# Calculate the Currents in a given circuit

1. Jul 24, 2011

### Vazer

The question involves Kirchoff's Rule.
I'm having a hard time trying to calculate the currents in the given circuit.
I'm trying to solve question number 3 in the attached image (the image on the bottom left hand corner). I have already measured the currents in lab, but I'm having trouble doing the calculation part.
I'd appreciate the help, thanks.

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Last edited: Jul 24, 2011
2. Jul 24, 2011

### Clever-Name

Are you comfortable solving a matrix?

Try setting up the equations in matrix form and solving.

3. Jul 24, 2011

### Vazer

It's been some time since I've done matrix. Is it an easier way to figure out the five currents? I'd be glad to learn this different approach to solving the problem.

4. Jul 24, 2011

### Clever-Name

That's always the method I use for solving so I would say yes, it's definitely an easier way to solve this type of problem. Using the equations given I would set it up as follows:

$$\left[ {\begin{array}{cc} 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 1 \\ -R_{1} & R_{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & -R_{4} & R_{5} \\ 0 & R_{2} & R_{3} & 0 & R_{4} \\ \end{array} } \right] \left[ {\begin{array}{cc} I_{1}\\ I_{2}\\ I_{3}\\ I_{4}\\ I_{5}\\ \end{array} } \right] = \left[ {\begin{array}{cc} 0\\ 0\\ 0\\ 0\\ E_{1}\\ \end{array} } \right]$$

Do you see where this comes from when using your equations? Do you remember how to row-reduce?

5. Jul 24, 2011

### Vazer

I'm trying to understand, but can't seem to figure how you placed them in matrix. No, don''t remember how to reduce.

6. Jul 24, 2011

### Vazer

I'll look over matrices online, I'm seeing the pattern. I appreciate the help CLever-Name, thank you.

7. Jul 24, 2011

### Clever-Name

Using matrices is a difficult concept to grasp at first but it's an extremely useful tool once you understand, for that reason I am going to try and explain it here.

I have set up your equations in the form
$$Ax = b$$
Where A is your matrix of known values, x is your column vector of unknowns, which is what you're trying to solve for, and b is also a column vector you know.

To expand out a system like this you have to multiply each ROW of A by the column vector x and set it equal to the corresponding ROW in b (which is just a single value, in this case we have 4 0's and one E_1)

For example, for the first row we have
$$\left[ {\begin{array}{cc} 1 & 1 & -1 & 0 & 0 \end{array} } \right] \left[ {\begin{array}{cc} I_{1}\\ I_{2}\\ I_{3}\\ I_{4}\\ I_{5}\\ \end{array} } \right] = 0$$

We multiply across the row and down the column.

$$1*I_{1} + 1*I_{2} - 1*I_{3} + 0*I_{4} + 0*I_{5} = 0$$

Simplifying this we get back the original first equation:

$$I_{1} + I_{2} = I_{3}$$

As an exercise try doing this with each row and you'll see that you get back your original equations. To be more explicit I am going to do the same thing with the last row so you see a more complicated example: ***I made a mistake in my previous post in the last row, the $R_{4}$ should be an $R_{5}$ *****

$$\left[ {\begin{array}{cc} 0 & R_{2} & R_{3} & 0 & R_{5} \end{array} } \right] \left[ {\begin{array}{cc} I_{1}\\ I_{2}\\ I_{3}\\ I_{4}\\ I_{5}\\ \end{array} } \right] = E_{1}$$

Expanding out we get:
$$0*I_{1} + R_{2}*I_{2} + R_{3}*I_{3} + 0*I_{4} + R_{5}*I_{5} = E_{1}$$

Simplifying we can get back the original 5th equation:

$$E_{1} - R_{2}I_{2} - R_{3}I_{3} - R_{5}I_{5} = 0$$

Now the way to solve this system is to row-reduce the matrix. Rather.. put it in the following form:

$$\left[ {\begin{array}{cc} 1 & 1 & -1 & 0 & 0 \\ 0 & 1 & X & X & X \\ 0 & 0 & 1 & X & X \\ 0 & 0 & 0 & 1 & X \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} } \right] \left[ {\begin{array}{cc} I_{1}\\ I_{2}\\ I_{3}\\ I_{4}\\ I_{5}\\ \end{array} } \right] = \left[ {\begin{array}{cc} 0\\ 0\\ 0\\ 0\\ X\\ \end{array} } \right]$$
The X's are there because the process of row-reducing MAY change what these values are.

By doing this process we end up being able to re-expand everything and get a something equivalent to our original equations but incredibly simple to solve for the currents. As you may see if we were to expand out the last row we would get $I_{5} = X$ The X will depend on the process of row reducing and simplifying.

I think it would be easier at this point to substitute in numbers in our matrix. The values I can read from your pictures produce the following (correct me if I'm wrong):
$$\left[ {\begin{array}{cc} 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 1 \\ -10.6 & 5.00 & 0 & 0 & 0 \\ 0 & 0 & 0 & -10.2 & 5.00 \\ 0 & 5.00 & 34.5 & 0 & 5.00 \\ \end{array} } \right] \left[ {\begin{array}{cc} I_{1}\\ I_{2}\\ I_{3}\\ I_{4}\\ I_{5}\\ \end{array} } \right] = \left[ {\begin{array}{cc} 0\\ 0\\ 0\\ 0\\ 9.9\\ \end{array} } \right]$$

You said you have worked with matrices in the past so I am going to stop here. Hopefully it has refreshed your memory on how to row-reduce. If not then I can continue on and give you a little more guidance as to how to procede

Last edited: Jul 24, 2011
8. Jul 24, 2011

### Vazer

On the I5 row of last matrix posted the 10.2 should be a 5.00 = R5.

So after reducing, I can solve for all five currents?
I1=-1A
I2=-1A
I3=5.6A
I4=5.2A
I5=-34.6A

9. Jul 24, 2011

### Clever-Name

*fixed*

And those currents don't look right. Plug you values back into the first equation
$$I_{1} + I_{2} = I_{3}$$

Are you able to show your steps? Maybe I can find where you went wrong.

10. Jul 24, 2011

### Vazer

Since the equations are set up in Ax=b form then:

For example on the Fifth row(I5):

Ax=b

[0+5.00+34.5+0+5.00] [I5] = [9.9]
44.5 I5= 9.9
I5=2.22x10-1A
when rounding it to three significant figures.

I1 through I4= 0
All are the same values when I try this on the other equations.

11. Jul 24, 2011

### Vazer

Sorry it's 5.00I2+34.5I3+5.00I5=9.9
for row five (I5)

12. Jul 24, 2011

### Clever-Name

The first way you wrote it isn't correct, but you corrected it in the second post so that's good.

With a system set up this way you need to perform some operations on the matrix first before you can evaluate for your missing quantities. I'll show you some of the steps for this situation:

$$\left[ {\begin{array}{ccccc|c} 1 & 1 & -1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 & 1 & 0 \\ -10.6 & 5.00 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -10.2 & 5.00 & 0 \\ 0 & 5.00 & 34.5 & 0 & 5.00 & 9.9 \\ \end{array} } \right]$$

Here I have set it up as an augmented matrix. So we have [A|b] in the above matrix, this is easier to work with but you need to recognize that the vertical line indicates Ax = b and not just A = b.

Continuing on we want to reorganize it so that we can easily row-reduce. So first I want to do the following:
Shift the 3rd row to the 2nd, the 5th row to the 3rd, the 2nd row to the 4th, and the 4th row to the 5th. Giving us:

$$\left[ {\begin{array}{ccccc|c} 1 & 1 & -1 & 0 & 0 & 0 \\ -10.6 & 5.00 & 0 & 0 & 0 & 0 \\ 0 & 5.00 & 34.5 & 0 & 5.00 & 9.9 \\ 0 & 0 & -1 & 1 & 1 & 0 \\ 0 & 0 & 0 & -10.2 & 5.00 & 0 \\ \end{array} } \right]$$

It now has this sort of 'step' form. Our next step is the following:
We need to modify the 2nd row, we cant have anything in the slot where the -10.6 is. So we have to 'modify' it using the 1st row. So the operation is: $10.6R_{1} + R_{2} \rightarrow R_{2}$
Giving us:
$$\left[ {\begin{array}{ccccc|c} 1 & 1 & -1 & 0 & 0 & 0 \\ 0 & 15.60 & -10.6 & 0 & 0 & 0 \\ 0 & 5.00 & 34.5 & 0 & 5.00 & 9.9 \\ 0 & 0 & -1 & 1 & 1 & 0 \\ 0 & 0 & 0 & -10.2 & 5.00 & 0 \\ \end{array} } \right]$$

We have now 'eliminated' that pesky -10.6 in the left-most column of row 2. Since it's row 2 we don't want any number other than 0 in the first column. For row 3 we don't want any number other than 0 in the first 2 columns, and so on.

The next step would be $R_{3} - \frac{5}{15.6}R_{2} \rightarrow R_{3}$ The reason we're doing this is to eliminate the pesky 5 in the 2nd column of row 3.

This gives us:

$$\left[ {\begin{array}{ccccc|c} 1 & 1 & -1 & 0 & 0 & 0 \\ 0 & 15.60 & -10.6 & 0 & 0 & 0 \\ 0 & 0 & 37.897 & 0 & 5.00 & 9.9 \\ 0 & 0 & -1 & 1 & 1 & 0 \\ 0 & 0 & 0 & -10.2 & 5.00 & 0 \\ \end{array} } \right]$$

Can you see what our next step would be?

13. Jul 24, 2011

### Vazer

Shouldn't the 37.8974 on the third row be 32.8974?
After this then the -1 on the fourth row needs to be eliminated then the -10.2 on the fifth row.

14. Jul 24, 2011

### Vazer

Nevermind, sorry another mistake. Figured how you did the elimination in row three:
(34.5+5.00) - (5.00/15.6)*5.00 = 37.8974

15. Jul 24, 2011

### Vazer

So, that means in the next row:
R4 + (1/37.897)R3 -> R4
(1+1)+(1/37.897)(34.5) = 2.910
R4=2.910

And the last row -10.2 needs to be eliminated.
R5 + (10.2/2.910)R4 -> R5
R5 = 50.952

Last edited: Jul 24, 2011
16. Jul 24, 2011

### Vazer

Is this correct? Learning how to post the matrices in code, don't know how to hide the left, right text from code.

left[{\begin{array}{ccccc|c}
1 & 1 & -1 & 0 & 0 & 0 \\
0 & 15.60 & -10.6 & 0 & 0 & 0 \\
0 & 0 & 37.897 & 0 & 5.00 & 9.9 \\
0 & 0 & 0 & 2.910 & 1 & 0 \\
0 & 0 & 0 & 0 & 50.952 & 0 \\
\end{array} }\right]

17. Jul 24, 2011

### Clever-Name

Almost, but not quite.

Our operation on Row 4 is $R_{4} + \frac{1}{37.897}R{3} \rightarrow R_{4}$

Giving us:

$$\left[ {\begin{array}{ccccc|c} 1 & 1 & -1 & 0 & 0 & 0 \\ 0 & 15.60 & -10.6 & 0 & 0 & 0 \\ 0 & 0 & 37.897 & 0 & 5.00 & 9.9 \\ 0 & 0 & 0 & 1 & 1.132 & 0.261 \\ 0 & 0 & 0 & -10.2 & 5.00 & 0 \\ \end{array} } \right]$$

Explicitly we have to assign the operation to EACH column of Row 4 and 3.

Thus we have 0 + 0 for the first column. 0 + 0 for the 2nd column. -1 + (1/37.897)(37.897) = 0 for the 3rd column. 1 + 0 = 1 for the 4th column. 1 + (1/37.897)(5) = 1.132 for the 5th column. And this operation also has to be applied to the outside column as well. Giving us 0 + (1/37.897)(9.9) = 0.261 outside.

18. Jul 24, 2011

### Vazer

Thanks for the help and the patience Clever-Name, appreciate the step by step procedure as to how to solve this problem using matrices. Seems a lot a work than I thought it would be, but I actually like this approach, just need to practice more. Thanks.

19. Jul 24, 2011

### Clever-Name

It does seem pretty daunting at first but once you get the hang of it it really is a nice method. We might as well go the next 2 steps to complete our solution. From my last matrix we do the following step: $R_{5} + 10.2R_{4} \rightarrow R_{5}$

Giving:

$$\left[ {\begin{array}{ccccc|c} 1 & 1 & -1 & 0 & 0 & 0 \\ 0 & 15.60 & -10.6 & 0 & 0 & 0 \\ 0 & 0 & 37.897 & 0 & 5.00 & 9.9 \\ 0 & 0 & 0 & 1 & 1.132 & 0.261 \\ 0 & 0 & 0 & 0 & 16.546 & 2.662\\ \end{array} } \right]$$

If we were to now write this again in it's full (Ax = b) form we would have:
$$\left[ {\begin{array}{cc} 1 & 1 & -1 & 0 & 0 \\ 0 & 15.60 & -10.6 & 0 & 0 \\ 0 & 0 & 37.897 & 0 & 5.00 \\ 0 & 0 & 0 & 1 & 1.132 \\ 0 & 0 & 0 & 0 & 16.546 \\ \end{array} } \right] \left[ {\begin{array}{cc} I_{1}\\ I_{2}\\ I_{3}\\ I_{4}\\ I_{5}\\ \end{array} } \right] = \left[ {\begin{array}{cc} 0\\ 0\\ 9.9\\ 0.261\\ 2.662\\ \end{array} } \right]$$

This is equivalent to the same setup we had at the very beginning. We can now expand this out to give us some linear equations and we end up with:

$$I_{1} + I_{2} - I_{3} = 0$$
$$15.60I_{2} - 10.6I_{3} = 0$$
$$37.897I_{3} + 5I_{5} = 9.9$$
$$I_{4} + 1.132I_{5} = 0.261$$
$$16.546I_{5} = 2.662$$

Now you can easily work from the bottom up and solve for I_5, then I_4, etc. etc. Getting all of your currents out of it.