# Is my solution or the solutions manual incorrect?

#### Aria Nemati

Joined Jul 19, 2019
27
I have the circuit:

where

According to the solutions manual. The answer is:

This is what I calculated:

As you can see. The only difference is that I did

$$\frac{1}{10}−\frac {j}{10}+\frac{1}{8+j6}$$ instead of $$\frac{1}{10}+\frac {j}{10}+\frac{1}{8+j6}$$​

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#### Papabravo

Joined Feb 24, 2006
15,750
Your solution is wrong. It should be:
$Y_C\;=\;\frac{1}{Z_C}\;=\;\frac{1}{\frac{1}{j\omega C}}\;=\;j\omega C$

#### Aria Nemati

Joined Jul 19, 2019
27
But Zc is negative. Where does the negative sign go?

#### Papabravo

Joined Feb 24, 2006
15,750
But Zc is negative. Where does the negative sign go?
When you take the reciprocal it disappears.
$\frac{1}{j}\;=\; -j$
Right?

#### Aria Nemati

Joined Jul 19, 2019
27
When you take the reciprocal it disappears.
$\frac{1}{j}\;=\; -j$
Right?
Ok. I'm getting confused now. I just did a problem that is almost the same and got the correct answer. So what is the difference that I have failed to see?

Now I did the same problem the way I am supposed to and I get a wrong answer:

The angle is wrong. It should be positive not negative. So what am I doing wrong here?

Last edited:

#### Aria Nemati

Joined Jul 19, 2019
27
Same here. I get the right angle, but I get positive while it should be negative

#### The Electrician

Joined Oct 9, 2007
2,832
How do you know what the correct solutions for post #5 and #6 are? Are those also worked out in a solutions manual? If so, post the solutions given in the solutions manual.