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March 6th, 2017, 08:31 AM   #1
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Open circuit?



Hey guys
I'm looking at this circuit here. This open voltage is confusing me. Should I just consider this to be a simple 30v Pd?

In this case should I simply use the voltage division rule?
The half way point on the potentiometer would be 100 ohms right?

$\displaystyle 30 * (60/(100+60))$
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March 6th, 2017, 08:47 AM   #2
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Hang on this is tricky it wouldn't be voltage division as the two are in parallel right? Hmmm... K now I'm even more confused.
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March 6th, 2017, 09:39 AM   #3
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March 6th, 2017, 10:13 AM   #4
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Hmmm so

R2 parallel RL

(60 * 100) / (60 + 100) = 37.5

37.5 + 100 = 137.5ohms

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30 / 137.5 = 0.218A

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0.218 * (100/ (60+ 100)) = 0.13625A

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0.13625 * 60 = 8.175V
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March 7th, 2017, 01:08 AM   #5
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You could have made life easier tho, since you have the current (0.218A) and have the resistance at the load nodes (37.5R) then the voltage is simply 0.218A x 37.5R = 8.182V

You also have a rounding error
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