
As we can read form the graph the current will be equal I = 120V/ 380Ω = 0.315A

Even this answer is too simple, I think. You cannot tell, without measurement, how an individual bulb will behave. The surface temperature will surely depend upon the actual filament and envelope design and not just on its power rating. I really don't like the idea of putting filament bulbs as 'examples' of Resistors - presumably in an attempt to make the topic more approachable so the original question is 'questionable'. Science teaching is full of dodgy examples, thought up by people who don't have a lot of practical experience or detailed knowledge of the example items they include in their question.

Proof of the formulae in the red squares is easy. Start with the formula for parallel resistors. The Total Power will be V²/R_p so multiply both sides of the parallel resistor equation by V² and that produces what you want. Likewise, write down the formula for series resistors and remember Power is I²R. The answer comes oput by multi[plying both sides of the resistance formula by I².

 

[Shivansh Mathur]

Yes perfect. Now I've got the answer and understood it completely.

Thanks to all for guiding.

 

[sophiecentaur]

Now I've got the answer and understood it completely.

Thanks for asking the 1/P question. To be honest, I'd never thought of it in that way before.

 

[William White]

Even this answer is too simple, I think. You cannot tell, without measurement, how an individual bulb will behave. .

no its not

Why insist on making things more complicated than they are?

The purpose of the question was to deal with a simplified 'ideal' system for a complete begineer to start to understand; not a real world system.

When the students here are asking about, for example, projectile motion, nobody is going over the top with air resistance, wind, the dimples and imperfections in the projectile; they solve simple SUVAT equations. How far do you go? You cannot model everything, its impossible.

There really is a tendency to make things more complicated for no gain.

 

[sophiecentaur]

no its not

It's halfway between a simple resistor and a real device. If you want to help someone get to the real world then your graph should be coupled with a caveat or they could take it as gospel. If you had added "one example of a 100W light bulb" I wouldn't have picked it up.
There are a whole lot of possible circuit problems that include non-Ohmic devices and they are good problem-solving fodder. Resistor/diode networks Resistor/Thermistor combinations etc etc. They all result in a nice equation to solve.
I'm all for starting simple but with the simplification being made clear.

 

[Neeraj Kumar]

Sir, i myself have not done any math. I just seek to find a short way to calculate equivalent power in a circuit where devices of equal power rating are connected.
The source VOLTAGE is kept constant as you said.

for eg-

When three bulbs of 60W-200V rating are connected in series to a 200V supply, the power drawn by them will be : (Ans= 20W)

Now, i am asking for the proof of the formula that i have highlighted.
View attachment 87396
Have provided the circuit diagram also this time :)
View attachment 87397

from where did you get this picture

which book it is?
give the name of the author

 

[sophiecentaur]

The question still makes no sense.

Absolutely.

Sir, i myself have not done any math. I just seek to find a short way to calculate equivalent power

There isn't a short way to get a reliable answer. Can you not accept that? The "short way" is only available for Resistors that do not change their value over their operating range. If you haven't done any Maths then I would advise you to avoid Electronic Engineering questions.

 

[padmaraju bapala]

since in a series circuit having e resistances r1, r2, r3, R = r1+r2+r3
then I²R = i1²*r1+i2²r2+i3²r3, (since I = i1 = i2 = i3)
then P = p1+p2+p3
but how actually it be 1/P = 1/p1+1/p2+1/p3

 

[sophiecentaur]

but how actually it be 1/P = 1/p1+1/p2+1/p3

I had a problem with the implications of that equation. Power must be conserved so there are only certain values for p1, p2, p3 that can satisfy both that equation and the equation P = p1+p2+p3
Another way of looking at is is that there is only a limited set of a,b,c for which the mean is the same as the harmonic mean - so that equation is just a bit of Maths jiggery pokery and one shouldn't struggle too much with interpreting it in physical terms.
It's good to get us thinking though.

 

[jim hardy]

Beginners should start simple and derive general equations for themselves only after they've enough experience to do so. The old "What before Why" teaching method.

Resistors in series have same current through them.
So total power = I² X R total = I² X (R1 +R2 + R3) = P1 + P2 + P3

Resistors in parallel have not the same current through but the same voltage across them.
So power is E2/Requivalent = E²/(1/(1/R1 + 1/R2 + 1/R3)) = E² X (1/R1 + 1/R2 + 1/R3) = P1 + P2 + P3

Can anyone really look at this picture and say total power is other than 180 watts?

that equation is just a bit of Maths jiggery pokery and one shouldn't struggle too much with interpreting it in physical terms.

I'd say the grad student who wrote that exercise is a nincompoop and so is the author of the book who didn't check up on him..
Mathematical tap-dancing doesn't impress me at all. Well, maybe Cyd Charisse could ...

 

[jim hardy]

I think i should put forward the question as under:

we have 2 identical resistors R(1) and R(2),
P(1) = Power of R(1)
P(2) = Power of R(2)
P(s) = Equivalent power of resistors R(1) and R(2) when connected in series.

P(s) = 1/P(1) + 1/P(2) HOW?

Okay

if you take POWER to define the resistance at some specific voltage, then that resistance R is E²/P

Connecting the resistances in series causes them to share voltage, ie lowers voltage across each.

It was not stated in original question that this was to be done.

If it were done, when it were done, then 'tis better it were done quickly...

Total resistance then becomes E²/P1 + E²/P2 = E² X ( 1/P1 + 1/P2 )

and power P becomes E²/R_total = E² / (E² X ( 1/P1 + 1/P2 ) )

P = 1 / ( 1/P1 + 1/P2 )
invert both sides and 1/P = 1/P1 + 1/P2 as in the work cited

which is seventh grade algebra. But only after you've defined resistance as f(power)
which in my experience just IS NOT done.

I haven't much patience with showoff authors who confuse beginners on purpose.

Have i missed something in my arithmetic?
Am i too harsh with author or is that just bad teaching ?

Here's some tap-dancing to admire. Let math be in the eye of the beholder.
sigh...

old jim

 

[sophiecentaur]

Cyd Charisse

Ha Ha - you old dear!

 

[ShoyebRC]

I think i should put forward the question as under:

we have 2 identical resistors R(1) and R(2),
P(1) = Power of R(1)
P(2) = Power of R(2)
P(s) = Equivalent power of resistors R(1) and R(2) when connected in series.

P(s) = 1/P(1) + 1/P(2) HOW?

and if

P(p) = Equivalent power of resistors R(1) and R(2) when connected in parallel.

then,

P(p) = P(1) + P(2) HOW?

Let's say we have two bulbs of 100W-220V rating. We can get their resistance by R = V^2/P. We get 484 ohm.

Connecting them in parallel, we get equivalent resistance = 1/ (1/484 + 1/484) = 242 ohms. So we have a circuit with equivalent resistance of 242 ohms and 220 Volt. So the equivalent power is P = V^2/R = 200 W. We get the same result by P(eq) = P1 + P2.

Connecting them in series, we get equivalent resistance of 968 ohm. Thus the power is 220^2/968 = 50 W. We get the same result as P(eq) = 1/ (1/P1 + 1/P2).

Thus,
For series, P(eq) = 1/ ( 1/P1 + 1/P2)
For parallel, P(eq) = P1 + P2

As, we get full power with parallel circuit and fraction of the full power with series circuit.

 

[sophiecentaur]

As, we get full power with parallel circuit and fraction of the full power with series circuit.

The following comment may have appeared higher up but here goes:
In the 'old days' the electric cooker hobs had a very simple Heat Control. There were two elements inside a metal disc. With both connected in parallel you had full power. With just one element you had half power and, with the two elements connected in series, you had quarter power. The elements didn't get too hot so the nominal quarter power was 'near enough. Try it with filament light bulbs and the resistance / temperature curve of tungsten meant that the quarter power was . . . . .(For the student: which way did it go?)

 

[sophiecentaur]

P(s) = 1/P(1) + 1/P(2) HOW?

This keeps turning up in the thread. It can't be correct because it is dimensionally wrong.

It's not P_(s). it's 1/ P_(s), Same as the parallel resistor formula and the series conductance formula.

But it's a very old post.