Resolving complex number equation - electrical engineering

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mrjoet

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Apr 10, 2021
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Hi,

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

VS1=Vs1eπ4j\displaystyle V_{S1} = V_{s1}e^{\frac{π}{4}j}

IS2=Is2e−π2j\displaystyle I_{S2} = I_{s2}e^{\frac{-π}{2}j}

ZL1=150j\displaystyle Z_{L1} = 150j

ZC1=−66.67j\displaystyle Z_{C1} = -66.67j

ZR2L2=100+50j\displaystyle Z_{R2L2} = 100 + 50j

ZR1C2=3.24−19.46j\displaystyle Z_{R1C2} = 3.24 - 19.46j

VY=ZR2L2VS1(ZL1+ZC1+ZR1C2)+IS2ZL1ZR1C2ZR2L2(ZL1+ZC1+ZR1C2)+ZL1(ZR1C2+ZC1)\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:

VY=3.5e2.3j\displaystyle V_Y = 3.5e^{2.3j}

Is anyone able to help solve this?
 
[MATH]V_{S1} = V_{s1} \cdot \frac{\sqrt{2}}{2}(1 + j)[/MATH]
[MATH]I_{S2} = I_{s2} \cdot (-j)[/MATH]
what are the values of [MATH]V_{s1} \text{ and } I_{s2}[/MATH] ?

(the lower case "s" subscript values vice the capital ones) ...
 
Ah yes sorry I forgot to add those:

Vs1=7\displaystyle V_{s1} = 7

Is2=25\displaystyle I_{s2} = 25
 
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
 
Hi,

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

Is2=25∗10−3\displaystyle I_{s2} = 25*10^{-3} or Is2=−0.025j\displaystyle I_{s2} = -0.025j

Does that make a difference as I still can't get it to work, I come out with 3.53e−0.88j\displaystyle 3.53e^{-0.88j}
 
Last edited:
OK I just realised I have the correct result but I need to take 3.53eπ−0.88j=3.53e2.26j\displaystyle 3.53e^{π-0.88j} = 3.53e^{2.26j}
 
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