Resistance in a series-parallel combination

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Homework Help Overview

The discussion revolves around a circuit problem involving two heating elements, each with a resistance of 20 Ohms. The original poster seeks to determine a resistor value (R) that maintains consistent total power output whether one or both heating elements are operational.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster explores various approaches to find the resistor value, including using formulas for power and resistance in series and parallel configurations. They question the implications of their calculations on power output when both heating elements are used versus when only one is used.

Discussion Status

Some participants have provided guidance on setting up equations for total power loss in different scenarios, suggesting equating these equations to solve for R. However, there is no explicit consensus on the best approach, and the original poster expresses uncertainty about the reasoning behind using power loss equations.

Contextual Notes

The original poster is working within the constraints of a homework assignment, which may impose specific rules on how to approach the problem. There is an ongoing exploration of assumptions related to power distribution and resistance in the circuit.

Icheb
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The circuit in question is this:
http://www.atnetzwerk.de/temp/wasserkocher2.gif

There are two heating elements with a resistance R_0 = 20 Ohm. I have to find R so that the total power output of the heating elements stays the same no matter if I only use one heating element or both.

I first thought of using [tex]R = \frac{R_0 * R_0}{R_0 + R_0}[/tex], but I think that would mean that the power output would be higher when both heating elements are in use. Am I right?

Then I tried solving [tex]R = \frac{U^2}{P}[/tex] for P and doing [tex]U^2 / R_1 = U ^2 / R_2[/tex] for R_1 being the resistance when only one heating element is in use and R_2 being the resistance when both heating elements are in use. Obviously that just leaves me with [tex]R_1 = R_2[/tex], which doesn't help with this problem.

Another approach I tried was just saying that R has to be equal to R_0, because then the maximum available power P would arrive at the resistor at R_0 when only one heating element is in use and P/2 would arrive at each of them when both of them are turned on. However I don't know how to prove this or even if this is the right idea at all.

Does anyone have an idea as to how I could solve this problem?
 
Last edited by a moderator:
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Icheb said:
The circuit in question is this:
http://www.atnetzwerk.de/temp/wasserkocher2.gif

There are two heating elements with a resistance R_0 = 20 Ohm. I have to find R so that the total power output of the heating elements stays the same no matter if I only use one heating element or both.

I first thought of using [tex]R = \frac{R_0 * R_0}{R_0 + R_0}[/tex], but I think that would mean that the power output would be higher when both heating elements are in use. Am I right?

Then I tried solving [tex]R = \frac{U^2}{P}[/tex] for P and doing [tex]U^2 / R_1 = U ^2 / R_2[/tex] for R_1 being the resistance when only one heating element is in use and R_2 being the resistance when both heating elements are in use. Obviously that just leaves me with [tex]R_1 = R_2[/tex], which doesn't help with this problem.

Another approach I tried was just saying that R has to be equal to R_0, because then the maximum available power P would arrive at the resistor at R_0 when only one heating element is in use and P/2 would arrive at each of them when both of them are turned on. However I don't know how to prove this or even if this is the right idea at all.

Does anyone have an idea as to how I could solve this problem?

Just work out the equation for total power loss for each scenario, equate them and solve for R.

[tex]P = I^2R_{eff}[/tex]

[tex]I = U/(R+R_0)[/tex]
(1)[tex]P = I^2R_0 = \frac{U^2}{(R+R_0)^2}R_0[/tex]

[tex]I = U/(R + R_0/2)[/tex]
(2)[tex]P = I^2R_0/2 = \frac{U^2}{(R+R_0/2)^2}R_0/2[/tex]

Equate (1) and (2) and solve for R. You will need to solve a quadratic equation.

AM
 
Last edited by a moderator:
You are the best! Thank you very much.
 
If you don't mind my asking: How did you arrive at the conclusion that you need the equations for total power loss? I simply can't figure this out.
 

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