Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN: 9781305632134
Author: J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher: Cengage Learning
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Textbook Question
Chapter 7, Problem 7.2MCQ
Even though the fault current is not symmetrical and not strictly periodic, the rms asymmetrical fault current is computed as the rms ac fault current times an "asymmetry factor," which is a function of _______.
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Chapter 7 Solutions
Power System Analysis and Design (MindTap Course List)
Ch. 7 - Even though the fault current is not symmetrical...Ch. 7 - The amplitude of the sinusoidal symmetrical ac...Ch. 7 - Equipment ratings for the four-bus power system...Ch. 7 - Equipment ratings for the five-bus power system...Ch. 7 - Prob. 7.22PCh. 7 - A three-phase circuit breaker has a 15.5-kV rated...Ch. 7 - Prob. 7.32P
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- Consider the circuit diagram below. Compute a single equivalent impedance for this circuit for a source frequency of F = 60 Hz. Express your final answer as a complex impedance with rectangular coordinates. You must show your all your work for the complex math. Include a diagram of the equivalent circuit as part of your solution.arrow_forwardConsider the circuit diagram below. Compute a single equivalent impedance for this circuit for a source frequency of f = 165 Hz. Express your final answer as a phasor with polar coordinates. You must show your all your work for the complex math. Include a diagram of the equivalent circuit as part of your solution.arrow_forwardConsider the circuit diagram below. Using mesh analysis, compute the currents (a) IR1, (b) IL1, and (c) IC1. Express your final answers as phasors using polar coordinates with phase angles measured in degrees. Your solution should include the circuit diagram redrawn to indicate these currents and their directions. You must solve the system of equations using MATLAB and include the code or commands you ran as part of your solution.arrow_forward
- use kvl to solvearrow_forwardR1 is 978 ohms R2 is 2150 ohms R3 is 4780 R1 is parallel to R2 and R2 is parallel to R3 and R1 and R3 are in seriesarrow_forwardQ7 For the circuit shown in Fig. 2.20, the transistors are identical and have the following parameters: hfe = 50, hie = 1.1K, hre = 0, and hoe = 0. Calculate Auf, Rif and Rof. Ans: 45.4; 112 KQ; 129. 25 V 10k 47k 4.7k Vo 150k w Vs 47k 4.7k W 22 5μF 33k 50uF 50μF 4.7k 4.7k R₁ Rof Rif R1000 Fig. 2.20 Circuit for Q7.arrow_forward
- Q6)) The transistors in the feedback amplifier shown are identical, and their h-parameters are.. hie = 1.1k, hfe = 50, hre=o, and hoe = 0. Calculate Auf, Rif and Rof. {Ans: 6031583; 4. Kor. Is 4 4.7 k www 4.7k 91k 4.7k 91k 10k 1k. 10k 21000 4.7k w 15k Fig. 2.19 Circuit for Q6.arrow_forwardQ5 For the circuit shown in Fig. 2.18, hie =1.1 KQ, hfe =50. Find Avf, Rif and Rof Ans: -3.2; 193 ; 728 N. Vcc Vs Rs=10kQ Re=4KQ RF - = 40ΚΩ www Fig. 2.18 Circuit for Qs.arrow_forwardSheet No.2 Qi For the source follower shown in Fig. 2.14, Ipss =16 mA, V₂ =-4V, and VGsQ=-2.86 V. Find Avf, Rif and Rof. Assume rd is high. Ans: 0.833; ∞0; 365.7 . VDD Vo Vs R = 2.2 k Fig. 2.14 Circuit for Qi.arrow_forward
- Q4 For the circuit shown in Fig. 2.17, he-100, he -1KQ. Find A, A, R and Rof- Ans:-100; -5; 100 K; 250K. Voc RB = 100 k R.=5k Vs Rs 500 R. = 1 kn Fig. 2.17 Circuit for Quarrow_forwardQ3 The circuit of Fig. 2.16 is to have Af = -1 mA/V, D=1+ BA=50, a voltage gain of -4, Rs = 1KQ, and hfe = 150. Find RL, Re, Rif and Rof. Ans: 4 KN; 980 ; 150 KN; ∞. Vcc RL Vs -OV +11 Fig. 2.16 Circuit for Q3.arrow_forwardQ2 For the circuit shown in Fig. 2.15 hfe =150, hie =1KQ. Find Avf and Rif. Ans: 0.986; 152 KN. Vee R=4k2 Rs 1kQ Vo V, VR=1 KQ Fig. 2.15 Circuit for Q2-arrow_forward
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