Prove that you are a genius?

Question

http://img407.imageshack.us/img407/2054/18number14zc8.jpg

Solve that problem and I will give ten points to the first person that gets it right. That person to me will be a genius. Thanks.

Answer 1

I'm not a genius, but I'll give this a shot.

The confusing part is that there are two point b's in the diagram:
one on the right battery terminal, and the other at the right of the bridge.

Given that this looks like a classic tester to find an unknown resistance (some ancient scientist has their name attached, although I don't recall the name), I will assume that the resistance measured across the battery terminal drops by half when switch S2 is closed.

Draw a diagram with the switch S2 open and a separate one with S2 closed. For parallel resistors, use the resistance reduction:
1/Rc=1/R1+1/R2+...+1/Rn

where Rc is the combined resistance of parallel resistors

and series:
Rc=R1+R2+...+Rn

So, with switch S2 open, the resistance around the circuit is:

The bridge has two branches with two resistors in series:
Rc=112+13

The two branches are in parallel, so
Rc=(112+13)*(112+13)/
((112+13)+(112+13))

or
Rc=(112+13)*(112+13)/
(2*(112+13))

so
Rc=(112+13)/2

Adding in the unknown R,
the total resistance across the battery is:
Rc=R+(112+13)/2

with switch S2 closed, the resistance around the circuit is:

Now the bridge has put the pairs of 112 and 13 ohm resistors in parallel

For either leg:
Rc=112*13/(112+13)
since the two legs are in series
Rc=2*112*13/(112+13)
Adding in the unknown R:
R+2*(112*13)/(112+13)

From the problem statement

The resistance when open is twice when closed
so
R+(112+13)/2=
2*(R+2*(112*13)/(112+13))

2*R-R=R=
(112+13)/2-4*(112*13)/(112+13)
R=15.9 ohms

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