Finding currents using the junction and loop rules

In summary, the conversation discusses a problem involving Kirchhoff's Voltage Law and solving for currents using matrices. The participants also discuss the correct signs for potential drops and the role of current direction in determining polarity. It is concluded that the given results are correct and the instructor's results are incorrect.
  • #1
Robb
225
8

Homework Statement


upload_2016-4-5_20-21-39.png


Homework Equations


i1+i2-i3=0
50i1+75i3=1
100i2-75i3=1

The Attempt at a Solution



matrix A=1 1 -1
0 100 -75[/B]
50 0 75
matrix B= 0
1
1

i1= -6ma
i2= 23ma
i3= 17ma

My instructor gets i1= 6ma, i2= 3ma, i3= 9ma. Please advise!
 
Physics news on Phys.org
  • #2
Your diagram does not show the 20 Ω resistor mentioned in the problem statement. Nor does is show points a and b.

Are you sure that this diagram fits the described problem?
 
  • #3
Sorry, we were told to omit that line.
 
  • #4
Check the signs of the terms in your KVL equations. Can you show some detail in your solution to the simultaneous equations? Are you using some software to solve them? Typing out a matrix doesn't tell us anything unless by some strange coincidence we happen to know what software you have and we have it too.
 
  • #5
I am using a TI 83+. Matrix A is a 3x3 and matrix B is a 3x1. A^-1*B. I have checked my signs many times so if they are wrong I am not seeing it.
 
  • #6
Write out your KVL equations again without combining the voltage terms or transposing the voltage terms to the RHS of the equation (so the RHS should be a zero). Say which direction you are "walking" around the loop and whether you are counting potential drops as positive or negative quantities.
 
  • #7
-50i1-75i3-1=0
100i2+75i3-1=0
i1+i2-i3=0

We were taught to place charges at each side of each component and the charge on the side you are exiting is the sign you use for your equation. The stuff I see online is a little different in terms of drops and gains. Not sure which to use?
 
  • #8
Okay. Those equations look correct. What values do you find with those equations?

Your circuit diagram is confusing because the potential drops that you've indicated for the resistors don't appear to obey the chosen current directions. For example, I1 is flowing upwards through the 50 Ω resistor but you marked its potential change as higher where the current exits the resistor (at its top).
 
  • #9
I see what you're saying would this be correct:
-50i1+75i3-1=0
100i2-75i3-1=0
i1-i2-i3=0

i1=-15.64
i2=-7.8
i3=-7.8
 
  • #10
Robb said:
I see what you're saying would this be correct:
-50i1+75i3-1=0
100i2-75i3-1=0
i1-i2-i3=0
No, your last set of equations were okay.

Here's your circuit re-drawn with appropriate potential changes indicated:
upload_2016-4-6_16-1-49.png


If you do a KVL walk clockwise around the loop on the left starting from the bottom left corner you encounter:

##-50 I_1 + 98 - 99 - 75 I_3 = 0##

##-50 I_1 - 75 I_3 - 1 = 0##

Can you do the same for the second loop?
 
  • #11
75i3+99-100+100i2=0
100i2+75i3-1=0

Not sure why the 75 ohm resistor doesn't have the signs opposite what you show. Wouldn't the top side be negative from the 99v source above it?
 
  • #12
Robb said:
75i3+99-100+100i2=0
100i2+75i3-1=0
Correct.
Not sure why the 75 ohm resistor doesn't have the signs opposite what you show. Wouldn't the top side be negative from the 99v source above it?
Look at the direction that you've chosen for I3. It flows down through the 75 Ω resistor, hence the potential drop in that direction.
 
  • #13
Ok. Now I get it. My current direction determines polarity!

I get:
i1=-15ma
i2=12ma
i3=-3ma

Still not the same answer instructor gets.
 
  • #14
Your results are correct for the circuit given. If your professor obtained different results from the same circuit then his results are incorrect.
 
  • #15
Gee, there's a shocker! Thanks for your help! Off to take the test!
 
  • #16
So I assume with a voltage source the polarity remains as whatever the source is? Meaning the current direction does not determine polarity through a source?
 
  • #17
Robb said:
So I assume with a voltage source the polarity remains as whatever the source is? Meaning the current direction does not determine polarity through a source?
Correct. A voltage source will produce or absorb any amount of current to maintain its specified potential difference.
 

1. What are the junction and loop rules in finding currents?

The junction rule states that the total current flowing into a junction must be equal to the total current flowing out of that junction. The loop rule, also known as Kirchhoff's voltage law, states that the sum of the voltage drops around a closed loop in a circuit must be equal to the sum of the voltage sources in that loop.

2. Why are the junction and loop rules important in circuit analysis?

The junction and loop rules are important because they provide a systematic approach to solving complex circuits. They allow us to determine the values of individual currents in a circuit based on the known values of voltage sources and resistors.

3. How do you apply the junction rule in finding currents?

To apply the junction rule, we first identify all the junctions in the circuit. Then, we write an equation for each junction, setting the sum of incoming currents equal to the sum of outgoing currents. We can then solve these equations to find the values of the individual currents.

4. Can the junction and loop rules be used for any type of circuit?

Yes, the junction and loop rules can be used for any type of circuit, including series and parallel circuits. These rules are based on the fundamental principles of conservation of charge and energy, and can be applied to any circuit regardless of its complexity.

5. Are there any limitations to using the junction and loop rules in finding currents?

While the junction and loop rules are powerful tools for circuit analysis, they do have limitations. They can only be used for circuits that are in a steady state, meaning that the currents and voltages do not change over time. They also assume ideal conditions, such as no resistance in the wires or perfect voltage sources, which may not always be the case in real circuits.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
619
  • Introductory Physics Homework Help
Replies
1
Views
633
Replies
8
Views
410
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
22
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
31
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top