
#1
Sep1011, 08:51 PM

P: 38

1. The problem statement, all variables and given/known data
Find the current at point I: 2. Relevant equations V=IR 3. The attempt at a solution It seems there should be no current at point I, since there is no voltage. The total resistance in the circuit is 77 ohms. 10.0 V  (10.0 V/77 ohms)*(77)=0. However, the answer is supposedly 0.13 A. What is wrong with my reasoning? Thanks in advance. 



#2
Sep1011, 08:56 PM

HW Helper
PF Gold
P: 3,444

OK, so the the total resistance is 77 Ohms. If you put a 10V battery across 77 Ohms, how much current is provided by the battery? That's what the problem is asking.




#3
Sep1011, 09:01 PM

P: 38

Oh, I see. But at that point, there would be no current, right? If there is no voltage, how can there be a current?




#4
Sep1011, 10:08 PM

Sci Advisor
P: 4,003

Current after passing all resistors
No.
The current leaves the positive terminal of the battery and travels around the circuit through all the resistors (splitting equally when equal resistors are in parallel) and then exactly the same current arrives back at the negative terminal of the battery. 



#5
Sep1011, 10:09 PM

HW Helper
PF Gold
P: 3,444

Voltage is a potential difference between two points in a circuit. Between which two points are you saying there is no voltage? Also, if you agree that the battery provides a current, where do the electrons that leave the battery go?




#6
Sep1011, 10:30 PM

P: 38

If there is no potential difference between these two points, how can there be a current (0=IR)? 



#7
Sep1011, 10:33 PM

Sci Advisor
P: 4,003

You have drawn these as two different points for convenience, but they are electrically the same point.
The current leaves the positive terminal of the battery and travels around the circuit through all the resistors (splitting equally when equal resistors are in parallel) and then exactly the same current arrives back at the negative terminal of the battery. 



#8
Sep1011, 11:34 PM

P: 38

Thank you, I think I understand now.




#9
Sep1111, 04:43 AM

P: 335

V = IR so I = V/R, on substituting the values, we get 10/77, which 0.129A, which is the answer.



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