# AC Assgnment 8

**Question description**

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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