Loop Currents for Circuits (using matrix)

[mneox]

Homework Statement

The question is just asking to find the current through each of the resistors.
Our teacher taught us this loop current method, and then we have to use a matrix to solve for the currents.

i1 denotes loop 1 and current, etc.

Homework Equations

Loop 1: 4i₁ - 2i₂ - i₃ = 10
Loop 2: -2i₁ + 7i₂ - 2i₃ = 0
Loop 3: -i₁ - 2i₂ + 3i₃ + E = 0

And we know i₃ = -2

The Attempt at a Solution

Alright so when I put these four equations as a matrix, I get:

4 -2 -1 0 | 10
-2 7 -2 0 | 0
-1 -2 3 1 | 0
0 0 1 0 | -2

Where first column goes is for i₁, second is for i₂, fourth is E.

When I put the matrix in rref, I get:

i₁ = 2
i₂ = 0
i₃ = -2
E = 8

Now I'm wondering if I did something wrong. How can the current be 0A or -2A?

And how would I calculate the current traveling through the bottom 1Ω resistor? Would I get negative currents again?

Thanks, I'm just wondering what I did wrong if anything, or how is it possible to have current of zero or below zero.

 

[mneox]

um, sorry to bump but anybody?
i'm still lacking clarity, any sort of input would be greatly appreciated...thanks

 

[ehild]

Your solution for the loop currents and E is good. Now you have to determine the real currents flowing through each resistor. In case of a common resistor of two loops, you have two loop currents which flow in opposite directions, so you have to subtract one from the other to get the real current. A negative result means that the current flows opposite to the assumed direction.

As for the bottom 1 ohm resistor, i1 current flows downward and i3 upward, so I=i1-i3=2-(2) = 4A downward.
As for the 2 ohm resistor on the right, the current flowing from right to left is i2-i3 =0-(-2)=2A.

You have i2 current flowing through the 3 ohm resistor, and it is zero. That means zero potential difference across its terminals.

ehild

 

[collinsmark]

When I put the matrix in rref, I get:

i₁ = 2
i₂ = 0
i₃ = -2
E = 8

Okay, so far so good!

Now I'm wondering if I did something wrong. How can the current be 0A or -2A?

You need to sum the individual currents through a given resistor to find the total current through that particular resistor.

And how would I calculate the current traveling through the bottom 1Ω resistor? Would I get negative currents again?

If you get a negative current, it just means the current is in the opposite direction of how you defined it. But since you will be arbitrarily defining the direction of currents anyway, getting negative currents is expected about half the time. More on that below.

Thanks, I'm just wondering what I did wrong if anything, or how is it possible to have current of zero or below zero.

It's easy to have a current of zero. It just means no current is flowing through that resistor. A current below zero just means its moving opposite to how you defined the direction.

You have all the information you need already, and have already used it, although you just might not know it yet. You took a shortcut when you created your loop equations (there isn't anything wrong at all with your shortcut, just make sure you know why it's valid). Let me explain with an example. Consider your first loop, Loop 1.

Loop 1: 4i₁ - 2i₂ - i₃ = 10

You create current loops by summing all of the voltage drops around the loop. According to one of Kirchhoff's laws, the sum of all the voltage drops around a given loop must equal zero.

So in this case,

Voltage drop across the top 1Ω resistor
+ Voltage drop across the vertical 2Ω resistor
+ Voltage drop across the bottom 1Ω resistor
+ Voltage drop across the 10 V supply
= 0​

where all the voltage drops are measured from + to - in the direction of i₁ (You could instead sum the voltage increases that go from - to +, rather than voltage drops [which go from + to -], if you wish. Just pick a convention and stick with it.)

So your Loop 1 equation becomes,

(1 Ω)i₁ + (2 Ω)(i₁ - i₂) + (1 Ω)(i₁ - i₃) -(10 V) = 0

Do you see how I did that? Check to make sure that this matches your version of Loop 1. Now you already know how much current goes through each resistor in that loop. For example, through the 2 Ω resistor it is i₁ - i₂, and in the direction of i₁.

Now do the same thing for your Loop 2. This time, you'll find that the current through the vertical 2 Ω resistor is i₂ - i₁, but this time in the direction of i₂. The calculation gives you the opposite sign. But since the current is in reference to something in the opposite direction it all works out.

 

[mneox]

Your solution for the loop currents and E is good. Now you have to determine the real currents flowing through each resistor. In case of a common resistor of two loops, you have two loop currents which flow in opposite directions, so you have to subtract one from the other to get the real current. A negative result means that the current flows opposite to the assumed direction.

As for the bottom 1 ohm resistor, i1 current flows downward and i3 upward, so I=i1-i3=2-(2) = 4A downward.
As for the 2 ohm resistor on the right, the current flowing from right to left is i2-i3 =0-(-2)=2A.

You have i2 current flowing through the 3 ohm resistor, and it is zero. That means zero potential difference across its terminals.

ehild

Thanks for clarifying ehild! I understood my answer much better through your reply.

Okay, so far so good!

You need to sum the individual currents through a given resistor to find the total current through that particular resistor.

If you get a negative current, it just means the current is in the opposite direction of how you defined it. But since you will be arbitrarily defining the direction of currents anyway, getting negative currents is expected about half the time. More on that below.

It's easy to have a current of zero. It just means no current is flowing through that resistor. A current below zero just means its moving opposite to how you defined the direction.

You have all the information you need already, and have already used it, although you just might not know it yet. You took a shortcut when you created your loop equations (there isn't anything wrong at all with your shortcut, just make sure you know why it's valid). Let me explain with an example. Consider your first loop, Loop 1.

Loop 1: 4i₁ - 2i₂ - i₃ = 10

You create current loops by summing all of the voltage drops around the loop. According to one of Kirchhoff's laws, the sum of all the voltage drops around a given loop must equal zero.

So in this case,

Voltage drop across the top 1Ω resistor
+ Voltage drop across the vertical 2Ω resistor
+ Voltage drop across the bottom 1Ω resistor
+ Voltage drop across the 10 V supply
= 0​

where all the voltage drops are measured from + to - in the direction of i₁ (You could instead sum the voltage increases that go from - to +, rather than voltage drops [which go from + to -], if you wish. Just pick a convention and stick with it.)

So your Loop 1 equation becomes,

(1 Ω)i₁ + (2 Ω)(i₁ - i₂) + (1 Ω)(i₁ - i₃) -(10 V) = 0

Do you see how I did that? Check to make sure that this matches your version of Loop 1. Now you already know how much current goes through each resistor in that loop. For example, through the 2 Ω resistor it is i₁ - i₂, and in the direction of i₁.

Now do the same thing for your Loop 2. This time, you'll find that the current through the vertical 2 Ω resistor is i₂ - i₁, but this time in the direction of i₂. The calculation gives you the opposite sign. But since the current is in reference to something in the opposite direction it all works out.

collinsmark, thanks for the superb post. It was incredibly thorough and clear.

Regarding your comment about my equation:
4i₁ - 2i₂ - i₃ = 10

Sorry about that. I actually did some other steps to reach that above equation but it was just too strenuous to type out all the work I did haha, so I settled for the final equations.

But again, thanks for your time and explanation. It most definitely helped!

Thanks to both you guys, cheers