# EE115. Chapter 19

19.4 a) 3A/2s = 1.5 A/s

b) 50 mA/5 microSec = 50*10^-3/5*10^-6 A/s = 10000 A/s

c) (150-100) mA/5 ms =50 mA/5ms =10 A/s

d) (100-150) mA/20 microSec = -50*10^-3/20*10^-6 = 2500 A/s

e) (35-30) mA/1 microSec =5*10^-3/10^-6 =5000 A/s

f) (96-80) mA/0.4 microS = 16*10^-3/4*10^-7 = 4*10^4= 40000 A/s

g) (11-10) A/ 1 s =1 A/s

19.6

U = L*dI/dt

75 =L*2500

L =0.03 H

19.8

L = miu*N^2*S/l N is the number of turns, miu(air) =4*pi*10^-7 H/m. l is the length

a) L(a) = L =4*pi*10^-7*20^2*3.14*10^-4/25*10^-2 =6.314*10^-7 H =631.4 nH

b) L = L(a)*5000 = 3157 microH =3.157 mH

c) L= L(a)*(200/20)^2 =631.4*100 nH =63140 nH =63.14 microH

d) L = L(a)/2 = 315.7 nH

e) L = 4*pi*10^-7*2000*100^2*5*10^-4/10^-2 = 1.256 H

19.10

Another name is RadioFrequency coil (for smaller inductances), or RadioFrequency choke (for higher inductances)

19.12.

U = L*dI/dt = 33*10^-3*1500 = 49.5 V

19.14.

k = fi1/fi2 = 50 microWb/200 microWb = 1/4 =0.25

19.16

LM = k*L1 =0.6*50 mH = 30 mH

19.18

For an iron core transformer k is in the interval, k = 0.95-0.998

19.20

a) secondary voltage Vs =Vp/10 = 120/10 =12 V

c) secondary power Ps = Vs^2/R = 12*12/12 =12 W

b) secondary current Is = Ps/Vs =12/12 =1 A

d) Primary power Ppri= Ps= 12 W for an ideal transformer

For a not ideal transformer Ppri = 1.2*Ps =1.2*12 =14.4 W

e) Ipri = Ppri/Vpri = 12/120 =0.1 A for an ideal transformer

19.22

We assume ideal transformer

Vs2 = Vpri*Ns2/Npri = 120*50/250 = 24 V

Ps2 = Vs2^2/R2 =24*24/24 =24 W

Ppri =24 W

Ipri = Ppri/Vpri = 24/120 =0.2 A

19.24

Vs = Vpri* Ns/Npri

Np/Ns = Vp/Vs

a) 120/60 =2

b)120/600 =0.2

c)120/420 =0.286

d)120/24 = 5

e)120/12.6 =9.524

19.26

Having an isolated secondary has as primary advantage a lower risk or electric choke. This is because allways the

transformers are designed for a maximum power hence, when a short circuit happens in the secondary there is a certain

limited maximum value of the current.

19.28.

To avoid the overloding of a transformer the following two rules are valid.

1. Never subject the secondary to a power higher that the power for wich it was designed.

It means the load resistance of the secondary should have a certain lower limit.

2. Allways take in the computation of the primary power a power higher than the seondary one with a

factor higher or equal to 1.2. Ppri = 1.2*Psec

19.30

The figure 19.15 necessary for this problem is missing.

19.32

Vab =12.6 V

Vac = Vab/2 =6.3 V

Vbc = Vab/2 =6.3 V

19.34.

The percentage voltage regulation = (voltage on no load-voltage on load)*100/voltage on no load

It is equal to 4% for usual transformers.

Therefore on no load

Vab =104%*12.6 =13.104 V

Vac = 104%*6.3 =6.552 V

Vbc= Vac =6.552 V

19.36.

(Zp/Zs) = (Np/Ns)^2

In the figure for maximum transfer of power in the secondary Zs=RL

Np/Ns = sqrt(Zp/RL)

a) 11.547

b) 2

c) 0.1

d) 2.23

e) 0.289

f) 4.47

19.38

answer to problem 19.37 is Np/Ns = sqrt(Zp/Zs) = sqrt(r/RL) =sqrt(500/4) = 11.18

a) Zp= ri = 500 ohms (for maximum power transfer in the primary)

c) Ppri = V^2/Zp = 100*100/500 = 20 W

b) Psec = Pri = 20 W for ideal transformer

Psec = Ppri/1.2 =16.67 W

19.40

1/L = (1/L1)+(1/L2)

a) 3.75 mH

b) 4 mH

c) 102.06 microH

d) 280.37 microH

19.42.

L = L1+ L2 -2*LM =100+300-2*130 = 140 mH

19.44.

L1+L1 -2*LM = 100 mH

L1+L1 +2*LM = 300 mH

substracting we have

4*LM =200 mH

LM =50 mH

k = LM/L1 =50/100 =0.5

19.46.

Energy E =L*I^2/2 =5*0.2^2/2 = 0.1 J

19.48

Z =omega*L = omega*miu*N^2*S/l

L is the inductance, omega the angular speed of the voltage.

omega, miu, S and l are the same for primary and secondary.

Zp/Zs = Np^2/Ns^2 = (NP/Ns)^2

19.50

Zp/Zs = (Np/Ns)^2

a) Zp/Zs =4^2 =16

Zp = 16*Zs =16*RL=16*50 = 800 ohm

b) Zp/Zs =2^2 =4

Zp = 4*Zs= 4*Rl =4*50 =200 ohm