Electrical Engineering (EE) Exam > Electrical Engineering (EE) Tests > Test: Sinusoidal Steady‐State Analysis - 2 - Electrical Engineering (EE) MCQ
Test: Sinusoidal Steady‐State Analysis - 2 - Electrical Engineering (EE) MCQ
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10 Questions MCQ Test - Test: Sinusoidal Steady‐State Analysis - 2
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Test: Sinusoidal Steady‐State Analysis - 2 - Question 1
The circuit is as shown in fig. P1.7.17-18
Q.i2(t) = ?
Q.i2(t) = ?
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Detailed Solution for Test: Sinusoidal Steady‐State Analysis - 2 - Question 2
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Detailed Solution for Test: Sinusoidal Steady‐State Analysis - 2 - Question 3
Test: Sinusoidal Steady‐State Analysis - 2 - Question 4
Determine the complex power for the given values in question.
Q. P = 269 W, Q = 150 VAR (capacitive)
Detailed Solution for Test: Sinusoidal Steady‐State Analysis - 2 - Question 4
Test: Sinusoidal Steady‐State Analysis - 2 - Question 5
Determine the complex power for the given values in question.
Q. Q = 2000 VAR, pf = 0.9 (leading)
Detailed Solution for Test: Sinusoidal Steady‐State Analysis - 2 - Question 5
Test: Sinusoidal Steady‐State Analysis - 2 - Question 6
Determine the complex power for the given values in question.
Q. S = 60 VA, Q = 45 VAR (inductive)
Detailed Solution for Test: Sinusoidal Steady‐State Analysis - 2 - Question 6
Test: Sinusoidal Steady‐State Analysis - 2 - Question 7
Determine the complex power for the given values in question.
Q. Vrms = 220 V, P = 1 kW, |Z| = 40 Ω (inductive)
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Test: Sinusoidal Steady‐State Analysis - 2 - Question 8
Determine the complex power for the given values in question.
Q. Vrms = 21∠ 200V, Vrms = 21∠200V, I rms = 8.5∠500A
Detailed Solution for Test: Sinusoidal Steady‐State Analysis - 2 - Question 8
Test: Sinusoidal Steady‐State Analysis - 2 - Question 9
Determine the complex power for the given values in question.
Q. Vrms = 120∠ 300V, Z = 40 + j 80Ω
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