AEE100 / MEC1151- Electrical Circuits and Analysis
Dr Hong Fan, University of Glasgow, 2022
Tutorial – Topic 6: AC Circuits
[Impedance]
6.1 If V = 100V∠0° is applied across each of the circuit elements of Figure 6.1:
(a) Determine each current in phasor form.
(b) Express each current in time domain form.
V
2) / R = ZR
100L 0°
=
50L 0°
2 , = 25290°
= 220° A "
212=5020
V 90°
It = Zc = to L -
2,
°
100 LO
=
°
25 2 90
= 4L - 90° A Figure 6.1
b) IR = 2 COS WE A
V
IL 4 COS I Wt 90° ) A
1C = = -
Ic
Ic = 10 COS ( Wt -190° ) A
°
100 L O #
= °
10 L -
90
°
= 10 2 90 A
#
6.2 If the current through each circuit element of Figure 6.1 is 0.5 A∠0°:
(a) Determine each voltage in phasor form.
(b) Express each voltage in time domain form.
a) Vp = IZR b) VR = 25 coswtv
= ( 0.520°) ( 5020°) VL = 12.5 COS ( Wt +90° ) V
Ve = 5 cos ( wt -
90° ) V
= 2520° V #
vi. = 12L
= 10.5 L 0° ) ( 25L 90° )
= 12.5 L 90° V
Vc = 12C
= ( 0.520° ) ( 10L -90° )
= 5 L GOV
#
-
1
Dr Hong Fan, University of Glasgow, 2022
Tutorial – Topic 6: AC Circuits
[Impedance]
6.1 If V = 100V∠0° is applied across each of the circuit elements of Figure 6.1:
(a) Determine each current in phasor form.
(b) Express each current in time domain form.
V
2) / R = ZR
100L 0°
=
50L 0°
2 , = 25290°
= 220° A "
212=5020
V 90°
It = Zc = to L -
2,
°
100 LO
=
°
25 2 90
= 4L - 90° A Figure 6.1
b) IR = 2 COS WE A
V
IL 4 COS I Wt 90° ) A
1C = = -
Ic
Ic = 10 COS ( Wt -190° ) A
°
100 L O #
= °
10 L -
90
°
= 10 2 90 A
#
6.2 If the current through each circuit element of Figure 6.1 is 0.5 A∠0°:
(a) Determine each voltage in phasor form.
(b) Express each voltage in time domain form.
a) Vp = IZR b) VR = 25 coswtv
= ( 0.520°) ( 5020°) VL = 12.5 COS ( Wt +90° ) V
Ve = 5 cos ( wt -
90° ) V
= 2520° V #
vi. = 12L
= 10.5 L 0° ) ( 25L 90° )
= 12.5 L 90° V
Vc = 12C
= ( 0.520° ) ( 10L -90° )
= 5 L GOV
#
-
1
