Resolving complex number equation - electrical engineering

mrjoet

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Apr 10, 2021
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8
Hi,

I am following a text book and I've got this far which I know is correct:

\(\displaystyle V_{S1} = V_{s1}e^{\frac{π}{4}j} \)

\(\displaystyle I_{S2} = I_{s2}e^{\frac{-π}{2}j} \)

\(\displaystyle Z_{L1} = 150j \)

\(\displaystyle Z_{C1} = -66.67j \)

\(\displaystyle Z_{R2L2} = 100 + 50j \)

\(\displaystyle Z_{R1C2} = 3.24 - 19.46j \)

\(\displaystyle V_Y = Z_{R2L2} \frac{V_{S1}(Z_{L1} + Z_{C1} + Z_{R1C2}) + I_{S2}Z_{L1}Z_{R1C2}}{Z_{R2L2}(Z_{L1} + Z_{C1} + Z_{R1C2}) + Z_{L1}(Z_{R1C2}+Z_{C1})} \)

This is where I'm stuck, as I need to get from here to:

\(\displaystyle V_Y = 3.5e^{2.3j} \)

Is anyone able to help solve this?
 

skeeter

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Dec 15, 2005
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\(\displaystyle V_{S1} = V_{s1} \cdot \frac{\sqrt{2}}{2}(1 + j)\)

\(\displaystyle I_{S2} = I_{s2} \cdot (-j)\)

what are the values of \(\displaystyle V_{s1} \text{ and } I_{s2}\) ?

(the lower case "s" subscript values vice the capital ones) ...
 

mrjoet

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Apr 10, 2021
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Ah yes sorry I forgot to add those:

\(\displaystyle V_{s1} = 7\)

\(\displaystyle I_{s2} = 25\)
 

skeeter

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Dec 15, 2005
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I stored all your given values in alphabetical registers on my calculator ... the result is attached as an image. Converting the rectangular form to exponential is not even close to what you want to see.

5F3BCB5C-4E8D-4101-8012-7AC998569C61.png
 

jonah2.0

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Apr 29, 2014
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385
Beer soaked query follows.
I stored all your given values in alphabetical registers on my calculator ... the result is attached as an image. Converting the rectangular form to exponential is not even close to what you want to see.

View attachment 26402
What calculator app is that?
 

mrjoet

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Apr 10, 2021
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Hi,

Sorry I was going through this again and realised I gave some incorrect information.

\(\displaystyle I_{s2} = 25*10^{-3} \) or \(\displaystyle I_{s2} = -0.025j \)

Does that make a difference as I still can't get it to work, I come out with \(\displaystyle 3.53e^{-0.88j} \)
 
Last edited:

mrjoet

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Apr 10, 2021
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OK I just realised I have the correct result but I need to take \(\displaystyle 3.53e^{π-0.88j} = 3.53e^{2.26j} \)
 

skeeter

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Dec 15, 2005
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\(\displaystyle (\pi -0.88)=2.26\)
 
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