Unknown resistor solving

Aug 16, 2006 22 Replies
J

Duh! What did I just say?

And, yes, "singular" was a slip, should have said, "single".

...Jim Thompson

| James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
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Y

Isn't R3 paralleled with R6 or have I missed something?

- YD.

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R

I'd just redraw it:

Vr | [R1] | +----+-----+--o Va | | | | | [?R4] [R3] [?R6] | | | +--o Vm | | | | | [?R5] +----+-----+--o Vb | [R2] | Vs

So the current through the whole thing is (Vr - Va)/R1, which should be equal to (Vb - Vs)/R2, or at least it had better be, and that current also goes through the parallel combination of R3 || R6 || (R4 + R5). You know R3, so the best you can solve for is the current through the parallel combination of R6 || (R4 + R5); which voltage drop is, of course, Va - Vb.

You're right. There are too many unknowns, unless somebody knows something that they haven't revealed yet.

Maybe it's just that a page full of arithmetic is enough to baffle the teacher. ;-)

Cheers! Rich

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