Solving for current in AC circuit with complex impedance

[VinnyCee]

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?

 

[cristo]

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!

 

[VinnyCee]

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):

$$0.0002358\,+\,0.000833j$$

 

[cristo]

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 $$\frac{a}{c}+\frac{b}{c}j$$

 

[VinnyCee]

Sweet, thanks again! Now I understand.