My Math Forum Open circuit?

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 March 6th, 2017, 07:31 AM #1 Member   Joined: Nov 2016 From: Ireland Posts: 72 Thanks: 2 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? $\displaystyle 30 * (60/(100+60))$
 March 6th, 2017, 07:47 AM #2 Member   Joined: Nov 2016 From: Ireland Posts: 72 Thanks: 2 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.
 March 6th, 2017, 08:39 AM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,721 Thanks: 1375 Thanks from romsek and Kevineamon
 March 6th, 2017, 09:13 AM #4 Member   Joined: Nov 2016 From: Ireland Posts: 72 Thanks: 2 Hmmm so R2 parallel RL (60 * 100) / (60 + 100) = 37.5 37.5 + 100 = 137.5ohms ________________________ 30 / 137.5 = 0.218A ________________________ 0.218 * (100/ (60+ 100)) = 0.13625A ________________________ 0.13625 * 60 = 8.175V
 March 7th, 2017, 12:08 AM #5 Senior Member   Joined: Apr 2014 From: UK Posts: 807 Thanks: 300 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 Thanks from Kevineamon

 Tags circuit, open