Chapter 9 Problems with Solution - Circuits I | EEE 202, Assignments of Microelectronic Circuits

Download Chapter 9 Problems with Solution - Circuits I | EEE 202 and more Assignments Microelectronic Circuits in PDF only on Docsity! 73 CHAPTER 9 PROBLEMS 9.1 Determine the average power supplied by each source in the circuit in Fig. 9.1. 1Ω +- j1Ω-j1Ω V010 °∠ A302 °∠ Fig. 9.1 9.2 Given the circuit in Fig. 9.2, determine the impedance ZL for maximum average power transfer and the value of the maximum average power transferred to this load. V06 °∠ 1Ω 1Ω -j1Ω V012 °∠ LZ+ +- - Fig. 9.2 9.3 Calculate the rms value of the waveform shown in Fig. 9.3. 1 t(s)765432 6 v(t) (V) Fig. 9.3 9.4 Determine the source voltage in the network shown in Fig. 9.4. SV rmsV0240 °∠ 40 kW 0.78 pf lagging 60 kW 0.85 pf lagging +- + - 0.1Ω j0.5Ω Fig. 9.4 9.5 A plant consumes 75 kW at a power factor of 0.70 lagging from a 240-V rms 60 Hz line. Determine the value of the capacitor that when placed in parallel with the load will change the load power factor to 0.9 lagging. 74 CHAPTER 9 SOLUTIONS 9.1 Because the series impedance of the inductor and capacitor are equal in magnitude and opposite in sign, from the standpoint of calculating average power the network can be reduced to that shown in Fig. S9.1. CSI A302 °∠ + V010 °∠ - +- 1Ω VSI 1V Fig. S9.1 The general expression for average power is ( )IVcosVI2 1P θ−θ= In the case of the current source V1 = 10V, ICS = 2A, θV = 0° and θI = 30°. Therefore, the average power delivered by the current source is ( ) ( ) ( ) W66.8 30cos210 2 1PCS = °−      = In order to calculate the average power delivered by the voltage source, we need the current IVS. Using KCL °∠==°∠+ 010 1 302 1VS VI or IVS = 8.33∠-6.9°A Now ( ) ( ) ( )( ) W34.41 9.60cos33.810 2 1PVS = °−−°= Therefore, the total power generated in the network is PT = PCS + PVS = 50 W 77 Hence, for maximum average power transfer * THL ZZ = or Ω+= 2 1j 2 1 LZ Therefore, the network is reduced to that shown in Fig. S9.2(c). Ι +-V56.7149.9 °∠ 2 1 2 1 2 1j+ 2 1j− Fig. S9.2(c) Then A56.7149.9 2 1j 2 1 2 1j 2 1 56.7149.9 °∠= ++− °∠ =I and the maximum average power transferred to the load is ( ) W90 2 149.9 2 1P 2L =       = 9.3 In order to calculate the rms value of the waveform, we need the equations for the waveform within each of the distinctive intervals. In the interval 0 ≤ t ≤ 2s, the waveform is a straight line that passes through the origin of the graph. The equation for a straight line in this graph is v(t) = mt + b 78 Where m is the slope of the line and b is the v(t) intercept. Since the line goes through the origin, b = 0. The slope m is 3 s2 V6m == Therefore, in the interval 0 ≤ t ≤ 2s, v(t) = 3t The waveform has constant values in the intervals 2 ≤ t ≤ 3s and 3 ≤ t ≤ 4s, i.e., v(t) = 6 2 ≤ t ≤ 3s v(t) = 0 3 ≤ t ≤ 4s Since the waveform repeats after 4s, the period of the waveform is T = 4s Now that the data for the waveform is known, ( ) 2 14 0 2 rms dttvT 1V     ∫= Therefore, in this case ( ) ( ) ( ) [ ] ( ) ( ) rmsV87.3 15 3624 4 1 t36t3 4 1 dt0dt6dtt3 4 1V 2 1 2 1 2 1 3 2 2 0 3 2 14 3 2 3 2 2 2 0 2 rms = =     +=     +=         ∫+∫+∫= 9.4 We begin our analysis by labeling the various currents and voltages in the circuit as shown in Fig. S9.4. SV 40 kW 0.78 pf lag 60 kW 0.85 pf lag +- + - 0.1Ω j0.5Ω SI 1I 2I rmsV0240 °∠=LV Fig. S9.4 79 Our approach to determining VS is straight forward: We will compute the currents I1 and I2; add them using KCL to find IS; determine the voltage across the line impedance and finally use KVL to add the line voltage and load voltage to determine the source voltage. The magnitude of the current I1 is ( ) ( ) ( ) .rmsA12.294 85.0240 000,60 pf P 1 1 = = = L 1 V I And the phase angle is ( ) °−= −=θ − 79.31 85.0cos 1I1 The negative sign is a result of the fact that the power factor is lagging. Thus I1 = 294.12∠-31.79° A rms. The magnitude of the current I2 is ( ) ( ) ( ) .rmsA68.213 78.0240 000,40 pf P 2 2 = = = L 2 V I And the phase angle is ( ) °−= −=θ − 74.38 78.0cos 1I2 Thus I2 = 213.68∠-38.74° A rms. Using KCL