Circuit problem with ammeter and unknown resistanceby Femme_physics Tags: ammeter, circuit, resistance, unknown 

#55
May1811, 12:13 PM

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I recognize the characters and have rudimentary vocabulary from selfstudy אולפן. But in this case, google translate came to the rescue. (caveat: for those wanting to try this, it requires setting up your keyboard for Hebrew. However, do this at your own peril , an accidental keystroke can suddenly change your keys from Latin to Hebrew and you'll spend the next hour trying to change it back. In fact, it is a good idea to solve these exercises using at least one more method. If you arrive at the same answer, there is a high likelihood your solution is correct. (ILS also alludes to this in step 4 of his last post). 



#56
May1811, 11:34 PM

PF Gold
P: 2,551

Hmm...does there have to be a point in a circuit that has 0 V potential? 



#57
May1811, 11:44 PM

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You seem to have missed my last reply (2 equations!): http://www.physicsforums.com/showpos...4&postcount=50 



#58
May1811, 11:45 PM

PF Gold
P: 2,551

Sorry, I lost the thread of this thread!
My worst pun to date ;) I'll get to it and post back 



#60
May1811, 11:49 PM

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Just try it! [edit] (You procrastinator you! ) [/edit] 



#61
May1911, 12:16 AM


#65
May1911, 12:37 AM


#66
May1911, 12:39 AM

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P: 9,812

Why minus??????????!!!!!!!!!!!!
ehild 



#67
May1911, 12:40 AM

PF Gold
P: 2,551

My bad, discount the minus. I was excited I was getting somewhere :P
Rx = R3R2/R1 Is that right? 



#68
May1911, 12:45 AM

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Wait! Sorry! No it is a mock! I was just wondering how Rx could be negative. What could it mean? Did you just discover negative resistors? But that's great! We could generate voltage out of nothing!!! With the method we used, we did introduce 1 unknown to many, but it didn't matter, because it would cancel out. 



#69
May1911, 12:48 AM

PF Gold
P: 2,551

Fixed. (as in my post above ehild!) Is that right? 



#70
May1911, 12:51 AM

P: 1,877

Of course, you're free to take my approach or not. In my mind the great thinking comes from realizing why you get the answer you do, why the circuit behaves as it does, and not in actually getting the answer. Hey, congrats on getting the problem right though! Now, try using the voltage divider at the left and rightmost points, setting them equal, and solving for Rx. Once you do, you will see how wonderful keeping it simple really is. Using the voltage divider ought to take 4 minutes at most :) 



#71
May1911, 01:02 AM

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ehild 



#72
May1911, 01:57 AM

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