Resistance in a series-parallel combination

In summary, the conversation discusses finding the resistance (R) for a circuit with two heating elements (R0 = 20 Ohm) so that the total power output remains the same when using one or both elements. Different equations and approaches are considered, but the final solution involves equating and solving the equations for total power loss in each scenario. This requires solving a quadratic equation.
  • #1
Icheb
42
0
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?
 
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  • #2
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
 
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  • #3
You are the best! Thank you very much.
 
  • #4
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.
 

What is a series-parallel combination?

A series-parallel combination is a circuit arrangement in which resistors are connected both in series and in parallel. In this configuration, some resistors are connected in a series, while others are connected in parallel. This type of combination is commonly used in circuits to achieve the desired resistance value.

How do you calculate the total resistance in a series-parallel combination?

To calculate the total resistance in a series-parallel combination, you can use the following formula: RT = R1 + R2 + (1/R3 + 1/R4 + ...)-1 This formula takes into account the resistors connected in series, as well as the inverse of the resistors connected in parallel.

What happens to the total resistance in a series-parallel combination when more resistors are added?

When more resistors are added to a series-parallel combination, the total resistance will decrease. This is because adding more resistors in parallel will decrease the overall resistance, while adding more resistors in series will increase the overall resistance. Therefore, the effect of adding more resistors in parallel will outweigh the effect of adding more resistors in series, resulting in a decrease in total resistance.

How does the total resistance in a series-parallel combination compare to the individual resistances?

The total resistance in a series-parallel combination will always be less than the individual resistances. This is due to the fact that connecting resistors in parallel decreases the overall resistance, while connecting resistors in series increases the overall resistance. Therefore, the total resistance will be less than the individual resistances in a series-parallel combination.

Why is a series-parallel combination preferred over a purely series or parallel circuit?

A series-parallel combination offers the advantages of both a series circuit and a parallel circuit. It allows for a variety of resistance values to be achieved, while also providing a lower total resistance than a purely series circuit. Additionally, it is more flexible and can handle a wider range of current and voltage levels compared to a purely parallel circuit. Therefore, a series-parallel combination is preferred for its versatility and efficiency in various circuit designs.

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