University AC Circuit Phase Relationships Analysis and Solutions

This assignment provides a detailed analysis of AC circuit phase relationships in various configurations. It explores the phase relationships between voltage and current in purely resistive, inductive, and capacitive circuits, both in series and parallel arrangements. The document covers the phase lead and lag characteristics, including 90-degree phase shifts in inductive and capacitive circuits. It further examines series and parallel LC circuits, detailing how inductive and capacitive reactance affect voltage and current relationships. The analysis extends to RLC circuits, illustrating how inductive and capacitive reactance interact. The document also includes calculations for net reactance, reactive branch currents, real power, apparent power, and power factor, offering a comprehensive understanding of AC circuit behavior. The solutions provide insights into how these parameters change with different circuit configurations and component values.

Solutions
23-2
This a purely resistive series AC circuit. The phase relationships between voltages and currents
in a series AC resistive circuit is 00,, i.e the current is in phase with the voltage.
a. VT and I: 00 phase- VT is in phase with I
b. V1 and I: 00 phase- V1 is in phase with I
c. V2 and I: 00 phase- V2 is in phase with I
23-4
This a purely resistive parallel AC circuit. The phase relationships between voltages and currents
in a parallel AC resistive circuit is 00, i.e the current is in phase with the voltage.
a. VA and I1: 00 phase - VA is in phase with I1
b. VA and I2: 00 phase - VA is in phase with I2
c. VA and IT: 00 phase - VA is in phase with IT
23-6
This is a purely inductive series AC circuit and the phase relationships between voltages and
currents in such a circuit is 900 lead
a. VT and I: 900 phase lead, VT leads I by 900
b. V1 and I: 900 phase lead, V1 leads I by 900
c. V2 and I: 900 phase lead, V2 leads I by 900
23-8
This is a purely inductive parallel AC circuit and the phase relationships between voltages and
currents in such a circuit is 900 lead.
a. VA and I1: 900 phase lead, VA leads I1 by 900.
b. VA and I2: 900 phase lead, VA leads I2 by 900
c. VA and IT: 900 phase lead, VA leads IT by 900
23-10
This is a purely capacitive series AC circuit and the phase relationships between voltages and
currents in such a circuit is 900 lag.
a. VT and I: 900 phase lag, VT lags I by 900
b. V1 and I: 900 phase lag, V1 lags I by 900
c. V2 and I: 900 phase lag, V2 lags I by 900
23-12
This is a purely capacitive parallel AC circuit and the phase relationships between voltages and
currents in such a circuit is 900 lag.

a. VA and I1: 900 phase lag, VA lags I1 by 900.
b. VA and I2: 900 phase lag, VA lags I2 by 900
c. VA and IT: 900 phase lag, VA lags IT by 900
23-14
This a series LC circuit and have the following phase relationships.
a. XL and Xc
XL is greater than XC
Therefore, XL leads VL by 900 and Xc lags VL by 900 and the phase difference between XL and Xc
is
90 ° +90 °=180 °
b. VL leads I by 900
c. VC lags I by 900
d. VT leads I by 900
e. VL and Vc
Inductive reactance is greater than capacitive reactance.
Therefore, VL leads I by 900 and Vc lags I by 900 and the phase difference between VL and Vc is
90 ° +90 °=180 °
f. If XL and Xc values are interchanged, then the capacitive reactance will be higher than
inductive reactance.
Therefore, VT lags I by 900
23-16
This a parallel LC circuit and have the following phase relationships.
a. IL and VA- IL lags VA by 900
b. IC and VA- IC leads VA by 900
c. IL and IC
Capacitive reactance is greater than inductive reactance.
Therefore, being a parallel LC circuit, IL leads I by 900 and Ic lags I by 900 and the phase
difference between IL and IC is
90 ° +90 °=180 °
d. VA leads IT by 900 since XC is greater than XL
e. If XL and XC values are interchanged, then the inductive reactance will be higher than
capacitive reactance.

Therefore, VA lags IT by 900
23-18
a. For series RLC circuit, XL and Xc are 1800 out of phase.
b. VL and I are 900 out of phase with VL leading I
c. VC and I are 900 out of phase with VC lagging I
d. VR is in phase with I
e. VL and VC are 1800 out of phase.
f. VT and I
X net= XC−X L
X net=125−50=75 Ω Capacitive
ZT = √752 +1002=125 Ω
θ=tan−1 R
ZT
=tan−1 100
125 =38.67°
VT lags the current I because capacitive reactance is higher than inductive inductance.
g. VT lags VR by -36.870 since VR is in phase with I
h. Interchanging XL and XC
X net= XL−XC
X net=125−50=75 Ω Inductive
ZT = √752 +1002=125 Ω
θ=tan−1 R
ZT
=tan−1 100
125 =38.67°
VT leads the current I because inductive reactance is higher than capacitive inductance.
23-20
a. The net reactance, X: For series XL and XC, the reactance subtract.
X net= XL−XC
X L=X L1 + X L2=1.5+1.2=2.7 k Ω
XC =XC 1 + XC 2=400+800=1.2 k Ω
X net=2.7−1.2=1.5 k Ω
b. ZT = √ Xnet
2+ R2
R=R1+R2=1+1=2 k Ω
ZT = √1.52 +22=2.5 k Ω

c. I = V T
ZT
= 25 V
2.5 k Ω =10 mA
d. V R 1=I R1=10 mA × 1k Ω=10 V
V R 2=I R2 =10 mA × 1 k Ω=10 V
e. V L1=I XL 1=10 mA ×1.5 k Ω=15 V
V L2=I XL 2=10 mA ×1.2 k Ω=12 V
f. V C 1=I XC 1=10 mA ×0.4 k Ω=4 V
V C 2=I XC 2=10 mA ×0.8 k Ω=8 V
g. θz=tan−1 X
R =tan−1 1.5
2 =36.87 °
20-22
a. VA is in phase with IR
b. VA lags IC by 900
c. VA leads IL by 900
d. IL and IC have 1800 phase difference since IL lags VA by 900 and IC leads VA by 900
e. I L= V A
X L
= 36
9 0 =0. 4 A
I C= V A
XC
= 36
6 0 =0. 6 A
I X =I C−I L=0.6−0.4=0.2 A
I R =V A
R = 36
240 =0.15 A
θ=tan−1 I X
I R
=tan−1 0.2
0.15 =53.13 °
Therefore, VA lags IT by 53.130 since capacitive current is higher than inductive current.
f. Interchanging values of capacitive and inductive reactance.
I L= V A
X L
= 36
6 0 =0. 6 A
I C= V A
XC
= 36
9 0 =0. 4 A
I X =I L−I C=0.6−0.4=0.2 A
I R =V A
R = 36
240 =0.15 A

θ=tan−1 I X
I R
=tan−1 0.2
0.15 =53.13 °
Therefore, VA leads IT by 53.130 since inductive current is higher than capacitive current.
20-24
a. I R 1= V A
R1
= 15
100 =0.15 A
I R 2= V A
R2
= 15
50 =0.30 A
b. I L 1= V A
XL 1
= 15
10 =1 .5 A
I L 2= V A
XL 2
= 15
60 =0.25 A
c. I C 1= V A
XC 1
= 15
40 =0.375 A
I C 2= V A
XC 2
= 15
20 =0.75 A
d. Net reactive branch current IX;
Inductive branch current: I L ¿ I L 1+ I L2=1.5+0.25=1.75 A
Capacitive branch currents: IC=IC 1+I C 2=0.375+0.75=1.125 A
Net reactive branch current:
I X =I L−I C=1.75−1.125=0.625 A
e. Resistive branch current:I R =I R 1 + IR 2=0.15+ 0.30=0.45 A
Total current, I T= √IR
2+ I X
2 = √0.452+ 0.6252=0.7701 A
f. Total impedance
ZEQ= V A
IT
= 15
0.7701 =19.48 Ω
20-26
a. RT =R1 +R2=12+18=30
I = V T
RT
= 15
30 =0.5 A

Real power, P=V T I cos θ=15× 0.5 ×cos (0 °)=7.5 W
Apparent power, S=V T I =15× 0.5=7.5 VA
Power factor, PF= P
S = 3
3 =1
b. XT =X L1 + XL 2=100+150=25 0 Ω
I= V T
XT
= 120
250 =0.48 A
Real power, P=V T I cos θ=120× 0.48 ×cos (90 °)=0 W
Apparent power, S=V T I =120× 0.48=57.6 VA
Power factor, PF= P
S = 0
57.6 =0
c. XT = XC 1 × XC 2
XC 1+ XC 2
=100 × 25
100+25 =20 Ω
I = V T
XT
= 1 0
20 =0. 5 A
Real power, P=V T I cos θ=10× 0.5 ×cos (−90 ° )=0 W
Apparent power, S=V T I =10× 0.5=5 VA
Power factor, PF= P
S = 0
5 =0
d. I L= V A
X L
= 1 8
6 0 =0. 3 A
I C= V A
XC
= 18
9 0 =0. 2 A
I T=I L−I C=0.3−0.2=0.1 A
Real power, P=V A I T cos θ=18 ×0.1 ×cos (90 ° )=0 W
Apparent power, S=V A IT =18 ×0.1=1.8VA
Power factor, PF= P
S = 0
1.8 =0
20-28

a. Real power, P=V A IT cosθ1=120 ×5 ×cos (−45° )=424.26 W
Apparent power, S=V A IT =120 ×5=600 VA
Power factor, PF= P
S =cos θ1=cos (−45 ° )=0.707 1
b. Real power, P=V A IT cosθ1=240 ×18 ×cos (−26.56 ° )=3864.1W =3.8641 kW
Apparent power, S=V A IT =240 ×18=4320 VA=4.32 kVA
Power factor, PF= P
S =cos θ1=cos (−26.56° ) =0.8945
c. Real power, P=V A IT cosθ1=100 ×3 ×cos (78 °)=62.37 W
Apparent power, S=V A IT =10 0 ×3=3 00VA
Power factor, PF= P
S =cos θ1=cos (78 °)=0.2079
d. Real power, P=V A IT cosθ1=120 ×8 ×cos (56 ° )=536.83 W
Apparent power, S=V A IT =120 ×8=9 60 VA
Power factor, PF= P
S =cos θ1=cos (56 °)=0.5591