# Thread: Loop Analysis via determinants

1. ## Loop Analysis via determinants

I will begin by saying hello! I will provide a picture and my objectives with the question.
.
I am wanting to present the following loops in the matrix for $Ax=B$ We will call that "a)". Additionally, I want to solve for b) $I_1$, $I_2$ and $I_3$, providing their sense. Lastly, I want to determine c) the power dissipated in the $6.8k\Omega$ resitor.

a)
$I_1: (2.2k\Omega+9.1k\Omega)I_1-(9.1k\Omega)I_2=18V$

$I_2: (9.1k\Omega+7.5k\Omega+6.8k\Omega)I_2-(9.1k\Omega)I_1-(6.8k\Omega)I_3=-18V$

$I_3: (6.8k\Omega+3.3k\Omega)I_3-(6.8k\Omega)I_2=-3V$

and hence, we have our values for putting it in the form of Ax=B

____11.3 -9.1 0
A= -9.1 23.4 -6.8
____0 -6.8 10.1

___ I1
x = I2
___ I3

____18
B= -18
____-3

I apologize but I don't yet know how to execute matrices via latex code. Regardless, that's part a).

b)

First, I will determine the determinate for A, since it applies to all of the currents.

det(A)=(11.3*23.4*10.1)-(11.3*-6.8*-6.8)+(-9.1*-9.1*10.1)-(-9.1*-6.8*0)+0-0
det(A)=2984.511

For I1, det(A1)=(18*23.4*10.1)-(18*-6.8*-6.8)+(-9.1*-18*10.1)-(-9.1*-6.8*-3)=5261.82

and hence I1=5261.82/2984.511=1.76A clockwise.

I2= -3939.24/2984.511=-1.32A= 1.32 A counter-clockwise.

I3= -1310.97/2984.511=-0.439A = 0.439 A counter-clockwise.

c) The power dissipated in the 6.8-kolm resistor will be affected by currents I2 and I3, which are I=1.32A+0.439A=1.76A
P=I^2*R=(1.76A)^2*(6.8*10^3 olm) =21.0 kJ

2. Originally Posted by profound
det(A)=(11.3*23.4*10.1)-(11.3*-6.8*-6.8)+(-9.1*-9.1*10.1)-(-9.1*-6.8*0)+0-0
Can't help, but I have a question (out of curiosity):
why do you show -6.8 * -6.8 instead of (-6.8)^2 ? Or simpler still: 6.8^2 ?

3. I wouldn't recommend Kramer's Rule on a 3 x 3. Gaussian Elimination with Back Substitution is the most efficient exact method known. In any case, I don't get what you get. I get I1 = 1.2 (clockwise), I2 = 0.48 (counter-clockwise), and I3 = 0.62 (counter-clockwise). Double-check your determinants.

Note on c): obviously, the numerical result will be affected by the correctness of the previous work. However, I would also comment that the units of power are watts, not joules. A watt is one joule per second.

4. @Wilmer

There is no real reason, it was merely so I could look back at my previous work if I messed up something with respect to a minus sign.

@Ackbeet

My teacher wants us to use determinants thus far in what we have covered. I will check my results and report back.

5. Nevermind, I figured it out.

6. No, I think your matrix is correct, with the mesh analysis and all. Here's WolframAlpha's answer for the system.

For the determinant A, WolframAlpha gives 1311.749.

For the A1 determinant, WolframAlpha gives 1581.78.

For the A2 determinant, WolframAlpha gives -630.48.

And finally, for the A3 determinant, WolframAlpha gives -814.11.

So it looks like you're consistently evaluating determinants incorrectly. Could you please show me the A determinant, step-by-step?

[EDIT]: Ah, got your message about figuring it out. Good.

7. Just noticed this now. Since the resistances are given in kilo-olm quantities, that means current will be in mJ and Power will by in mJ (ie. small power).

8. I think you're confused about units. Joules are not the units for either current or power! Current is in amps or some multiple thereof (perhaps mA), and power is in watts or some multiple thereof (perhaps mW).