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Resistance of a network

Posted: Mon Dec 15, 2008 5:57 pm
by btilly
I suggested this one as a PE problem but never got a response. Perhaps they thought it required too much physics, even though I stated all of the necessary physics.

Eddie the electrician gets bored with circuits wired in series and parallel, and decides to try something more complicated. He takes the numbers from 1 to 100 and makes the following connections with wires with 1 ohm resistors in the middle. First he connects 1 to 2, 2 to 3, ..., 99 to 100. Then he connects 2 to 4, 4 to 6, ..., 98 to 100. Then he connects 3 to 6, 6 to 9, ..., 96 to 99. And so on up to his final wire connecting 50 to 100.

He then measures the resistance of the whole network from 1 to 100. What resistance does he find?

Physics Reminder: In an electrical circuit, V = IR where V is the voltage in volts, I is the current in amps, and R the resistance in ohms. Furthermore current is neither created nor destroyed in simple circuits.

Re: Resistance of a network

Posted: Tue Dec 16, 2008 6:13 pm
by harryh
2.266378... ohms
And when Eddie's brother got more ambitious and repeated the trick with numbers 1..1000, he found 2.222733... ohms.
Furthermore current is neither created nor destroyed in simple circuits.
This doesnot look like a simple circuit to me, but I assumed that the same premise holds :)

Re: Resistance of a network

Posted: Tue Dec 16, 2008 11:50 pm
by btilly
harryh wrote:2.266378... ohms
And when Eddie's brother got more ambitious and repeated the trick with numbers 1..1000, he found 2.222733... ohms.
Yup!
harryh wrote:
Furthermore current is neither created nor destroyed in simple circuits.
This does not look like a simple circuit to me, but I assumed that the same premise holds :)
There are no capacitors and we don't have alternating current, so yes it is a simple circuit. Albeit a complicated one.

I included that fact because I didn't want to rely on any physics knowledge that people might not know, but highlighting that fact does make it easier.