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CHAPTER 11 ERRATA
Text
Examples and Practice Problems
P.P.11.4
The value of the power absorbed by the 40-V source should be –60W and
the value of the power absorbed by the j20-V source should be –40W.
Example 11.6
The equation for the current is incorrect, the value of the resistance
in the denominator needs to be 33.66 (24.25 + 9.412). The angle of I
should be 100.42˚.
P.P.11.6
In the answer, the maximum average power absorbed is actually 6.863W.
P.P.11.12
The correct answer for apparent power delivered: + 10.35 kVA
In addition, the units of Reactive power should be kVAR, not kVA.
P.P.11.13
It needs to be pointed out that the indicated answers, assume that the
current through the 60-ohm resistor has no phase shift. We cannot
uniquely determine the phase angle associated with this current. Thus, the
most correct answer is:
V = 240.7(21.45˚+φ) V
where φ is the angle of the current through the 60-ohm resistor.
Problem
Problem 11.47 (d)
The units on I must be A not V.
Problem 11.64
The value for the term in the middle box should be 2 kW at a
leading power factor of 0.707.
Problem 11.64
The values for the resistor and the inductor should be as shown
below.
8
+

Is
1200º V
j12Ω
Problem 11.92
The value of the lighting load on the right side of the circuit should
be 20 kW.
Problem 11.97
The value of the inductor on the right side of the circuit should be
j20 ohms.