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GR0177 #85
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Alternate Solutions |
94709
2007-10-29 23:04:05 | You can calculate current although one doesn't have to. This is just a voltage divider.
 ,  ,  Volts.
Therefore,
Since the base is  V, the potential at the dividing segment is 2.4V. |  |
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Comments |
grae313
2007-11-01 10:17:36 | Alternately, (and this is essentially equivalent 94709's solution, but involves simpler calculations, IMO), once you have the two resistances for the two wires, you can realize that  which means that the ratio of the voltage drops in each wire segment has to be 1:4, or:
 and also
easy math to arrive at the answer.
grae313
2007-11-01 10:17:59 |
ack, I meant to classify this as an alternate solution
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|  | 94709
2007-10-29 23:04:05 | You can calculate current although one doesn't have to. This is just a voltage divider.
, , Volts.
Therefore,
Since the base is V, the potential at the dividing segment is 2.4V.
94709
2007-10-29 23:14:15 |
eh, you might not be checking this site any more, but thanks a lot, yosun.
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| | keflavich
2005-11-10 22:38:30 | There are a few typos here. }) , you left out the 5. That 0 at the end should be a ')'.
But also... where does the 1 come from in  ?
yosun
2005-11-10 22:49:04 |
keflavich: thanks for the typo alert; the typo has been corrected.
the 1 comes from the fact that the potential at the end of the fatter wire is 1V, and thus the voltage across the junction is just the voltage at the end of the fat wire plus the current times the resistance of the fatter wire. (i tend to leave off the units to avoid confusion with variables)
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Richard
2007-10-29 10:07:52 |
A perhaps conceptually easier approach would be to simply find the potential difference across the first wire, and then subtract it from the potential at the end of the thin wire. In my view, this is consistent with how you went about finding the current...
=2.4)
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jmason86
2009-07-09 23:54:25 |
Alternatively, you could have done    yielding the same value for I, but negative.
Then following along Yosun's line (where he implied  ), we now have instead,  where I is now negative. Plug in to get 
I was kind of confused by the above response that Yosun gave using the fatter wire for everything... so I decided to try flipping it around to make sure it still works... and it does :)
Cheers
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