March 6th, 2017, 08:31 AM  #1 
Member Joined: Nov 2016 From: Ireland Posts: 67 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, 08:47 AM  #2 
Member Joined: Nov 2016 From: Ireland Posts: 67 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, 09:39 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,549 Thanks: 1260  
March 6th, 2017, 10:13 AM  #4 
Member Joined: Nov 2016 From: Ireland Posts: 67 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, 01:08 AM  #5 
Senior Member Joined: Apr 2014 From: UK Posts: 765 Thanks: 288 
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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