Recent content by 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.

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 &...

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

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

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.

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

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.

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

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

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

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.

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...