Text 3.2
a) The dots go on the upper left and lower right, respectively.
b) See previous assignment, text 3.2b.
c)
lambda_1 = L11*i1  L12*i2
lambda_2 =L21*i1 + L22*i2
where L11 = 3.125 mH, L12=L21= 0.938 mH, L22 = 0.781 mH,
Text 3.10
a)
If the transformer is ideal, then there are no losses. Since the load is resistive,
the power factor is unity.
b)
Simply divide the voltage by current, which are both real in this case:
Text 3.12
(N1/N2)
a) The ideal transformer has infinite impedance. One way to determine the primary self
inductance is to disconnect the secondary (i2=0) and measure the impdance.
In this case, we see that we have 45 H in series with 15 H.
Therefore,
Likewise, if we set i1=0 and measure the impedance looking in from v2 terminals, we see
an ideal transformer with a 15 H inductor on the opposite side. We need to reflect the
impedance through the transformer to the 2side. This means we multiply by (N2/N1)^2:
To get the mutual indutance, we need to determine the flux linked by coil 2 due to current
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 Fall '08
 SAUER
 Alternating Current, Inductor, ideal transformer, primary self inductance

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