# AC Assgnment 8

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Here is what my instructor said to fix

Resubmit the assignment. Clearly indicate the final answers for all the work, such as: 16 a) the answer, b) another answer, c) etc.

23-2
R1=12 Ω
R2=18 Ω
Vt=15V
Rt=R1+R2
Rt=12+18=30 Ω
I=Vt/Rt=15/30=0.5A
V1=IR1=0.5*12=6V
V2=IR2=0.5*18=9V
(a)Vt and I are in phase
(b)V1 and I are in phase
(c)V2 and I are in phase

23-4:
R1=12 Ω
R2=18 Ω
Va=36V
Rt=R1+R2
12+18=30 Ω
I1=Va/R1=36/12=3A
I2=Va/R2=36/18=2A
It=I1+I2=3+2=5A
(a)Va and I1 are in same phase
(b)Va and I2 are In same phase
(c)Va and It are in same phase

23-6:
Vt=120V
XL1=100 Ω
XL2=150 Ω
XLT=XL1+XL2
100+150=250 Ω
I=Vt/Xlt=120/250=0.48A
V1=IXlt=0.48*100=48
V2=IXlt=0.48*150=72
The ohms of XL are just as effective as ohms of R in limiting the current or producing a voltage drop.
XL has a phasor quantity with a 90° phase angle.

23-8:
Va=120V
Xl1=100 Ω
Xl2=400 Ω
I1=Va/Xl1=120/100=1.2A
I2=Va/Xl2=120/400=0.3A
It=I1+I2=0.3+1.2=1.5A
(a)Va and I1 are
(b)Va and I2 are
(c)Va and It are
The ohms of XL are just as effective as ohms of R in limiting the current or producing a voltage drop.
XL has a phasor quantity with a 90° phase angle.

23-10
Vt=18V
Xc1=220 Ω
Xc2=680 Ω
V1=IXc1=4.4V
V2=IXc2=13.6V
Xc=220+680
I=Vt/Xc=18/900=0.02A
Since there is no R or XL, the series ohms of XC can be combined directly.

23-12:
Vt=10V
Xc1=100 Ω
Xc2=25 Ω
I1=Va/Xc1=10/100=0.1A
I2=Va/Xc2=10/50=0.2A
It=0.2+0.1=0.3A
Since there is no R or XL, parallel IC currents can be added.

23-14:
Vt=24V
XL=180 Ω
Xc=120 Ω
Xl=180-120=60 Ω
I=Vt/Xl=0.4A
VL=IXL=0.4*180=72V
VC=IXc=0.4*120=48V

23-16:
It=1A
IL=Va/XL=0.3A
Ic=Va/Xc=0.2A

23-18:
Z=((R)2+(X)2)1/2
Z=125
I=V/Z=125/125=1A
Tan-1 – X/R=-36.86
Vr=IR=100V
VL=IXL=50V
Vc=IXc=125V
X net=75
Vt=125V

23-20:
R1=1000 Ω
Xc1=400 Ω
XL1=1500 Ω
Xc2=800 Ω
R2=1000 Ω
XL2=1200 Ω
Vt=25V
Vt/Rt=It=1/80A
Vr1=1205V
Vr2=12.5V
VL1=18.75V
VL2=15V
Vc1=5V
Vc2=10
Zt=2500
Angle=-53.130

23-22:
Va=36V
R=240 Ω
XL=60 Ω
Xc=90 Ω
It=((Ir)2+(Ix)2)1/2
Ir=Va/R=36/240=0.15A
It=0.25A
Zeq=Va/It=36/0.25=144
Tan-1 – Ix/Ir=-53.130

23-24:
Va=15V
XL1=10 Ω
XL2=60 Ω
R1=100 Ω
R2=50 Ω
Xc1=40 Ω
Xc2=20 Ω
Ir1=Va/R1=0.15A
Ic1=0.375A
Il1=1.5A
Ir2=0.3A
Il2=0.25A
Ic2=0.75 Ω
Ix=0.625
It+((Ir)2+(Ix)2)^1/2
It=0.770
Tan-1 – Ix/Ir=-39.065
Zeq=Va/It=19.48

23-26
Fig 23-17:
Real power=I^2R
=(0.5)^2(30)
=7.5
Apparent power=VI
=15*0.5=7.5
Power factor=real power/apperent power=7.5/7.5=1

Fig 23-19:
Real power=I^2R=(0.48)^2(250)=57.6
Apparent power=VI=120*0.48=57.6
p.f=1

fig 23-22:
Real power=I^2R=(0.3)^2(125)=10.8

Apparent power=VI=10(0.3)=3
PF=3.6

Fig 23-24:

Real power=I^2R=0.2^2(150)=6
Apparent power=VI=18*0.2=3.6
PF=1.667

23-28:

(a)
Real power=VIcos theta=120*5*cos(-45)=424.26
Apparent power=VI=120*5=600
Power factor=0.7071
(b)
Realpower=240*18*cos(-26.56)=3864.09
Apperent power=240*18=4320
PF=0.8944
(c)
Real power=100*3*cos78=62.0373
Apparent power=100*3=300
PF=0.206
(d)
Real power=120*8*cos56=536.82
Apparent power=960
PF=0.559

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