# Homework Help: Load Impedences - Rectangular to Polar

1. Nov 12, 2013

### bmed90

1. The problem statement, all variables and given/known data

Can you simplify rectangular expression

1/j30 + 1/10 + 1/(15-j25)

.134 angle(28.07)

2. Relevant equations

3. The attempt at a solution

I got

= 1/j30 + 1/10 + 1/(15-j25)

= -j.03+.1+.06+j.04

= .16 + j.01

= .16 angle(3.57)

The 1/(15-j25) term is kind of throwing me off a bit. Am i on the right approach?

2. Nov 13, 2013

### The Electrician

The first problem I see that you have is that 1/(15-j25) ≠ .06+j.04

Also, you should carry more than just 2 digits in your calculations.

3. Nov 13, 2013

### bmed90

Do you know what the correct answer is by any chance and how to get it? I just can't seem to get it.

4. Nov 13, 2013

### Staff: Mentor

You can avoid decimals entirely by working with fractions, giving you exact results.

You should know how to clear imaginary values from denominators by multiplying top and bottom by the complex conjugate of the denominator. For example:
$$\frac{A}{B + jC}\cdot\frac{B - jC}{B - jC} = \frac{AB - jAC}{B^2 + C^2}$$

So, (1) clear the denominators of imaginaries; (2) put everything over a common denominator; (3) collect reals and imaginaries in the numerator; (4) (optional) split into separate real and imaginary terms and reduce the fractions to lowest terms.

5. Nov 13, 2013

### bmed90

I hate to be inadequate but can you guys show me a detailed step by step on how to get through this from beginning to end? I have a quiz in a little while and if I can just see the steps it would clear up a lot of things.

6. Nov 13, 2013

### Staff: Mentor

Very sorry, but that would be against the Forum rules; we can provide hints, suggestions, and corrections, but not solutions. You have to do the work.

Take a look at the expression I provided which shows the method for clearing imaginary values from the denominators. Assign some numbers to the constants A, B, and C and try it out.

7. Nov 13, 2013

### bmed90

Ok so I think I got it.

1/(15-j25) = A/B-jC

=> A/B-jC * B-jC/B-jC = (15-j25)/850 = 15/850-j25/850 = .0176-j.0294

into original problem

-j.03+.1+ .0176-j.0294

=.1176-j.0594

sqrt[(.1176)^2 + (.0594)^2] angle(tan-1(.0594/.1176))

=.132 angle(26.79)

You know, the solutions my TA posts don't have all these steps written out. Is there someway to shortcut this whole process to make it faster come test time?

8. Nov 13, 2013

### Staff: Mentor

Nooo. You must use the complex conjugate of the denominator. The complex conjugate is formed by negating the imaginary term; if it was positive it bec9omes negative, if it was negative it becomes positive.

Here your denominator is 15 - j25, so its complex conjugate is 15 + j25.

Not really. The trick is practice, so you can spend more time doing than thinking about it

Note that you have the choice of doing multiplications and divisions in rectangular form as you've just done, or converting the values to polar form first and doing the multiplications and divisions that way (can be speedier after the conversions). But additions and subtractions of complex numbers still need to be done in rectangular form. A lot of switching back and forth from polar to rectangular can be tedious and errors can creep in. Whatever you do, keep several extra digits of accuracy in all intermediate steps.