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MechEngJordan
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Homework Statement
Calculate the primary current, and hence the voltage at the transformer input winding, V1.
Transformation ratio, a = N1/N2 = 0.1
R1 = 0.12 Ω; R2 = 12 Ω
X1 = 0.4 Ω ; X2 = 40 Ω
RC = 560 Ω
V2 = 2300 V
RL = 1 kΩ
'Xm = 800 Ω'
Homework Equations
R1eq = R1 + a2R2
X1eq = j(X1 + a2X2)
V2' = aV2
Current divider equation.
The Attempt at a Solution
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Approximate equivalent circuit:
R1eq = 0.12 + (0.1)2(12)
R1eq = 0.24 ΩX1eq = j(0.4 + (0.1)2(40))
X1eq =j0.8 Ω
V2' = 0.1(2300)
V2' = 230 V
a2ZL = (0.1)2(1000)
a2ZL = 10 Ω
Ip = V2'/a2ZL
Ip = 23 AFrom here is where I believed that there is likely a more efficient way to solve the problem -- particularly because the value of Xm was not actually given in the paper, but told to us during the tutorial, more-or-less made up on the spot.
Here is the outline of the given solution:
Ip= I1*Z2/(Z1+Z2)
⇒I1 = Ip*(Z1+Z2)/Z2
⇒I1 = 23*(Rc // Xm + X1eq + R1eq + a2ZL)/ (Rc // Xm)I0 = I1 - Ip
∴ V1 = I0(Rc // Xm)
I'd be thankful for any input.