6th Edition

ISBN: 9781260104639

Author: Alexander

Publisher: MCG CUSTOM

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Alternating Current (AC) is a type of electric power in which the electric current changes, both in magnitude and direction, at regular intervals. This AC Circuits can deliver power in either a Single Phase system or a Three Phase system.

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Textbook Question

Chapter 12.10, Problem 14PP

Let the line voltage V_L = 208 V and the wattmeter readings of the balanced system in Fig. 12.35 be P₁ = −560 W and P₂ = 800 W. Determine:

(a) the total average power
(b) the total reactive power
(c) the power factor
(d) the phase impedance

Is the impedance inductive or capacitive?

Answer: (a) 240 W, (b) 2.356 kV AR, (c) 0.1014. (d) 18.25 Chapter 12.10, Problem 14PP, Let the line voltage VL = 208 V and the wattmeter readings of the balanced system in Fig. 12.35 be , example 1 , inductive.

Chapter 12.10, Problem 14PP, Let the line voltage VL = 208 V and the wattmeter readings of the balanced system in Fig. 12.35 be , example 2

Figure 12.35

Two-wattmeter method applied to a balanced wye load.

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Chapter 12.10, Problem 13PP

Chapter 12.10, Problem 15PP

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Chapter 12 Solutions

FUNDAMENTALS OF ELECTRIC...(LL)>CUSTOM<

Ch. 12.2 - Given that Vbn=22030V, find Van and Vcn, assuming...Ch. 12.3 - A Y-connected balanced three-phase generator with...Ch. 12.4 - One line voltage of a balanced Y-connected source...Ch. 12.5 - A positive-sequence, balanced -connected source...Ch. 12.6 - In a balanced -Y circuit, Vab=44015 and ZY = (12 +...Ch. 12.7 - For the Y-Y circuit in Practice Prob. 12.2,...Ch. 12.7 - Calculate the line current required for a 30-kW...Ch. 12.7 - Assume that the two balanced loads in Fig....Ch. 12.8 - The unbalanced -load of Fig. 12.24 is supplied by...Ch. 12.8 - Find the line currents in the unbalanced...

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Solution Summary: The author analyzes the two-wattmeter method applied to a balanced wye load.

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Chapter 12 Solutions