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Two batteries in the same circuit

  • Thread starter Mårten
  • Start date
  • #1
126
1

Homework Statement


Find the voltage between A and B in the circuit below:

2rog6mg.gif


Homework Equations


Don't know really, except for Ohm's law and Kirchhoff's laws, in some combination.

3. The Attempt at a Solution (rather the solution that the key gives, which I don't understand)
(12-6)/150 = 0.04 Ampere
U_AB = 6 V + 0.04*50 = 8.0 V

4. Questions
The main problem I have here, is that I get confused when there are two batteries in one circuit. So:

a) Is there any standard approach to deal with serveral batteries, e.g. build some sort of battery equivalent, like you do when you have several resistors?

b) They seem to substract one of the battery's voltages from the other. Why's that? On one hand, you can see that one battery is "driving" the current in opposite direction to the other. But on the other hand, you could say that in the upper part of the circuit, the batteries' plus terminals form one common conductor, so it ought to be 12+6=18 V.

c) Then they seem to treat the resistors like they were in series. Well, in the left circle of the diagram, you can say they are in series, so maybe I'll agree with that.

d) Okey, now you have the current I. But now they do 6 V + 0.04*50 to get U_AB. Strange... Normally, you do U - RI to get the potential left... :confused:
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
12
Assuming these are ideal (hw question type) batteries with no internal resistance how much current is flowing in the circuit?
 
  • #3
LowlyPion
Homework Helper
3,090
4
The main problem I have here, is that I get confused when there are two batteries in one circuit. So:

a) Is there any standard approach to deal with serveral batteries, e.g. build some sort of battery equivalent, like you do when you have several resistors?

b) They seem to substract one of the battery's voltages from the other. Why's that? On one hand, you can see that one battery is "driving" the current in opposite direction to the other. But on the other hand, you could say that in the upper part of the circuit, the batteries' plus terminals form one common conductor, so it ought to be 12+6=18 V.

c) Then they seem to treat the resistors like they were in series. Well, in the left circle of the diagram, you can say they are in series, so maybe I'll agree with that.

d) Okey, now you have the current I. But now they do 6 V + 0.04*50 to get U_AB. Strange... Normally, you do U - RI to get the potential left... :confused:
The way that you want to treat this is as a loop.

Ohm's law says that the sum of the voltages around a loop - start at a point and return to it - since it is the same voltage as you started, then everything has to add up to zero.

So choose a direction. Either direction and follow element by element around the loop. BUT YOU MUST BE CAREFUL OF SIGNS. Since the batteries in this configuration are opposing your equation should reflect that one of them is negative relative to the other. When you add everything up you get the voltage from the current in the resistors and the difference in the batteries. The answer is correct it is .04 amps. Then choose either path between A and B and sum up the voltage of whichever battery and the voltage contribution from the current through which ever resister and you have your answer.
 
  • #4
126
1
So choose a direction. Either direction and follow element by element around the loop. BUT YOU MUST BE CAREFUL OF SIGNS. Since the batteries in this configuration are opposing your equation should reflect that one of them is negative relative to the other. When you add everything up you get the voltage from the current in the resistors and the difference in the batteries. The answer is correct it is .04 amps.
Okey, I did this, and it worked. That's what you call a potential walk, according to Kirchhoff 2, is it?

Then choose either path between A and B and sum up the voltage of whichever battery and the voltage contribution from the current through which ever resister and you have your answer.
Hm... Now you're adding the different elements, and above (with potential walk) you were subtracting. Confusing... Why's that? When am I supposed to add and when am I supposed to subtract? :confused:

And if you do this in the leftmost loop you get 12 V + 100*0.04 = 16 V. So it only works if I go through the battery to the right in the circuit. :confused:
 
  • #5
LowlyPion
Homework Helper
3,090
4
And if you do this in the leftmost loop you get 12 V + 100*0.04 = 16 V. So it only works if I go through the battery to the right in the circuit. :confused:
Beware the direction of the current. One path is 6V + .04(50) = 8V
The other path has current flowing opposite. That path is 12V - (.04)(100) = 8V.

Didn't I already tell you to Be Careful of signs?
 
  • #6
126
1
Okey, I got it now, rereading the Kirchhoff's laws... :redface: Thanks!
 

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