EBK ELECTRIC CIRCUITS
EBK ELECTRIC CIRCUITS
10th Edition
ISBN: 8220100801792
Author: Riedel
Publisher: YUZU
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Question
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Chapter 11, Problem 1P

a)

To determine

Find the phase sequence for the given set of voltages.

a)

Expert Solution
Check Mark

Answer to Problem 1P

The phase sequence is abc sequence.

Explanation of Solution

Given data:

Consider the set of voltages.

va=137cos(ωt+63°)V        (1)

vb=137cos(ωt57°)V        (2)

vc=137cos(ωt+183°)V        (3)

Calculation:

Convert the voltage in Equation (1) from cosine voltage form to phasor form.

Va=13763°V{Vmcos(ωt+θ)=Vmθ}

Convert the voltage in Equation (2) from cosine voltage form to phasor form.

Vb=13757°V

Convert the voltage in Equation (3) from cosine voltage form to phasor form.

Vc=137183°V

Subtract the phase angle of a-phase from all the phase angles of Va, Vb, and Vc.

Va=63°63°

Va=0°        (4)

Vb=57°63°

Vb=120°        (5)

And

Vc=183°63°

Vc=120°        (6)

From Equations (4), (5), and (6), the voltage at phase-b lags a-phase voltage by 120° and c-phase voltage leads a-phase voltage by 120°. Therefore, this phase relationship is abc phase sequence or positive phase sequence.

Conclusion:

Thus, the phase sequence is abc sequence.

b)

To determine

Find the phase sequence for the given set of voltages.

b)

Expert Solution
Check Mark

Answer to Problem 1P

The phase sequence is acb sequence.

Explanation of Solution

Given data:

Consider the set of voltages.

va=820cos(ωt36°)V        (7)

vb=820cos(ωt+84°)V        (8)

vc=820sin(ωt66°)V

vc=820cos(ωt156°)V        (9)

Calculation:

Convert the voltage in Equation (7) from cosine voltage form to phasor form.

Va=82036°V{Vmcos(ωt+θ)=Vmθ}

Convert the voltage in Equation (8) from cosine voltage form to phasor form.

Vb=82084°V

Convert the voltage in Equation (9) from cosine voltage form to phasor form.

Vc=82066°V

Subtract the phase angle of a-phase from all the phase angles of Va, Vb, and Vc.

Va=36°(36°)

Va=0°        (10)

Vb=84°(36°)

Vb=120°        (11)

And

Vc=156°(36°)

Vc=120°        (12)

From Equations (10), (11), and (12), the voltage at phase-b leads a-phase voltage by 120° and c-phase voltage lags a-phase voltage by 120°. Therefore, this phase relationship is acb phase sequence or negative phase sequence.

Conclusion:

Thus, the phase sequence is acb sequence.

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Chapter 11 Solutions

EBK ELECTRIC CIRCUITS

Ch. 11 - Prob. 4PCh. 11 - Repeat Problem 11.4 but assume that the...Ch. 11 - Prob. 6PCh. 11 - Find I0 in the circuit in Fig. P11.7. Find...Ch. 11 - Find the rms value of Io in the unbalanced...Ch. 11 - The time-domain expressions for three...Ch. 11 - Prob. 10PCh. 11 - The magnitude of the line voltage at the terminals...Ch. 11 - A balanced Δ-connected load has an impedance of...Ch. 11 - A balanced, three-phase circuit is characterized...Ch. 11 - Prob. 15PCh. 11 - In a balanced three-phase system, the source is a...Ch. 11 - Prob. 17PCh. 11 - The impedance Z in the balanced three-phase...Ch. 11 - For the circuit shown in Fig. P11.20, find the...Ch. 11 - A balanced three-phase Δ-connected source is shown...Ch. 11 - Prob. 22PCh. 11 - Fine the rms magnitude and the phase angle of ICA...Ch. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - The line-to-neutral voltage at the terminals of...Ch. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Calculate the complex power in each phase of the...Ch. 11 - Prob. 31PCh. 11 - Prob. 32PCh. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Three balanced three-phase loads are connected in...Ch. 11 - Prob. 37PCh. 11 - Prob. 38PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - Prob. 42PCh. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Prob. 47PCh. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Prob. 50PCh. 11 - Prob. 51PCh. 11 - Find the reading of each wattmeter in the circuit...Ch. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55PCh. 11 - Prob. 56PCh. 11 - Prob. 57PCh. 11 - Prob. 58PCh. 11 - Prob. 59PCh. 11 - Assume in Problem 11.59 that when the load drops...
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