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Chapter 31--Examples

Chapter 31--Examples. 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?. Calculate inductive reactance.

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Chapter 31--Examples

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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 a sinusoidal voltage with amplitude of 12 V is applied across the inductor. What is the frequency required?

  3. 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

  4. 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. • What is the impedance of the circuit? • What is the current amplitude? • What are voltage amplitudes across the resistor and the inductor and the capacitor? • What is the phase angle of the source voltage w.r.t. to the current? Does the voltage lead or lag the current? • Construct a phasor diagram.

  5. Impedance 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)

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

  7. Voltages across resistor, inductor, capacitor? • VR =IR=.05*200=10 V • VL =IXL=.05*100=5 V • VC =iXC =.05*666 = 33 V

  8. Phase Angle Voltage lags current because f is negative!

  9. 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 • What is the power factor? • What is the average power delivered by the source?

  10. Power Factor • Power factor = cos f =R/Z • Z=sqrt(R^2+XL2) • XL=wL=2*p*60*5.2=1960 W • Z=sqrt(360^2+1960^2)=1992 • cos f =360/1992 =0.181

  11. Average Power • Pav = ½ (V2 /Z)*cos f • Pav = ½ (240^2/1992)*.181 • Pav=2.62 W

  12. 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 • What is the ratio of secondary to primary turns of the transformer? • What power must be supplied to the transformer when the rms secondary current is 8.5 mA? • What current rating should the fuse in the primary have?

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

  14. 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

  15. 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.

  16. 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)

  17. 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

  18. Solution Cont’d • f=tan-1((XL-XC)/R)=-83.70 • Since phase angle is nearly -90o, the capacitor will actually dominate the circuit.

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