Load Impedences - Rectangular to Polar

[bmed90]

Homework Statement

Can you simplify rectangular expression

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

The answer is

.134 angle(28.07)

Homework Equations

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?

 

[The Electrician]

Homework Statement

Can you simplify rectangular expression

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

The answer is

.134 angle(28.07)

Homework Equations

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?

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.

 

[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.

 

[gneill]

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.

 

[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.

 

[gneill]

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.

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.

 

[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?

 

[gneill]

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

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.

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?

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.