Three impedances are connected in parallel. Z1=2-j, Z2=j+2, Z3=-2j. Find the equivalent admittance ?

[Swerve15]

 

[Deleted member 4993]

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If I were to do this I'll first calculate: from Z₁=2-j, Z₂=j+2, Z₃=-2j.

1/Z₁ = 1/5 * [2 + j].................................edited

and continue......

 

[Cubist]

1/Z₁ = 1/\(\displaystyle \sqrt{5}\) * [2 + j]

Good advice, but there's no need for the root in this case...

[math] \frac{1}{Z_1} = \frac{1}{\left(2-j\right)} = \frac{\color{red}{2+j}}{\left(2-j\right)\left(\color{red}{2+j}\right)} = \frac{2+j}{2^2+1}= \frac{2+j}{5} [/math]

 

[Deleted member 4993]

Good advice, but there's no need for the root in this case...

[math] \frac{1}{Z_1} = \frac{1}{\left(2-j\right)} = \frac{\color{red}{2+j}}{\left(2-j\right)\left(\color{red}{2+j}\right)} = \frac{2+j}{2^2+1}= \frac{2+j}{5} [/math]

Oooops ... going to the corner for 5 minutes with Jomo (permanent resident) and the spider.