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Resistors (easy, but tricky question)

by DivGradCurl
Tags: resistors, tricky
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DivGradCurl
#1
Oct11-05, 03:58 PM
P: 351
You are given a number of 8 [tex]\Omega[/tex] 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 8 [tex]\Omega[/tex] resistance that is capable of dissipating at least 5.6 W?

My approach is probably wrong, and so I need someone to help me find the mistake:

[tex]5.6 \mbox{ W} \leq n \cdot m \cdot 1.0 \mbox{ W}[/tex]

The minimum case is when [tex]n=m[/tex], and so

[tex]5.6 \mbox{ W} \leq n^2 \cdot 1.0 \mbox{ W} \Rightarrow n \geq \sqrt{5.6} \Rightarrow n = 3[/tex]

The minimum number of resistors is then 9.

Any help is highly appreciated.
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Gokul43201
#2
Oct11-05, 04:56 PM
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This is not an area that I'm all too familiar with, but I'll throw in some comments - use them at your own risk.

It looks like you are calculating for a rectangular array consisting of m rows in parallel, each row having n resistors in series. In such a case, n=m is the only configuration that will give R(eff) = R = 8 ohm (ie: it is not just the minimal possibility, it is the ONLY possibility).

Proof : Resistance of one row of n resistors (each of resistance R) = nR. Now if you have m such rows in parallel, then

[tex]\frac {1}{R_{eff}} =\sum_{i=1}^{m} \frac{1}{nR} = \frac {m}{nR} [/tex]

But we want R(eff) = R (= 8 ohms), so :

[tex]\frac{1}{R_{eff}} =\frac{1}{R} = \frac {m}{nR} [/tex]

[tex]\implies m/n = 1 [/tex]

--end of proof--

This however, assumes that each of the m rows has exactly the same n number of resistors. That need not be required to generate R(eff)=R. For instance, I can have 3 rows containing 4, 4 and 2 resistors each (1/4+1/4+1/2 = 1). But clearly, this uses more resistors than the optimal 9 and that's where, I suspect, the minimization has a role.

So, in this case, the minimum happens at n=3 or a total of 9 resistors as you correctly figured.

However, I'm not conviced that any arrangement of resistors can, in fact, be reduced to the above configuration (but this may just be something I'm unaware of, or haven't thought sufficiently about).

What if you have n bunches in series, each bunch having m resistors in parallel ? Is there an equivalent circuit for this configuration that reduces it to the above one ? This may very well be possible, using some clever star-delta conversion trick, but it's something I'm not sure about. If it CAN be reduced to the previous configuration, then I'd surmise that 9 is indeed the least number of resistors (but I just don't feel right about it).

It seems odd that the question will ask specifically about 5.6W when the best configuration that will work dissipates over 50% more than that.

<<...this is when someone comes along and posts the obvious 3 line solution >>
DivGradCurl
#3
Oct11-05, 05:25 PM
P: 351
9 is, in fact, the right answer. Thanks for your comments. I can picture it better now! I think that works the same way for the generic arrangement you have just mentioned. The value 5.6 W is probably just a random number my instructor came up with to keep my question different from that of my classmates.


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