ECN Forum Forums General Topics of the Trade General Discussion Area Conundrum

NJ,/Kenbo: No offence intended, so please accept my apology if my remark was taken the wrong way.

regards

Alan

Alan,

Think of it like this.

Because the circuit is symmetrical, and the resistors identical the current flowing through each of the resistors (R1, R2, R3) leaving point A will be equal.

Since this is true, the voltage at the far end of each of the three resistors in section 1 will be the same, so for the purposes of analysis you can connect them all together (i.e. in parallel) without affecting the operation of the circuit.

The same can be said with the resistors connected to point B, which means that they can also be connected in parallel.

Now, if the top three resistors are all in parallel, and so are the bottom 3, then by default, the middle 6 are also in parallel.


And for resistors in parallel, their resistance is

(1/((1/1ohm)+(1/1ohm)+(1/1ohm)+(1/1ohm)+(1/1ohm)+(1/1ohm))) = 1/6 ohm
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Another way you can look at it, is since the middle section resistors have half as much current going through them (per resistor) as the top and bottom section (per resistor), they will have half the voltage drop, so by ohm's law this section must have half the resistance of the top or bottom section.

Alan no offence taken I was trying to be funny (smiley at end)

Bob took second look at your drawing before I realised you had the same as Scott.

Well done. I was going to offer prize of a free holiday in Scotland. At one of "Her Majesties" prime holiday camps. HMP Glenochil but I do not want to upset the governor (or get sacked) so the offer has expired

The secrete to solving the problem is to recognize that since nodes 1,2 and 3 and 4,5 and 6 are at the same potential. This allows
you to short circuit these terminals without changing the ckt and re-draw the circuit into a simpler form.


[This message has been edited by Bob (edited 09-22-2006).]