Determining the total current of many batteries

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Homework Help Overview

The discussion revolves around a circuit problem involving multiple batteries and resistors, where participants are tasked with determining the total current and potential differences in various parts of the circuit. The original poster expresses confusion regarding the total current and the behavior of the batteries in the circuit.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Ohm's law and Kirchhoff's laws, questioning the calculations for potential differences and the interpretation of circuit elements. There are attempts to clarify the correct identification of potential gains and drops across components.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning and calculations. Some guidance has been offered regarding the interpretation of potential differences and the roles of internal resistance in the circuit. Multiple interpretations of the problem setup are being explored, particularly concerning the polarity of the potential difference.

Contextual Notes

There is some ambiguity in the problem statement regarding the polarity of the potential difference between nodes, which has led to different interpretations among participants. The original poster has acknowledged potential mistakes in their calculations and is seeking clarification on specific points.

Asmaa Mohammad
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Homework Statement


The figure represents a part of a circuit, and the potential difference between the f and d is 12Volts, depending on the information shown in the figure, determine:
1-the reading of the Ammeter
2-the value of (Vb)₃
3- the potential difference between (c , b)
vEhMp.jpg

Homework Equations


Ohm's law: Vb=I (R+r)
Kirchhoff's law: ΣI=0
ΣV=ΣIR

The Attempt at a Solution


Seriously, I got confused with all these batteries, and I considered that the total current is 3A, since Vfd=12.
bFHZJ.jpg
So, I know there may be a lot of mistakes in my solution, and I would appreciate any help or explanation of how those whole batteries would work and produce the total current.
 
Last edited:
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Answers to Q1 and Q2 look good. In Q3, you are asked to find the potential difference between c and b. I think instead you have calculated Vcd (which is also not correct).

(Also, I've never seen Kirchhoff's circuit laws named as 'first law' and 'second law'. They are called Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL).)
 
cnh1995 said:
Answers to Q1 and Q2 look good. In Q3, you are asked to find the potential difference between c and b. I think instead you have calculated Vcd (which is also not correct).

(Also, I've never seen Kirchhoff's circuit laws named as 'first law' and 'second law'. They are called Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL).)
I made a mistake, I wrote Vcd, but I meant Vcb, I edited the photo. And in my textbook the Kirchhoff's current law can be the first law and the voltage law is named the second law.
So please, is the answer correct now?
 
Asmaa muhammad said:
So please, is the answer correct now
You got 3A flowing through the 3 ohm resistor towards right and the 7V battery has an internal resistance of 1 ohm. This means, between c and b, you have 3A current flowing towards right through a 4 ohm resistor and a 7V source in series with that. That doesn't give 13V between b and c.
 
cnh1995 said:
You got 3A flowing through the 3 ohm resistor towards right and the 7V battery has an internal resistance of 1 ohm. This means, between c and b, you have 3A current flowing towards right through a 4 ohm resistor and a 7V source in series with that. That doesn't give 13V between b and c.
The inner resistance of the battery would be used in Ohm's law, so the PD between the terminals of the battery is (V=Vb-Ir) so it equals 4V and this voltage is in series with the voltage on the 3Ω resistance which I would use Ohm's law to determine it (V=IR) so it equals 9V.
Then the total voltage would be (9+4=13V), Is that wrong, according to what you see I think that my mistake may be in the way I calculate them, so if you see any mistake please clarify it for me!
Where : r is the internal resistance, V is the potential difference, I is the current and Vb is the emf.
 
7V is the emf of the battery and 1 ohm is its internal resistance which is in series with the 3 ohm resistance. So, you have a 4 ohm resistor in series with an emf of 7V.
Asmaa muhammad said:
The inner resistance of the battery would be used in Ohm's law, so the PD between the terminals of the battery is (V=Vb-Ir) so it equals 4V.
Right, but what will be the polarity of that voltage? Is it a potential drop or a potential gain?
 
cnh1995 said:
7V is the emf of the battery and 1 ohm is its internal resistance which is in series with the 3 ohm resistance. So, you have a 4 ohm resistor in series with an emf of 7V.

Right, but what will be the polarity of that voltage? Is it a potential drop or a potential gain?
It is a potential drop, so it would be like this:
V=IR=3*4=12V
And then should I add them to the emf of the battery, so the total PD would be 19V, or add them to the PD between tne two terminals of the battery and then the total PD would be 16V?
Sorry if I understand slowly or ask a lot of question:biggrin:!​
 
Asmaa muhammad said:
And then should I add them to the emf of the battery, so the total PD would be 19V, or add them t the PD between tne two terminals of the batteryand then the total PD would be 16V?
No.
Just redraw the circuit from c to b.
Draw a 4 ohm resistor (3 ohm + 1 ohm) in series with the battery of 7V emf. Note the polarity of the battery. A current of 3A is flowing through the c-b branch from c to b. So, when going from c to b
i) What is the potential change across the 4 ohm resistor? Is it a gain or a drop?
ii) What is the potential change across the battery? Is it a gain or a drop?
Add the answers of i) and ii) and you'll get Vcb.
Asmaa muhammad said:
Sorry if I understand slowly or ask a lot of question:biggrin:!
Please don't be!
 
  • #10
I note that the problem statement leaves it a bit vague as to the polarity of the 12 V potential difference between nodes f and d. To me that means that there are two valid solutions to the problem.
 
  • #11
cnh1995 said:
No.
Just redraw the circuit from c to b.
Draw a 4 ohm resistor (3 ohm + 1 ohm) in series with the battery of 7V emf. Note the polarity of the battery. A current of 3A is flowing through the c-b branch from c to b. So, when going from c to b
i) What is the potential change across the 4 ohm resistor? Is it a gain or a drop?
ii) What is the potential change across the battery? Is it a gain or a drop?
Add the answers of i) and ii) and you'll get Vcb.
Ok,
MRjx2.jpg

i) V= IR=4*3=12V (a drop)
ii) V= Vb-Ir=7-3=4 ( a gain because the current of the battery is going out from the positive to the negative) I use the conventional direction.
So 12+4=16V
Or what did you mean by the potential change across the battery?
 
  • #12
gneill said:
I note that the problem statement leaves it a bit vague as to the polarity of the 12 V potential difference between nodes f and d. To me that means that there are two valid solutions to the problem.
Could you please explain how would the polarity of 12 V PD between f and d change the case? I didn't understand what do you mean?
And it is not given in the peoblem statement so I considered the current would flow from f to d.
 
  • #13
Asmaa muhammad said:
i) V= IR=4*3=12V (a drop)
Right.
Asmaa muhammad said:
V= Vb-Ir=7-3=4 ( a gain because the current of the battery is going out from the positive to the negative) I use the conventional direction.
No. The internal resistance is taken care of already in i). So in ii), you must consider the emf of the battery, which is 7V.

You are right about the potential gains and drops in i) and ii).
So, what is Vcb now?
 
  • #14
Asmaa muhammad said:
Could you please explain how would the polarity of 12 V PD between f and d change the case? I didn't understand what do you mean?
And it is not given in the peoblem statement so I considered the current would flow from f to d.
Consider if the current flowed from d to f instead.

Anyways, you should finish up your analysis of your version of the problem first before considering this second version.
 
  • #15
cnh1995 said:
Right.

No. The internal resistance is taken care of already in i). So in ii), you must consider the emf of the battery, which is 7V.

You are right about the potential gains and drops in i) and ii).
So, what is Vcb now?
So it would be 19V?
 
  • #16
Asmaa muhammad said:
So it would be 19V?
No. You are adding a drop and a gain.
 
  • #17
cnh1995 said:
No. You are adding a drop and a gain.
So what should I do?, you are going to make me cry.
 
  • #18
Asmaa muhammad said:
So what should I do, you are going to make me cry.
From c to d, you encounter a 12V "drop" in potential across the resistor and a 7V "gain" across the battery.

You spent 12 dollars and earned 7 dollars. How many dollars did you effectively earn or spend?
I don't know if I can make it any simpler.
 
  • #19
cnh1995 said:
From c to d, you encounter a 12V "drop" in potential across the resistor and a 7V "gain" across the battery.

You spent 12 dollars and earned 7 dollars. How many dollars do you have now?
I don't know if I can make it any simpler.
I lost 5 dollars so the potential difference would be 5V and that is a drop in the voltage.
Is this correct?
 
  • #20
Asmaa muhammad said:
I lost 5 dollars so the potential difference would be 5V and that is a drop in the voltage.
Is this correct?
Right!

Now you can think about the second version of the problem as gneill mentioned earlier.
 
  • #21
gneill said:
Consider if the current flowed from d to f instead.

Anyways, you should finish up your analysis of your version of the problem first before considering this second version.
Oh yes! That will change things:( Vb)3 would be 28V, and the ammeter still reads 3A, and the potential difference between c and b would be 19V ( as I understand it because I have a problem with Q3) are these answers correct?
 
  • #22
Asmaa muhammad said:
Oh yes! That will change things:( Vb)3 would be 28V, and the ammeter still reads 3A, and the potential difference between c and b would be 19V ( as I understand it because I have a problem with Q3) are these answers correct?
Yes, that looks good.
 
  • #23
2KbrA.jpg
Is this correct, cnh1995!
 
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  • #24
gneill said:
Yes, that looks good.
Thank you very much,gneill!
 
  • #25
Asmaa muhammad said:
2KbrA.jpg
Is this correct, cnh1995!
Yes. Good job!
 
  • #26
cnh1995 said:
Yes. Good job!
Thank you very much, you helped me a lot.
 
  • #27
Since the OP has correctly solved the problem (both cases) I 'll present a summary of the work in the form of two annotated circuit diagrams. In both diagrams the battery internal resistances have been lumped together with the external resistors.

Case 1: the given PD causes current to flow from f to d:

upload_2016-12-18_8-21-22.png


Case 2: the given PD casues current to flow from d to f:

upload_2016-12-18_8-22-7.png
 
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  • #28
Asmaa muhammad said:
Thank you very much, you helped me a lot.
You're welcome!:smile:
 

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