Power of resistors in parallel vs series circuit

In summary: Yes, in this problem, the voltage across each resistor is less than the voltage across the two resistors in series.
  • #1
crostud123
3
0

Homework Statement


Two Ohm's resistors in parallel consume power of 76 W and 24 W.
What power will each of them consume if transfer them to the serial circuit with the same voltage of source.

Sorry for my bad english translations I m not use to writing questions in English.. So what confuses me is that on this forum similar question was asked and it stated that Power in serial circuit is 1/P=1/P1+1/P2. (https://www.physicsforums.com/threa...series-and-parallel-combination.827649/page-2 ) Should I use that formula or not? I was thinking that both in parallel and serial power is just P=P1+P1, and power in parallel is therefore 100W, but power of serial is less because total serial resistance is greater than individual resistance of 1 resistor, while in parallel total resistance is less than individual resistance of 1 resistor, so total power would be greater in parallel circuit than in serial circuit?

Homework Equations



The Attempt at a Solution

[/B]
In parallel 1/R=1/R1+1/R2 and power is P=V^2/R
IN series R=R+R2 but power here is less since we have the same R1 and R2 but greater R(total) in serial makes it that P(total) in serial is less thatn P(total) in parallel?
Does this make sense or should it be like the previously asked question here where P in series is 1/P=1/P1+1/P2 ?
 
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  • #2
Hello crostud, :welcome:

It would be better if you understood the characteristics of parallel and series circuits. The proper formulas to use follow from that understanding.

If the resistors are in parallel, the voltage over each of them is the same.

You already have ## P=V^2/R ## They give the power ratio, so $${V^2/R_1 \over 76 } = {V^2/R_2 \over 24 }\ . $$Now what is asked is the power for the serial circuit, with resistance ##R_1 + R_2##. The expression for that power is ##V^2\over R_1+R_2##. From now on it is math: express e.g. ##R_2## in terms of ##R_1## and calculate the power.
 
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  • #3
Thank you very much but I got the relation between R2 and R1 and I still can't calculate P1 and P2 in serial (power of each resistant like it's asked from me). I don't know where to go from after I got the relation between R1 and R2 since I don't know what's the V, I only know that it's the same in parallel and serial.
 
  • #4
crostud123 said:
Thank you very much but I got the relation between R2 and R1 and I still can't calculate P1 and P2 in serial (power of each resistant like it's asked from me). I don't know where to go from after I got the relation between R1 and R2 since I don't know what's the V, I only know that it's the same in parallel and serial.
You know the power with V across each resistor.
With the resistors in series and their ratios known, you can calculate the voltage across each in terms of V e.g. V-x,x.
You can then calculate the power with V-x or x across each resistor compared to V.
You can't calculate and don't need the actual values of V or R1 or R2 or the current.
 
  • #5
Carrock said:
You know the power with V across each resistor.
With the resistors in series and their ratios known, you can calculate the voltage across each in terms of V e.g. V-x,x.
You can then calculate the power with V-x or x across each resistor compared to V.
You can't calculate and don't need the actual values of V or R1 or R2.

Sorry, still quite don't understand. I always get to a dead end :(
 
  • #6
crostud123 said:
R1 and R2 since I don't know what's the V
Show your work. Just shouting I don't know doesn't help you further.
You have ##V^2/R_1##; what doe you have for ##R_2## in terms of ##R_1## ? Why is it so difficult to calculate ##V^2 \over R_1+R_2## ?
 
  • #7
BvU said:
Why is it so difficult to calculate ##V^2 \over R_1+R_2## ?
I would find it difficult to do this; as it is not the question asked I won't try.This is correct.
In parallel 1/R=1/R1+1/R2 and power is P=V^2/R
IN series R=R1+R2 but power here is less since we have the same R1 and R2 but greater R(total) in serial makes it that P(total) in serial is less than P(total) in parallel?
From #2
$${V^2/R_1 \over 76} = {V^2/R_2 \over 24 }\ . $$
You can work out from this the numeric ratio ##{R_1 \over R_2}##.
I don't know what's the V, I only know that it's the same in parallel and serial.
WRONG! or English error... the voltage across each resistor in series (not serial) is less than V.

Hint: a related simpler problem:
If you have (say) two resistors in series, one twice as big as the other, connected across any nonzero voltage (say X), can you say anything about the ratio of the voltages across those resistors and express the voltages in terms of X? What must be the sum of those two voltages?I expect you to also get a similar expression for the main problem; you may then see how to solve it.

Your original attempt was a good start; this is as far as I can presently go without (I hope) breaching forum rules.
 
  • #8
crostud123 said:
What power will each of them consume if transfer them to the serial circuit with the same voltage of source
Once you have the sum of powers, it's easy to find the separate ones.
Carrock said:
I would find it difficult to do this; as it is not the question asked I won't try.
You don't have to: it's crostud's exercise.
 

1. What is the difference between resistors in parallel and series circuits?

In a series circuit, the resistors are connected end to end, creating a single current path. In a parallel circuit, the resistors are connected side by side, creating multiple current paths.

2. How does the power of resistors differ in parallel vs series circuits?

In a series circuit, the power of the resistors is equal to the sum of all the individual resistors. In a parallel circuit, the power of the resistors is divided between each resistor, with the total power being equal to the sum of the individual powers.

3. Which circuit configuration is more efficient for power dissipation?

In most cases, a parallel circuit is more efficient for power dissipation. This is because the resistors in a parallel circuit have their own individual current paths, allowing for more current to flow through each resistor and more heat to be dissipated.

4. How does the voltage affect the power of resistors in parallel and series circuits?

In a series circuit, the voltage is divided between each resistor, resulting in less voltage across each individual resistor and therefore less power dissipation. In a parallel circuit, the voltage is the same across each resistor, resulting in more power dissipation for each resistor.

5. Can the power of resistors in parallel ever be greater than the power of resistors in series?

No, the power of resistors in parallel can never be greater than the power of resistors in series. This is because the total power in a parallel circuit is always equal to or less than the total power in a series circuit with the same resistors.

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