Tricky resistor network problem?
Exercise 1-2. Beginning with four 1k resistors, synthesize a resistor with a value of 3/5 k and one with 5/3 k. Use no more than 4 1k resistors in each case.
Now, this would be an easy question with more resistors, but I can't see any way to make 1/5 k (let alone 3/5k) with four resistors; it'd be easy with 15 (three sets of 5 parallel resistors wired in series for 3/5 k, or five sets of 3 parallel resistors for 5/3 k). But *four* 1k resistors? I'm baffled.
P.S. I'm not enrolled at MIT and this is not a credit course, I'm taking it solely for my own intellectual curiosity and there are no professors I can ask.
Answer:
3/5K = 0.6K = Two 1K in parallel plus that combination in series with 1K plus that combination of 3 in parallel with 1K.
5/3K = 1.667K = Two 1K in series plus that combination in parallel with 1K plus that combination of 3 in series with 1K.
How about I just give you every possible combination and resistance value:
4 in series = 4k
4 in parallel = 1/4k
2 in parallel with 2 more in series = 2 1/2k
3 in parallel with 1 more in series = 1 1/3k
2 in parallel then 2 in parallel(in series) = 1k
That's for all 4. Now using 1,2 or 3:
1 resistor = 1k
2 in series = 2k
2 in parallel = 1/2k
3 in series = 3k
3 in parallel = 1/3k
1 in series with 2 in parallel = 1 1/2k
Well, I can do 5/3 for you - 1 resister in series with a parallel network of 1 resister in one leg, and 2 in the other. Total Resistance = 5/3 ohm...
Ron.
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