arg Z
load
(
29
60

deg
=
Z
load
8.501
Ω
=
Z
load
4.25
7.362i

Ω
=
Z
load
V
ph
I
ph
=
d. The impedance is found by dividing the voltage by the current in phasor notation
I = 14.14 at 30 degrees
V = 120.2 at 90 degrees
c. The cosine leads the sine by 90 degrees, so changing the reference to cosine will
subtract 90 degrees from each phasor in part b.
I
ph
14.14 e
j 60
⋅
deg
⋅
⋅
amp
⋅
=
I = 14.14 at 60 degrees
I
rms
14.142
=
I
rms
I
pk
2
=
I
pk
20
=
V
ph
120.2 e
j 0
⋅
deg
⋅
⋅
volt
⋅
=
V = 120.2 at 0 degrees
V
rms
120.208
=
V
rms
V
pk
2
=
V
pk
170
=
b.
Phasors are alway rms for the magnitude.
If sin(
ϖ
t) is the reference, then the phase
angles are the same as in part a.
So the phasors are:
i(t) = 20 sin (
ϖ
t +
π
/3) amps
The current has a peak value of 20 amps and is leading the voltage by 60 degrees or
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 Spring '12
 Piller
 Trigonometry, Alternating Current, Volt

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