Finding the current through the battery

[Vanessa Avila]

Homework Statement

I am confused what to do but I have to find the current through the battery:

Homework Equations

V=IR

The Attempt at a Solution

I decided to find the equivalent resistance and do V/R = I using the 14V, but I believe my Req is wrong. What is throwing me off is the horizontal 1 ohm resistor.

 

[gneill]

Show the details of your attempt. How did you try to reduce the network to a single equivalent resistance?

What's your backup plan?

 

[berkeman]

Homework Statement

I am confused what to do but I have to find the current through the battery:

Homework Equations

V=IR

The Attempt at a Solution

I decided to find the equivalent resistance and do V/R = I using the 14V, but I believe my Req is wrong. What is throwing me off is the horizontal 1 ohm resistor.

There may be a trick to simplify it because of the symmetry of the resistance values, but it's probably just easiest to write the KCL equations and solve them. Can you show us that work?

EDIT -- Oops, beaten out by gneill again!

 

[Vanessa Avila]

Would this be right?

14V - I₁(1Ω) - I₁(2Ω) = 0
So that would give me I₁ = 4.67 A

 

[gneill]

Would this be right?

14V - I₁(1Ω) - I₁(2Ω) = 0
So that would give me I₁ = 4.67 A

You'll have to show us what $i_1$ is on your diagram, but I would say that it's very unlikely. Any current through R3 would "break" any attempt to consider the vertical paths as isolated series connected resistors.

 

[Vanessa Avila]

You'll have to show us what $i_1$ is on your diagram, but I would say that it's very unlikely. Any current through R3 would "break" any attempt to consider the vertical paths as isolated series connected resistors.

My bad. Here's what I was trying to do:

I realized that the I₁ splits as soon as it reaches the junction and i can either keep going to the right or down. So that means my equation for my loop has to involve I₂ right?

 

[berkeman]

My bad. Here's what I was trying to do:

I realized that the I₁ splits as soon as it reaches the junction and i can either keep going to the right or down. So that means my equation for my loop has to involve I₂ right?

You can either write the multiple KVL equations, or the multiple KCL equations. I personally prefer KCL, but either will work. Are you familiar with how to write them?

 

[Vanessa Avila]

You can either write the multiple KVL equations, or the multiple KCL equations. I personally prefer KCL, but either will work. Are you familiar with how to write them?

Yes. How my professor taught it to us is that we have to set directions for the current and create a clockwise or a counter-clockwise loop. My biggest problem with it is keeping it to only 3 unknown currents (I₁, I₂, and I₃) as he advised. I get confused when I am past assigning where I₃ is.

 

[Vanessa Avila]

I think how i labeled the diagram was wrong. Would the current going through R3 be (I₂ + I₃) ?

 

[gneill]

One way to minimize the number of currents you "invent" while labeling the circuit is to only ever add the minimum required number of currents at any junction you come across while proceeding systematically through the circuit. If you can, label new paths using mathematical combinations of existing currents.

For example, if you go into a junction with i1 and two unlabeled paths diverge from there, make one current i2 and the other i1-i2. Thus only one "new" current is introduced at that junction rather than two.

 

[Vanessa Avila]

One way to minimize the number of currents you "invent" while labeling the circuit is to only ever add the minimum required number of currents at any junction you come across while proceeding systematically through the circuit. If you can, label new paths using mathematical combinations of existing currents.

For example, if you go into a junction with i1 and two unlabeled paths diverge from there, make one current i2 and the other i1-i2. Thus only one "new" current is introduced at that junction rather than two.

Okay. I relabeled everything. Can you check to see if i did it right?

 

[gneill]

Check the KCL sum at the center right node. Everything else looks okay.

 

[Vanessa Avila]

Check the KCL sum at the center right node. Everything else looks okay.

would it be -I₃-I₂?

 

[gneill]

would it be -I₃-I₂?

No. remember, whatever is flowing out of the node must also be flowing into the node. You have I₂ and I₃ both flowing out, so what must flow in on the remaining path?

 

[Vanessa Avila]

I₂ + I₃ ?

Sorry, I'm a little confused about this stuff

 

[gneill]

I₂ + I₃ ?

Yup.

Sorry, I'm a little confused about this stuff

No worries. After a bit of practice it'll suddenly "cick" and you'll wonder what all the fuss was about

 

[Vanessa Avila]

Yup.

No worries. After a bit of practice it'll suddenly "cick" and you'll wonder what all the fuss was about

Oh okay! Thanks

So if I were to assign three loops like this

would these be the right eqns?

Eqn #1: (loop with 14V, starting after the 14V)
14V - (I₂+I₁)*(1Ω) - (I₁+I₂+I₃)*(2Ω) = 0

Eqn #2: (upper loop, starting above I₂+I₁)
(I₂+I₁)*(1Ω)+(I₂)*(2Ω) - (I₃)*(1Ω) = 0

Eqn #3: (lower loop, starting above I₁+I₂+I₃)
(I₁+I₂+I₃)*(2Ω) + I₃*(1Ω) + (I₃+I₂)*(1Ω) = 0

 

[gneill]

I'd say that they look good. I don't spot any problems.

 

[Vanessa Avila]

I'd say that they look good. I don't spot any problems.

:) Awesome! So finding the current through the battery, i have to find I₁, I₂ and I₃ and sum them together to use in the V=IR equation, where V = 14V?

 

[gneill]

:) Awesome! So finding the current through the battery, i have to find I₁, I₂ and I₃ and sum them together to use in the V=IR equation, where V = 14V?

Your I₁ flows through the battery. Find I₁ and you're done (of course you need to mostly solve everything to get there... unless you use something like Cramer's Rule to find just the one current).

 

[Vanessa Avila]

Okay thanks a lot! :) I really appreciate it!

Your I₁ flows through the battery. Find I₁ and you're done (of course you need to mostly solve everything to get there... unless you use something like Cramer's Rule to find just the one current).