Answer:
A) 282.34 - j 12.08 Ω
B) 0.0266 + j 0.621 / unit
C)
A = 0.812 < 1.09° per unit
B = 164.6 < 85.42°Ω
C = 2.061 * 10^-3 < 90.32° s
D = 0.812 < 1.09° per unit
Explanation:
Given data :
Z ( impedance ) = 0.03 i + j 0.35 Ω/km
positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km
A) calculate Zc
Zc =
=
=
= 282.6 < -2.45°
hence Zc = 282.34 - j 12.08 Ω
B) Calculate gl
gl =
d = 500
z = 0.03 i + j 0.35
y = j4.4*10^-6 S/km
gl = 
= 
= 0.622 < 87.55 °
gl = 0.0266 + j 0.621 / unit
C) exact ABCD parameters for this line
A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )
C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )
D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
where : cos h (gl) = 
sin h (gl) = 
Answer:
why you doin this
Explanation:
is this so we get free points?
It is to be noted that it is impossible to find the Maclaurin Expansion for F(x) = cotx.
<h3>What is
Maclaurin Expansion?</h3>
The Maclaurin Expansion is a Taylor series that has been expanded around the reference point zero and has the formula f(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!
<h3>
What is the explanation for the above?</h3>
as indicated above, the Maclaurin infinite series expansion is given as:
F(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!
If F(0) = Cot 0
F(0) = ∝ = 1/0
This is not definitive,
Hence, it is impossible to find the Maclaurin infinite series expansion for F(x) = cotx.
Learn more about Maclaurin Expansion at;
brainly.com/question/7846182
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