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1 Chapter 31--Examples. 2 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.

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Presentation on theme: "1 Chapter 31--Examples. 2 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."— Presentation transcript:

1 1 Chapter 31--Examples

2 2 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?

3 3 Calculate inductive reactance X L =  L but don’t know  ! V=IX L =I  L where V=12 L=0.45 mH I=2.6 mA Solve for  =1.03 x 10 7 rad/s But we want f and we know that  =2  f so f=  2  F=1.63 x 10 6 Hz or 1.63 MHz

4 4 Problem You have a 200  resistor, H inductor and a 600  F 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.

5 5 Impedance Note that X C is larger than X L thus the load is a capacitive load and ELI the ICEman says that the current should lead voltage ( or voltage Lags current)

6 6 Current Amplitude V=IZ I=V/Z =30/600=0.05 mA

7 7 Voltages across resistor, inductor, capacitor? V R =IR=.05*200=10 V V L =IX L =.05*100=5 V V C =iX C =.05*666 = 33 V

8 8 Phase Angle Voltage lags current because  is negative!

9 9 Problem A large 360 , 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?

10 10 Power Factor Power factor = cos  =R/Z Z=sqrt(R^2+X L 2 ) X L =  L=2*  *60*5.2=1960  Z=sqrt(360^2+1960^2)=1992 cos  =360/1992 =0.181

11 11 Average Power P av = ½ (V 2 /Z)*cos  P av = ½ (240^2/1992)*.181 P av =2.62 W

12 12 Problem A transformer is connected to a 120 V (rms) ac line is to supply 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?

13 13 Ratio of Turns= Ratio of Voltages 13000/120=108

14 14 Power In=Power Out P out =13000*8.5E-3=110.5 W In the primary, P=Vi or 110.5=120*i So i=0.920

15 15 Problem An L-R-C series circuit has R=500 , L=2 H, and C=0.5  F and V=100 V. For f=60Hz, calculate Z,V R, V L,V C and the phase angle.

16 16 Solution f=60 Hz,  =2*  *60=377 rad/s R=500 X L =  *L=377*2=754 X C = 1/(  *L)=1/(377*0.5E-6)=5305 X L -X C = =-4551 Based on X L -X C, we expect the load to be more capacitive (ICE, Current leads Voltage)

17 17 Solution Cont’d Z=Sqrt(R 2 +(X L -X C ) 2 )=sqrt( (-4551) 2 ) Z=4578  I=V/Z=100/4578=21.8 mA V R =IR=(.021)*500=10.9 V V C =IX C =(.021)*5305=111 V V L =IX L =(.021)*754=16.4 V

18 18 Solution Cont’d  =tan -1 ((X L -X C )/R)= Since phase angle is nearly -90 o, the capacitor will actually dominate the circuit.


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