# EE115 Homework, Chapter 21             and so on……

21.2

a) Vinduced is equal and contrary to the applied Voltage.

Vi = -L*dI/dt = -Va

a) Vi=-10 V  or Vi=+10 V depending if i is increasing or decreasing through 0 mA

b) Vi = 0V

c) Vi = 0 V

21.4

a) I(R) = V(R)/R = 60/15 = 4 A

b) I(L) = V(L)/XL = 80/20 =4 A

c) I(T) = I(R)= I(L) = 4 A,  this is because it is a series circuit

21.6.

V(T) = 100 V = sqrt(V(R)^2 +V(L)^2) = sqrt(60^2+80^2)

21.8

a) V(R)= I*R =1.6*10^3*30*10^-3 =1.6*30 = 48 V

b) V(L) = 30mA *1.2K =30*1.2 = 36 V

c) V(T) = sqrt(V(R)^2+V(L)^2) =sqrt(48^2+36^2) =60 V

21.10.

V(T) = sqrt(V(r)^2+V(L)^2)

a) sqrt(12^2+6^2) = 13.42 V

b) 47.17 V

c) 18.36 V

d) 56.57 V

21.12 See attached figure 21-12.jpg

21.14.

Z(T) = R+i*XL = 20+i*60 ohms,   i = sqrt(-1)
module(Z(T)) = sqrt(20*20+60*60) = 63.24 ohms

I= V(T)/Z = 12/(20+i*60)  A

module(I) = V(T)/module(Z) = 12/63.24 =0.19 A

V(L)^2+V(R)^2 = V(T)^2

V(R)/V(L)= R/X(L) = 1/3

V(L) =3*V(R)

9*V(R)^2 +V(R)^2 =144

V(R) = sqrt(14.4) = 3.79 V

V(L) = 3*V(R) = 11.38 V

tan(theta) = XL/R = 3, theta = 71.56 deg

21.16.

Z(T) = R+i*XL

module(Z) = sqrt(R^2+XL^2)

a) Z =40+i*30 ohms

module(Z) = 50 ohms

b) Z =50+i*50 ohms

module(Z) = 70.71 ohms

c) Z = 100+i*10 ohms

module(Z) = 100.5 oh ms

d) Z = 10+i*100 ohms

module(Z) = 100.5 ohms

I=V(T)/Z

module(I) = V(T)/module(Z)

a) I =50/(40+30*i)

module(I) = 50/50 =1 A

b) I = 141.4/(50+i*50)

module(I) =141.4/70.71 = 2 A

c) I = 10/(100+10*i)

module(I) = 10/100.5 =0.0995 =0.1 A

d) I = 10/(10+100*i)

module(I) = 0.0995 =0.1 A

V(L)^2+V(R)^2= V(T)^2

V(L)/V(R) = XL/R

or simpler

V(L) = module(I) * XL

V(R) = module(I)*R

a) V(L)=1*30= 30 V

V(R) =1*40= 40 V

b) VL= 2*50 =100 V

VR =2*50 =100 V

c)VL=0.1*10 =1 V

VR= 0.1*100 =10 V

d) VL= 0.1*100 =10 V

VR= 0.1*10 =1 V

tan(theta) = XL/R

a) tan(theta) = 3/4,    theta = 36.87 deg
b) tan(theta) =1 ,      theta = 45 deg
c)  tan(theta) = 1/10,    theta = 5.71 deg
d) than( theta) = 10,     theta = 84.29 deg

21.18
a) XL= 2*pi*F*L   increseas with F
b)  ZT = R+i*XL   increases with F (because XL increases with F)
module(ZT) = sqrt(R^2+XL^2) increases with F

c) I = VT/ZT    decreases with F
d) V(R) = I*R  decreases with F (because I decreases with F)

e) V(L) = I*XL = VT*XL/ZT = VT*(2*pi*F*L)/(R+i*2*pi*F*L)  stays the same with F

d) tan(Theta) = XL/R = 2*pi*F* (L/R)  increases with F hence theta increases with F

21.20.

a and b) V(R) = V(L) = VA = 12 V because it is a parallel circuit

21.22

IR = VA/R = 12/30 = 0.4 A

IL = VA/XL = 12/40 = 0.3 A

IT = sqrt(IR^2+IL^2) = 0.5 A

module(Zeq) = Va/IT =12/0.5 = 24 ohms

Zeq = R*XL/(R+XL) = 30*40*i/(30+40*i)

tan(theta) = XL/R = 4/3

theta =53.13 deg

21.24

IR = VA/R =36/2K = 18 mA

IL = VA/XL = 36/1K = 36 mA

IT= sqrt(IR^2+IL^2) = 40.25 mA

module(Zeq) = VA/IT = 36/40.25 mA = 894 ohms

tan(theta) = XL/R = 0.5,    theta = 26.56 deg

21.26
IR = VA/R =24/10 =2.4 A

IL = VA/XL =24/4 =6 A

IT= sqrt(IR^2+IL^2) = 6.46 A

module(ZEq) = VA/IT = 24/6.46 = 3.71 ohms

tan(theta) = XL/R =2/5,    theta =21.8 degree

21.28.
IR= VA/R     IL= VA/XL   IT =sqrt(IR^2+IL^2)     module(ZEQ) = VA/IT     tan(theta) = XL/R

a) IR = 1A,   IL= 1A    IT =1.41 A    Zeq =50/1.41 = 35.46 ohms    tan(theta) = 1  theta = 45 deg
b) IR=2 A,   IL =0.2 A   IT =2.01 A   Zeq =9.95 ohms   tan(theta) = 10,   theta =84.29 deg
c) IR =0.2 A,   IL =2 A, IT =2.01 A   Zeq =9.95 ohms   tan(theta) = 0.1,  theta =  5.71 deg

21.30
XL= 2*pi*F*L =2*pi*3183*0.05 = 1000 ohm

IR = VA/R = 15/1.5K =10 mA

IL = VA/XL = 15/1K =15 mA

IT = sqrt(IR^2+IL^2) = 18.02 mA

Zeq = VA/IT = 15/18.02 mA = 832 ohm

tan(theta) = XL/R =1/1.5     theta =33.69 deg

21.32
a) IR = VA/R  stays the same with F

b) IL = VA/XL  increases if F decreases

c) IT = sqrt(IR^2+IL^2)  increases if F decreases

d) Zeq = V/IT   decreases if F decreases (because IT increases)
e) than(theta) = XL/R = 2*pi*F *(L/R) decreases if F decreases, hence theta decreases

21.34

By DEFINITION the quality factor Q = XL/R = omega*(L/R) therefore the increasing of the quality factor at higher

frequencies is limited by the resistance of the coil.

21.36

Q = XL/R = 2*pi*F*L/R

R = 2*pi*F*L/Q =2*pi*5*10^6*350*10^-6/25 =439.8 ohm