Calculating the currents in each resistor

[n387g]

Homework Statement

Calculate the currents in each resistor in this figure:
http://session.masteringphysics.com/problemAsset/1084690/2/GIANCOLI.ch26.p32.jpg

Homework Equations

Kirchhoff's rules
In a closed loop: the sum of voltage is 0
A a junction: the sum of current is 0

The Attempt at a Solution

Would all these resistors be considered in series? Therefore the current would be the same across each resistor?

I have the closed loop equations:
58V-I1(120)-I2(82)-I3(64)=0
3V-I4(25)-I3(64)-I5(110)=0

However, I do not know how to formulate the junction equations over multiple resistors and I know I need more equations for the amount of unknowns that I have. Suggestions?

 

[gneill]

Components in series share the same current; you don't need to define separate currents for each component.

Not all the components are in series with each other in this circuit; the 64 Ohm resistor introduces additional connections to two of the junctions. If all component junctions had only two connections, everything would be in series. That's not the case here.

For the node equations, treat resistors that are in series in a branch as though they were a single resistor whose value is the sum of the individual resistances. This works because they share the same current.

 

[n387g]

I'm still at a lose of what to exactly do, when approaching this problem mathematically.

I went ahead and tried to calculate the current of just R1=64, R2=82, R3=64 and the battery voltage of 58V since if you just consider those three resistors they are in series and therefore share the same current. Using V=IR, I found that I of R1,R2,R3 was 0.218A.

I used this same method, but using R4=25, R3=64, R5=110 and the battery 3.0V as a second closed loop. Using V=IR, I found that I of R3,R4,R5 was 0.015A.

I know I'm missing something, but don't know what. Possibly something that links these two together (which would be R3=64?) but I don't know how to apply it.

 

[gneill]

I'm still at a lose of what to exactly do, when approaching this problem mathematically.

I went ahead and tried to calculate the current of just R1=64, R2=82, R3=64 and the battery voltage of 58V since if you just consider those three resistors they are in series and therefore share the same current. Using V=IR, I found that I of R1,R2,R3 was 0.218A.

Ah. Those resistors may be in series in that loop, but in the larger picture your R1 is also in the other loop as well, and the current produced through it by the actions of the other battery are going to change the voltage drop across it.

I used this same method, but using R4=25, R3=64, R5=110 and the battery 3.0V as a second closed loop. Using V=IR, I found that I of R3,R4,R5 was 0.015A.

I know I'm missing something, but don't know what. Possibly something that links these two together (which would be R3=64?) but I don't know how to apply it.

Yes. You've hit the nail on the head; the loops will influence each other because they share components (in this case your R1, the 64 Ohm resistor). If R1 wasn't there you'd have only one loop, all around the outside.

You can write two loop equations, specifying a separate current in each. Say, I1 and I2. Let I1 go clockwise around the loop on the left side, and let I2 go clockwise around the loop on the right. I1 and I2 will both flow through R1 (note: in opposite directions). When you "walk" around each loop to write its KVL loop equation, when you get to the shared component R1 include contributions from both currents towards the voltage drop. Thus you should have a term like (I1-I2)R1 in the first loop, and (I2 - I1)R1 in the second loop.

You can choose the directions for the loop currents in any way you want, but I myself find that it avoids confusion to always make them clockwise, and just let the mathematics sort out the signs of the currents.

 

[n387g]

I'm still having trouble.
Based on your last post I formulated these two equations based on the two loops (I went counterclockwise, but that should only affect sign not quantity):

For loop 1:
58V-(I1*120)-(I1*82)-[(I1-I2)*64]=0

For loop 2:
3-(I2*25)-[(I2-I1)*64]-(I2*110)=0

I then used these two equations to solve for the two unknowns, finding that I1=0.283A and I2=0.106A

Did I do something incorrectly? Where did I go wrong?

 

[gneill]

I'm still having trouble.
Based on your last post I formulated these two equations based on the two loops (I went counterclockwise, but that should only affect sign not quantity):

For loop 1:
58V-(I1*120)-(I1*82)-[(I1-I2)*64]=0

For loop 2:
3-(I2*25)-[(I2-I1)*64]-(I2*110)=0

I then used these two equations to solve for the two unknowns, finding that I1=0.283A and I2=0.106A

Did I do something incorrectly? Where did I go wrong?

Your loop equations look fine. Must be an algebra slip-up somewhere.

 

[n387g]

Okay, thanks!

I just want to double check one more thing. When solving for the currents for each resistor, The current over R1=64 would be I1-I2? And then the current over R=120 and R=82 would be the same, I1? While the current over R=25 and R=110 would be the same, I2?

 

[gneill]

Okay, thanks!

I just want to double check one more thing. When solving for the currents for each resistor, The current over R1=64 would be I1-I2? And then the current over R=120 and R=82 would be the same, I1? While the current over R=25 and R=110 would be the same, I2?

That's right. You got it.