Resistance in a series-parallel combination

1. May 28, 2006

Icheb

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

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 $$R = \frac{R_0 * R_0}{R_0 + R_0}$$, 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 $$R = \frac{U^2}{P}$$ for P and doing $$U^2 / R_1 = U ^2 / R_2$$ 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 $$R_1 = R_2$$, 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: May 2, 2017
2. May 28, 2006

Andrew Mason

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

$$P = I^2R_{eff}$$

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

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

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

AM

Last edited by a moderator: May 2, 2017
3. May 28, 2006

Icheb

You are the best! Thank you very much.

4. May 29, 2006

Icheb

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.