Engineering Maximizing Battery Voltage in a Complex Combination Circuit

[john2013]

The problems says "In the circuit shown in the figure all the resistors are rated at a maximum power of 1.20W"
Then it asks "What is the maximum emf that the battery can have without burning up any of the resistors?"

Really, The only part I think I'm having trouble with is finding what the resistance is when they are all put together, because I treat the top right part as if the two 20s are in parallel with each other, then that is in series with the 10 Ohm resistor, so I add that. Then the 3 at the top right are parallel, so I add them accordingly (with the 1/r thing)...and I think that is where my problem lies. Especially when I add the 20 ohms as if they are in parallel.

Here is the picture:

http://session.masteringphysics.com/problemAsset/1054826/4/YF-26-68.jpg


I think the Voltage I keep getting is 11.4 or so.

 

[Borek]

Your approach sounds correct to me, but second opinion won't hurt.

 

[shikobe]

So dude, was your answer right? I'm sort of in the same predicament as you right now lols

 

[lewando]

You can find the equivalent resistance using your stated approach. This is the first step. It allows you to define the source current in terms of E and R_eq. The problem is--now what do you do? You cannot simply apply the power rating of 1.2W to R_eq because the power rating of a resistor combination depends on the combination. For example, if you had N resistors in parallel, the power rating of this R_eq would be N*1.2W.

What you are looking for is the weakest link[the hottest link, actually]-- the resistor with the highest I²R product.

 

[NascentOxygen]

The two 20's in parallel amount to a 10. And that 10 is in series with another 10, making it 20. So in the top right corner there are 3 parallel branches: that 20 I just described in parallel with 15 in parallel with 25. The rest of the circuit you can add up in your head because they are in series.

Once you have the total loop resistance, you can work out an equation for the current drawn from the source, E volts.

Then, as lewando stated, you need to look for a resistor with a high resistance AND carrying a high current, so it has the highest I²R product.

 

[NascentOxygen]

There is not much working involved. They could have made the question a lot trickier by swapping two of the resistor values around, but they have settled on an easy arrangement.

 

[shikobe]

The Hightest I²R would be the resistor with 50\Omega right? I got an equivalent resistance of the whole circuit of 5940/47=126.38\Omega. I found the current in which the power of highest resistance would be less than 1.2W. So,
I²R<1.2
I<\sqrt{}1.2/50=0.155A
Then I found the \epsilon using \epsilon=IR=0.155x126.28=19.57V

Is that right :$

 

[Borek]

The Hightest I²R would be the resistor with 50\Omega right?

I don't think so. This is the highest resistance, but only half the current.

 

[NascentOxygen]

There isn't a resistor of 50 ohms. There are two resistors of 50 ohms and they are in parallel, which means the current divides and only half goes through each. (And because power = I²R when the current is halved, the power dissipated is one-quarter what it could otherwise be.)

So simply looking for the highest value resistor is not good enough. You have to calculate the I²R product for likely candidates before you can identify which resistor will get hottest.