In which direction does the current flow?

[bolzano95]

Homework Statement

We bind two batteries with voltage 2V and 1.7V with inner resistance 0.2Ω and 0.1Ω in parallel series in such way that we bind each positive poles and negative poles together. What current runs trough the circle? What is the voltage between contact points of the batteries?
In the circle we include resistor R=2Ω.
What is the voltage between contact points of the batteries now?
What current flows through the circle?

Homework Equations

The Attempt at a Solution

First, I draw the electrical circle. Then I noticed that the wire is connecting + from first battery to + from first battery. I looked if the voltages of batteries are different and they are- therefore the current will flow from bigger to smaller electrical potential.
Question: From where does the current flows: From - pole of 1.7V to - pole of 2 or vice versa? Why? How does this work?

I solved this problem like this: What is the potential difference between the two batteries? 0.3V. I applied Ohm's Law and got the current.

In solution sheet it was written $$I= I_{1} -I_{2}$$

For the voltage between the contact points of batteries I would instantly answer 0.3V, because that is the potential of which can current actually flow. But the correct answer is 1.7V. Is my thinking even correct?
$$U= U_{1} -IR_{1}=U_{2} +IR_{2}$$

 

[tnich]

But the correct answer is 1.7V. Is my thinking even correct?
$$U= U_{1} -IR_{1}=U_{2} +IR_{2}$$

I think the question "What is the voltage between contact points of the batteries?" refers to the voltage across the positive and negative terminals of each battery. Since there is no resistance between the positive terminals of the two batteries, or between the negative terminals, this voltage must be the same for each battery. I came up with the same equation you did and it led to a different answer than 1.7V. To me, it seems that in addition to the EMF of battery #2, there must be a voltage drop due to the internal resistance in series with the EMF, which is what your equation expresses. So the voltage between the contact points of the batteries would have to be more than 1.7V.

 

[CWatters]

The answer for the first part can't possibly be 1.7V because one of the batteries is 1.7V and the other is higher. So the book answer is wrong.

The OP should draw the equivalent circuit and label the components and voltages.

Conventional current always flows from high to low voltage so from the 2V EMF in the "2V battery" to the 1.7V EMF in the "1.7V battery".

 

[Dr Dr news]

Set up a loop equation 2 V - 0.7 i - 0.1 i - 1.7 V = 0, i = 0.375 A, so the terminal voltage on both batteries will be 1.74 V. In case 2, the loop equation would be 2 V - 0.7 i - 2 i - 0.1 i - 1.7 V = 0, i = 0.107 A, Now it depends on which side of the 2 Ω resistor you connect your voltmeter. On the side toward the 2 V battery it should read 1.93 V and on the side toward the 1.7 V battery, 1.71 V. This is another case where the textbook answer will frustrate the student.

 

[haruspex]

- 0.7 i

-0.2i?

 

[Dr Dr news]

I think I need a new glasses prescription when I read a 0.2 as 0.7. Thank you. Back to the loop equations: Case 1 would look like: 2 V - 0.2 i - 0.1 i - 1.7 V = 0, i = 1.0 A so the terminal voltage on both batteries would be 1.80 V. Case 2 would look like 2 V - 0.2 i - 2 i - 0.1 i - 1.7 V = 0, i = 0.13 A, so the terminal voltage on the 2 V battery would be 1.97 V. and on the 1.7 V battery 1.71 V, with a 0.26 V drop across the 2 ohm resistor. The question might be how can the terminal voltage of a battery with an emf of 1.7 V ever be greater than 1.7 V?