Two batteries and two resistors circuit - wrong possible answers ?

[mfratczak88]

Homework Statement

In the figure below, two batteries with emf E1 and E2 and internal resistances r1 and r2 forms a circuit together with two identical resistors each with resistance R. Find the current passing through each resistor.

Relevant Equations

KVL

It's ABCD question with answers attached.
Problem is that, I don't think that neither of those are correct. From KVL the equation seems to be i = (E2-E1)/(2r1+2r2 + R). I've attached images of the circuit, answers, and my solution.
Can someone please point me if I am wrong, and if yes, where's the error ?

Many thanks :)

Edit: It seems that for some reason Images of the solutions cannot be uploaded. None the less here's my solution:

Initial current i1 will produce two currents i2 and i3 so that from KCL we have i1 = i2+i3.
From KVL
1) -E1 + E2 - i1r1 -i1r2 - i2R = 0
2) -i3R+i2R = 0 => i2 = i3

using the fact that i2=i3 and i1 = i2+i3 we have
-E1+E2 - 2*(i2r1) - 2*(i2r2) - i2R = 0 => i2 = (E2-E1)/(2r1+2r2+R).

 

[berkeman]

2r1+2r2 + R

I don't get their answers either, but I don't necessarily agree with your use of the "2" terms above. Can you say in more detail how you got your solution?

 

[mfratczak88]

I don't get their answers either, but I don't necessarily agree with your use of the "2" terms above. Can you say in more detail how you got your solution?

I've edited my post, because I couldn't post the image. There you will find my solution.

 

[berkeman]

Sorry, I'm not able to follow your logic. The two voltage sources are in series, so their voltages subtract and their internal resistances add. What is the total resistance of the external R//R?

 

[mfratczak88]

Sorry, I'm not able to follow your logic. The two voltage sources are in series, so their voltages subtract and their internal resistances add. What is the total resistance of the external R//R?

From KVL, if I go with the direction of the current in the inner loop on the left, with E2 there's a voltage drop as I am going from + to minus, with E1 there's a voltage increase as I am going from - to +. So it should be E1 - E2 not E2 - E1, the rest should be fine.

 

[cnh1995]

From KVL, if I go with the direction of the current in the inner loop on the left, with E2 there's a voltage drop as I am going from + to minus, with E1 there's a voltage increase as I am going from - to +. So it should be E1 - E2 not E2 - E1, the rest should be fine.

Is the given answer c)?

I agree none of the answers is correct. I suspect they made a printing mistake with parantheses in option c).

 

[mfratczak88]

Is the given answer c)?

I agree none of the answers is correct. I suspect they made a printing mistake with parantheses in option c).

Can you write down KVL ? Whatever I can think of, the my KVL equations seems to be the same.

 

[berkeman]

You didn't answer what R//R is...

 

[mfratczak88]

You didn't answer what R//R is...

(1/2)(R)

 

[berkeman]

Right! And that R/2 is in series with the other two resistors, and in series with the difference in the two supply voltages. Can you see how a missing "/2" in one of the answers might make it correct?

 

[mfratczak88]

Right! And that R/2 is in series with the other two resistors, and in series with the difference in the two supply voltages. Can you see how a missing "/2" in one of the answers might make it correct?

So what should be the answer then ? I still don't get it why my KVLs is wrong ?

 

[berkeman]

So what should be the answer then ? I still don't get it why my KVLs is wrong ?

Only one KVL is needed if you combine the R//R into on R/2. Can you write that KVL and use it to solve for the current? Don't worry about the +/- sign for the current I, that just depends on the choice of current direction in your KVL loop.

 

[mfratczak88]

Only one KVL is needed if you combine the R//R into on R/2. Can you write that KVL and use it to solve for the current? Don't worry about the +/- sign for the current I, that just depends on the choice of current direction in your KVL loop.

It will be I = E1-E2/(r1+r2+(1/2)R). But this is the current for equivalent resistor. The question was what was the current flowing through each of the resistors. So in this case, if they have the same resistance the current will be half of the value above, which is E1-E2/(2r1+2r2+R)

 

[cnh1995]

@mfratczak88 , after going through your solution, it appears to me that you have assumed i1, i2 and i3 to be flowing anticlockwise. I believe your answer is correct as per your choice of current direction.

I chose i1, i2 and i3 to be clockwise and got E1-E2 in the numerator.

 

[mfratczak88]

@mfratczak88 , after going through your solution, it appears to me that you have assumed i1, i2 and i3 to be flowing anticlockwise. I believe your answer is correct as per your choice of current direction.

I chose i1, i2 and i3 to be clockwise and got E1-E2 in the numerator.

did you get (2r1+2r2+R) in the denominator ?

 

[cnh1995]

did you get (2r1+2r2+R) in the denominator ?

Yes.
The difference in our numerators is simply because of the choice of current direction.

 

[berkeman]

The question was what was the current flowing through each of the resistors.

Ah, apologies, I thought they wanted the overall loop current. So you are right, cut it in half to get the current in each of the R resistors.