Current through resistors in circuit with TWO ideal emfs

In summary, the circuit has two ideal batteries with emfs of 17.0 V and 0.5001 V, and resistances of 1.22 Ω each. There is one main loop with a wire connecting the bottom and the top piece. The two emfs are placed on the left and right side, both pointing towards the top. R_1 is located on the top left side, R_2 on the top right side, and R_3 in the middle piece of the connecting wire. Using the junction rule, i_3 is equal to the sum of i_1 and i_2. Using the loop rule in the right loop, we get -i_1(R_2)-i_3
  • #1
kevlar94
10
0

Homework Statement


In the figure, the ideal batteries have emfs 1 = 17.0 V and 2 = 0.5001 V, and the resistances are each 1.22 Ω.

I tried to include a link but I am unable.

As best as I can describe: There is one main loop with a wire connecting the bottom and the op piece. The two emfs are on the left and right side BOTH pointing towards the top (positive is both towards the top of the circuit/each other). R_1 is on the left side top and R_2 is on the right side top. R_3 is in the middle piece of connecting wire.

I hope that explains it well enough.

a) What is the current in resistor 2?
b) What is the current in resistor 3?

Homework Equations


ƩΔV=0 for a complete circuit
Ʃi in a junction must be equal. This is to conserve charge.

The Attempt at a Solution


I am getting close but the double emfs are confusing me. I have another problem similar to this that I am also not able to do. I have the test Monday and I am certain there will be a double emf circuit problem because our professor is just that sort of guy. He also never covered this in class.

i_2 goes out of the positive terminal of ε_2 into R_2, i_1 goes out of the positive terminal of ε_1 into R_1, and i_3 goes straight down.

Using the junction rule I get:

i_3=i_1 + i_2

I tried using the loop rule for the right loop going in a ccw direction assuming the current is going in the same direction -ccw.

I get -i_1*(R_2)-i_3*(R_3)+ε_2=0

left loop: -i_1*(R_1)-i_3*(R_3)+ε_1=0

I then subbed the three equations into each other using wolfram alpha to check my work before I did it by hand and they are wrong. I am not sure what I am doing incorrect and I tried to follow an example from the book but I could not figure out how to apply that material to this case.

Thanks for the help. I will post more later as I attempt it for the third time today.
 
Physics news on Phys.org
  • #2
kevlar94 said:

Homework Statement


In the figure, the ideal batteries have emfs 1 = 17.0 V and 2 = 0.5001 V, and the resistances are each 1.22 Ω.

I tried to include a link but I am unable.

As best as I can describe: There is one main loop with a wire connecting the bottom and the op piece. The two emfs are on the left and right side BOTH pointing towards the top (positive is both towards the top of the circuit/each other). R_1 is on the left side top and R_2 is on the right side top. R_3 is in the middle piece of connecting wire.

I hope that explains it well enough.

a) What is the current in resistor 2?
b) What is the current in resistor 3?

Homework Equations


ƩΔV=0 for a complete circuit
Ʃi in a junction must be equal. This is to conserve charge.



The Attempt at a Solution


I am getting close but the double emfs are confusing me. I have another problem similar to this that I am also not able to do. I have the test Monday and I am certain there will be a double emf circuit problem because our professor is just that sort of guy. He also never covered this in class.

i_2 goes out of the positive terminal of ε_2 into R_2, i_1 goes out of the positive terminal of ε_1 into R_1, and i_3 goes straight down.

Using the junction rule I get:

i_3=i_1 + i_2

I tried using the loop rule for the right loop going in a ccw direction assuming the current is going in the same direction -ccw.

I get -i_1*(R_2)-i_3*(R_3)+ε_2=0
Did you mean i1 in this equation, or was it supposed to be i2?

left loop: -i_1*(R_1)-i_3*(R_3)+ε_1=0

I then subbed the three equations into each other using wolfram alpha to check my work before I did it by hand and they are wrong. I am not sure what I am doing incorrect and I tried to follow an example from the book but I could not figure out how to apply that material to this case.

Thanks for the help. I will post more later as I attempt it for the third time today.
 
  • #3
vela said:
Did you mean i1 in this equation, or was it supposed to be i2?

That is supposed to be i2, sorry. The rest of the equations look correct.

Thanks.
 
  • #4
Your equations sound okay to me then, if I'm picturing the circuit correctly. When I plug the numbers in, I find i2=-4.37 A and i3=4.78 A. What did you get?
 
  • #5
vela said:
Your equations sound okay to me then, if I'm picturing the circuit correctly. When I plug the numbers in, I find i2=-4.37 A and i3=4.78 A. What did you get?

I wrote a long explanation of my work and then i went to submit and it lost it.

Basically I worked out the three loops again and simplified using the fact that R is equivilant to get the following system:

(-i2-i3)=6.97
(i1+i3)=13.9
(i1-i2)=-6.97

Wolfram alpha says that is impossible to solve. I would really appreciate any advice.

I have a picture of the circuit but I can not post it. That is making this really difficult to explain.
 
  • #6
I'm really sorry I did not notice this before. emf1=17V emf2=.5*emf1 or 8.5V. All the resistance is 1.22ohms

My apologies that I did not catch that, it did not copy correct. I have been using 8.5V for the problem.

Again, sorry about my mistake.
 
  • #7
You seem to have made a sign error somewhere, and you're missing an equation. If you subtract the first equation from the third, you end up with ##i_1 + i_3 = -13.9##, which is clearly inconsistent with equation 2.

The three equations you should have started with are
\begin{align*}
1.22 i_1 + 1.22 i_3 &= 17 \\
1.22 i_2 + 1.22 i_3 &= 8.5 \\
i_1 + i_2 &= i_3
\end{align*} which are just the equations you wrote above with the numbers plugged in.
 
  • #8
vela said:
You seem to have made a sign error somewhere, and you're missing an equation. If you subtract the first equation from the third, you end up with ##i_1 + i_3 = -13.9##, which is clearly inconsistent with equation 2.

The three equations you should have started with are
\begin{align*}
1.22 i_1 + 1.22 i_3 &= 17 \\
1.22 i_2 + 1.22 i_3 &= 8.5 \\
i_1 + i_2 &= i_3
\end{align*} which are just the equations you wrote above with the numbers plugged in.

I guess I was making everything to complicated. It works. I'm not sure why I was so confused.

There is no current through R_2. Why is that? If there is a connection between the emfs should there not be current everywhere in the circuit?

Thanks
 
  • #9
Usually there would be a current through R2, but it just happens to work out that the potential drop in this circuit across R2 is 0, so no current will flow through it.
 
  • #10
Interesting. Thanks for the help.
 

FAQ: Current through resistors in circuit with TWO ideal emfs

What is the formula for calculating current through resistors in a circuit with two ideal emfs?

The formula is I = (E1 - E2) / R, where I is the current, E1 and E2 are the emfs, and R is the resistance.

How do I determine the direction of current in this type of circuit?

The direction of current is determined by the direction of the higher emf. In this case, the current will flow from the higher emf towards the lower emf.

Can I use Ohm's law to calculate the current in this circuit?

Yes, Ohm's law can be used to calculate the current in this circuit. However, it is important to note that the resistance in this equation is the total resistance of the circuit, including both resistors.

What happens to the current if the resistors have different values?

If the resistors have different values, the current through each resistor will be different. The larger resistor will have a smaller current, while the smaller resistor will have a larger current.

How does the presence of two ideal emfs affect the total current in the circuit?

The presence of two ideal emfs affects the total current in the circuit by creating a potential difference between the two emfs. This potential difference causes the current to flow through the circuit, resulting in a non-zero current value.

Similar threads

Replies
13
Views
549
Replies
16
Views
1K
Replies
6
Views
978
Replies
2
Views
1K
Replies
6
Views
2K
Replies
6
Views
2K
Back
Top