In the middle of an AC analysis problem, I am stuck -

AI Thread Summary
The discussion revolves around solving a complex number equation for the current i_L in an AC analysis problem. The user initially arrives at an incorrect answer by misapplying complex number division rules, specifically by incorrectly splitting the fraction involving a complex denominator. Another participant clarifies that the correct approach is to multiply by the complex conjugate of the denominator, which simplifies the expression and leads to the correct result. The user acknowledges the mistake and gains a clearer understanding of how to handle complex fractions in this context. The conversation highlights the importance of proper techniques in complex number calculations for accurate results.
VinnyCee
Messages
486
Reaction score
0
OK, I have this complex number equation:

5\,V\,=\,\left[\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)\right]\,i_L

Now I try to solve for i_L:

i_L\,=\,\frac{5\,V}{\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)}

i_L\,=\,\frac{5\,V}{-20000j^2\,-\,6000j\,+\,1200}

Since j^2 is just -1:

i_L\,=\,\frac{5\,V}{21200\,-\,6000j}

And since \frac{1}{j} = -j, the final complex numbered answer I get is:

0.0002358\,+\,0.00083333j

However, this is incorrect! I have the answer for the problem, step-by-step, given by the prof. and I have double checked the answer using the Symbulator for the TI-89.

I should be getting:

0.00021836\,+\,0.0000618j

What am I doing wrong?:frown:
 
Physics news on Phys.org
VinnyCee said:
OK, I have this complex number equation:

5\,V\,=\,\left[\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)\right]\,i_L

Now I try to solve for i_L:

i_L\,=\,\frac{5\,V}{\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)}

i_L\,=\,\frac{5\,V}{-20000j^2\,-\,6000j\,+\,1200}

Since j^2 is just -1:

i_L\,=\,\frac{5\,V}{21200\,-\,6000j}\;\;\; (1)

And since \frac{1}{j} = -j, the final complex numbered answer I get is:

0.0002358\,+\,0.00083333j \;\;\; (2)

What've you done to get from (1) to (2)? I would first divide top and bottom by 5 to get
i_L\,=\,\frac{V}{4240-1200j}

Now, you need to get this into the form V(x+jy), so I would multiply top and bottom by the complex conjugate of the denominator

i_L\,=\,\frac{V}{4240-1200j}\cdot\frac{4240+1200j}{4240+1200j}

Now expand out the denominator, and you should obtain a purely real number, and the result will follow from dividing the real and imaginary parts of the numerator by the denominator.
I should be getting:

0.00021836\,+\,0.0000618j
That's what I get too!
 
Last edited:
Wow, thanks!

Why does the way I was trying not work?

\frac{1}{j}\,=\,-j <----- Right?

\frac{5}{21200}\,=\,0.0002358

and for the complex part:

\frac{5}{-6000}\m\left(-j\right)\,=\,0.000833j

That is where I get the answer I was getting (wrong):

0.0002358\,+\,0.000833j
 
Last edited:
VinnyCee said:
Wow, thanks!

Why does the way I was trying not work?

\frac{1}{j}\,=\,-j <----- Right?

\frac{5}{21200}\,=\,0.0002358

and for the complex part:

\frac{5}{-6000}\m\left(-j\right)\,=\,0.000833j

That is where I get the answer I was getting (wrong):

0.0002358\,+\,0.000833j

Because you made the error of assuming that \frac{a}{b+cj} = \frac{a}{b}+\frac{a}{cj}. This is not true (in general) and you cannot split up a fraction like this.

The trick for solving such a question is to always multiply the fraction by the conjugate of the denominator-- this makes the denominator real, and so the fraction becomes one of the form \frac{a+bj}{c} which of course can be simplified to obtain \frac{a}{c}+\frac{b}{c}j
 
Sweet, thanks again! Now I understand.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top