Transcript Chapter 31-

Chapter 31--Examples
1
Problem
You want the current amplitude though a
0.45 mH inductor (part of the circuitry for
a radio receiver) to be 2.6 mA when a
sinusoidal voltage with amplitude of 12 V
is applied across the inductor. What is
the frequency required?
2
Calculate inductive reactance
XL=wL but don’t know w!
 V=IXL =IwL where





V=12
L=0.45 mH
I=2.6 mA
Solve for w=1.03 x 107 rad/s
But we want f and we know that w=2pf so
f=w/2p
 F=1.63 x 106 Hz or 1.63 MHz

3
Problem
You have a 200 W resistor, 0.400 H inductor and a 600
mF capacitor. Suppose you connect these
components in series with a voltage source that has
an amplitude of 30 V and an angular frequency of
250 rad/s.
a)
What is the impedance of the circuit?
b)
What is the current amplitude?
c)
What are voltage amplitudes across the resistor and
the inductor and the capacitor?
d)
What is the phase angle of the source voltage w.r.t.
to the current? Does the voltage lead or lag the
current?
e)
Construct a phasor diagram.
4
Impedance
Z  R 2  ( X L  X C )2
X L  wL  250 * 0.4  100W
1
1
 666W

XC 
wC 250 * 6e  6
X L  X C  566
Z  200 2  (566) 2
Z  600W
Note that XC is larger than XL thus the load is a capacitive load and
ELI the ICEman says that the current should lead voltage ( or voltage
Lags current)
5
Current Amplitude
V=IZ
 I=V/Z =30/600=0.05 mA

6
Voltages across resistor, inductor,
capacitor?
VR =IR=.05*200=10 V
 VL =IXL=.05*100=5 V
 VC =iXC =.05*666 = 33 V

7
Phase Angle
X L  X C  567
tan  

R
200

  70
Voltage lags current because  is negative!
8
Problem
A large 360 W, 5.2H electromagnetic coil is
connected across the terminals of a
source that has voltage amplitude 240
V and frequency of 60 Hz
a) What is the power factor?
b) What is the average power delivered by
the source?
9
Power Factor
Power factor = cos  =R/Z
 Z=sqrt(R^2+XL2)

XL=wL=2*p*60*5.2=1960 W
 Z=sqrt(360^2+1960^2)=1992


cos  =360/1992 =0.181
10
Average Power
Pav = ½ (V2 /Z)*cos 
 Pav = ½ (240^2/1992)*.181
 Pav=2.62 W

11
Problem
A transformer is connected to a 120 V (rms) ac line is to
supply 13000 V (rms) for a neon sign. To reduce the
shock hazard, a fuse is to be inserted in the primary
circuit; the fuse is to blow when the rms current in
the secondary exceeds 8.5 mA
a)
What is the ratio of secondary to primary turns of the
transformer?
b)
What power must be supplied to the transformer
when the rms secondary current is 8.5 mA?
c)
What current rating should the fuse in the primary
have?
12
Ratio of Turns= Ratio of Voltages

13000/120=108
13
Power In=Power Out

Pout=13000*8.5E-3=110.5 W

In the primary, P=Vi or 110.5=120*i

So i=0.920
14
Problem
An L-R-C series circuit has R=500 W, L=2
H, and C=0.5 mF and V=100 V. For
f=60Hz, calculate Z,VR, VL,VC and the
phase angle.
15
Solution
f=60 Hz, w=2*p*60=377 rad/s
 R=500
 XL=w*L=377*2=754
 XC= 1/(w*L)=1/(377*0.5E-6)=5305
 XL-XC=754-5305=-4551
 Based on XL-XC, we expect the load to
be more capacitive (ICE, Current leads
Voltage)

16
Solution Cont’d

Z=Sqrt(R2+(XL-XC)2)=sqrt(5002+(-4551)2)

Z=4578 W
I=V/Z=100/4578=21.8 mA
 VR=IR=(.021)*500=10.9 V
 VC=IXC=(.021)*5305=111 V
 VL=IXL=(.021)*754=16.4 V

17
Solution Cont’d
=tan-1((XL-XC)/R)=-83.70
 Since phase angle is nearly -90o, the
capacitor will actually dominate the
circuit.

18