MHB Find Solution to Power & Resistance w/ 10 Ohm Resistors

[dstorm]

Here is the question: You are given a number of 10 ohm resistors, each capable of dissipating only 1.0 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 10 ohm resistance that is capable of dissipating at least 5.0 W?

When I attempted to do this problem, I had thought that power added in parallel, so I assumed that it would take 5 resistors. However, after checking to see if I was correct in the back of the book, I realized that I was doing something wrong. So now I'm left with the rules of resistance in parallel, being Rp(total) = Sum(Rp(i)) and the rule of resistance in series, being 1/(Rs(total)) = Sum(1/(Rs(i))). However, from here, I'm not sure how to manipulate these equations. How would I go about this problem? I'm sorry that I'm not very far along on it. It just stumped me.

 

[I like Serena]

Here is the question: You are given a number of 10 ohm resistors, each capable of dissipating only 1.0 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 10 ohm resistance that is capable of dissipating at least 5.0 W?

When I attempted to do this problem, I had thought that power added in parallel, so I assumed that it would take 5 resistors. However, after checking to see if I was correct in the back of the book, I realized that I was doing something wrong. So now I'm left with the rules of resistance in parallel, being Rp(total) = Sum(Rp(i)) and the rule of resistance in series, being 1/(Rs(total)) = Sum(1/(Rs(i))). However, from here, I'm not sure how to manipulate these equations. How would I go about this problem? I'm sorry that I'm not very far along on it. It just stumped me.

Hi dstorm! Welcome to MHB! :)

If you put 2 resistors in series, the total resistance doubles.
And if you put 2 resistors in parallel, the total resistance is halved.

So suppose we put 2 in series and put them parallel to another 2 in series.
Then we will still have 10 Ohm as the total resistance.
What will the power dissipation in each resistor be?

 

[dstorm]

Hi dstorm! Welcome to MHB! :)

If you put 2 resistors in series, the total resistance doubles.
And if you put 2 resistors in parallel, the total resistance is halved.

So suppose we put 2 in series and put them parallel to another 2 in series.
Then we will still have 10 Ohm as the total resistance.
What will the power dissipation in each resistor be?

Thanks! So the total power would be 4 W, because each can produce 1 W of power, right? So in order to get at least 5 W of power, we would need 9 resistors in a 3x3 form because the parallel ones would divide the resistance by 3 and the series ones would triple the resistance, thus equalizing it back to 10 ohms. So you can kinda combine the equations?

 

[I like Serena]

Thanks! So the total power would be 4 W, because each can produce 1 W of power, right? So in order to get at least 5 W of power, we would need 9 resistors in a 3x3 form because the parallel ones would divide the resistance by 3 and the series ones would triple the resistance, thus equalizing it back to 10 ohms. So you can kinda combine the equations?

Yeah.
Theoretically a more convoluted circuit could be set up.
But then you get fractions that simply won't come out as 10 Ohm any more.
And anyway, the current has to be distributed evenly to avoid resistors from blowing up.