V
C
I X
c
⋅
=
Note the magnitude of the voltage across the inductance is larger than the source voltage
magnitude.
Also, the angle is 90
o
ahead of the voltage across the resistance.
arg V
L
(
29
22.62deg
=
V
L
231.17volt
=
V
L
213.387
88.911j
+
volt
=
V
L
I X
l
⋅
=
arg V
R
(
29
67.38

deg
=
V
R
46.234volt
=
V
R
17.782
42.677j

volt
=
V
R
I R
⋅
=
c.
To find the voltage across each element, multiply current times impedance:
arg I
( )
67.38

deg
=
I
9.247amp
=
Note:
"arg" is the angle of a
complex number
I
3.556
8.535j

amp
=
I
V
Z
=
Z
5
12j
+
ohm
=
Z
R
X
l
+
X
c
+
=
X
c
13j

ohm
⋅
=
X
l
25j ohm
⋅
=
R
5 ohm
⋅
=
b.
The current is the voltage divided by the impedance.
Since the three elements are in series
we can add them directly:
We can let the voltage be our reference, so the angle is zero
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