# Homework Help: Impedance Question

1. Mar 25, 2016

### Enochfoul

1. The problem statement, all variables and given/known data
Hi would somebody please be able to check to see if my workings look good?

2. Relevant equations
1/Z=1/Z1+1/Z2

IF Z1 = 3 + J2 and Z2 = 1 - j3, calcualte Z giving your answer in the form r(cosθ + jin son θ) where θ is in radians

3. The attempt at a solution
Attached as a scanned image

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2. Mar 25, 2016

### Joppy

Are you sure its Z1 = 3 + j2 and not Z1 = 3 + 2j ?

Either way, your working looks good.

3. Mar 26, 2016

### Enochfoul

Thanks for the reply here is a screen shot of the question

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• ###### Impedance Question.png
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4. Mar 26, 2016

### Joppy

Thanks. While your method looks correct, you've made a simple error in translation.

That is, in your working you've taken Z1 as 3 + j2 and not 3 + 2j. Similarly with Z2, you've written it down as 1 - j3 instead of 1 - 3j.

Try again but this time with Zeq = ( (3 + 2j) (1 - 3j) / (1 - 3j) + (3 + 2j) )

Excuse the brackets, having some trouble with Latex.

5. Mar 27, 2016

### Enochfoul

Hi Joppy on careful inspection of my work I have written J^2 and J^3, however I have calculated the solution based on J2 and J3. I have altered the values and have come up with the same answer as in my original scanned image. Do you deem the answers to be correct?

Ive just realised my answer in radians should have been negative. I have changed the solution to
Z= 2.765(cos(-0.416) - Jsin(-0.416)) Does that seem correct?

Last edited: Mar 27, 2016
6. Mar 27, 2016

### Joppy

So you have! I was wondering why things didn't reduce sooner in your working : ).

Yes that's correct. Well done!

7. Mar 28, 2016

### Enochfoul

For my final answer I have a negative tangent, hence the solution will be in the 2nd and 4th quadrant of the circle where the tangent is negative.
My solution of -23.83 deg is correct but in positive format the answer would be 156.17 deg or 336.17 deg.

In the context of the question I suspect the correct solution would be 156.17° or its radian equivalent.

i.e

Z= 2.765(cos(2.735) - Jsin(2.735))

or do you believe I got it right as Z= 2.765(cos(-0.416) - Jsin(-0.416))

Thanks

8. Mar 28, 2016

### Staff: Mentor

The atan() function takes a single number as an argument, so when using it to find the angle associated with a Cartesian vector (or the angle of a complex number) it cannot distinguish whether a negative argument should place the angle in quadrants 2 or 4, or 1 or 3. In other words an argument comprised of (-y)/(+x) is indistinguishable from (+y)/(-x), and (+y)/(+x) is indistinguishable from (-y)/(-x). So the signs of the sine and cosine of an angle can be confused when using atan. (There is another function called atan2(y,x) which takes two arguments and always returns a result in the correct quadrant, but you can research that separately).

The best approach when looking for the angle associated with a vector or complex number is to make a quick sketch of the vector or point to locate the correct quadrant. Then adjust the angle returns by atan() if required.

As you can see in the figure your result of -23.8° is good. 336° would also work, but by convention these angles are always normalized to lie within the range -180° ≤ θ ≤ +180° . 156° would be incorrect because it would place the point or vector in the second quadrant rather than the third.

9. Mar 28, 2016

### Enochfoul

Thanks again an excellent answer I will stick with my solution surrounding -23.8