Solution’s Manual for Analog Fundamentals A Systems Approach 1st Edition Thomas L. Floyd David M. Buchla (Complete And Verified Study material) (152pages) LEARNEXAMS

8. V pp = 0.354 = 0.354 = 9.90 V V rms 0.707 Vp 9. Vavg = 0.636 Vp = 1.11 10. The 5th hannonic is (5)(500Hz) = 2.5 kHz 11. Odd harmonics Section 1-3 12. For the circuit shown in Figure 1-25 (text): ( 5.6 kQ) Vrn= Vol = (12 V) 8.9 kQ = 7.55 V Rrn= 1.5 kQ + (5.6 kQ 113.3 kQ) = 3.58 kQ The equivalent circuit is shown in Figure 1-2. Rm = 3.58 kCl Vm=7.S5V~ T FIGURE 1-2 13. For the circuit in Figure 1-25 (text), Rrn = 3.58 kQ and Vrn = 7.55 V (see problem 12). For RL = 1.0 kQ: V L = (7.55 V)( 1508 kk~) = 1.65 V For RL = 2.7 kQ: VL = (7.55 V)(i.i8 kk~) = 3.25 V For RL = 1.0 kQ: VL = (7.55 V)( lt8 ~~) = 3.79 V 14. From problem 12, Rrn = 3.58 kQ and Vrn = 7.55 V RN = Rrn = 3.58 kQ Vrn 7.55 V IN= Rrn = 3.58 kQ = 2.11 rnA 15. The load line (for Figure 1-26 of the text) crosses the y-axis at: