# Test: Sinusoidal Steady State Analysis - Electrical Engineering (EE)

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## 20 Questions MCQ Test GATE Electrical Engineering (EE) 2024 Mock Test Series - Test: Sinusoidal Steady State Analysis

Test: Sinusoidal Steady State Analysis for Electrical Engineering (EE) 2023 is part of GATE Electrical Engineering (EE) 2024 Mock Test Series preparation. The Test: Sinusoidal Steady State Analysis questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Sinusoidal Steady State Analysis MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Sinusoidal Steady State Analysis below.
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Test: Sinusoidal Steady State Analysis - Question 1

### In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66sin (10 t) V, i(t) = 2.3sin (103t + 68.3o) A.The nature of the elements would be

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 1

R-Ccauses a positive phase shift in voltage Test: Sinusoidal Steady State Analysis - Question 2

### vo (t) = ? Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 2

10 sin (t + 30°) = 10 cos (t - 60°)   Test: Sinusoidal Steady State Analysis - Question 3

### Vo = ? Test: Sinusoidal Steady State Analysis - Question 4

The circuit is as shown in fig. i1( t) = ?

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 4    Test: Sinusoidal Steady State Analysis - Question 5

The circuit is as shown in fig i2(t) = ?

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 5 Test: Sinusoidal Steady State Analysis - Question 6

Ix = ? Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 6  Test: Sinusoidal Steady State Analysis - Question 7

Vx = ? Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 7

Let Vo be the voltage across current source Vo(20 + j10) - (20 + j40) Vx = j600    Test: Sinusoidal Steady State Analysis - Question 8

Determine the complex power for hte given values in question.

P = 269 W, Q = 150 VAR (capacitive)

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 8

S = P-jQ = 269-j150 VA

Test: Sinusoidal Steady State Analysis - Question 9

Determine the complex power for hte given valuesin question.

Q = 2000 VAR, pf =09. (leading)

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 9

pf = cos θ = 0.9 ⇒ θ = 25.84°
Q = S sin θ ⇒   Test: Sinusoidal Steady State Analysis - Question 10

Determine the complex power for hte given values in question.

S = 60 VA, Q = 45 VAR (inductive)

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 10

Q = S sin θ ⇒  Test: Sinusoidal Steady State Analysis - Question 11

Determine the complex power for hte given values in question.

Vrms = 220 V, P = 1 kW, |Z| = 40Ω (inductive)

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 11  = 0.8264 or θ = 34.26°, Test: Sinusoidal Steady State Analysis - Question 12

Determine the complex power for hte given values in question

Vrms = 21∠20°V, Vrms = 21∠20°V, Irms = 8.5∠-50°A

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 12

S = Vrms I*rms = (21∠20°)(8.5∠50°)

= 61+j167.7VA

Test: Sinusoidal Steady State Analysis - Question 13

Determine the complex power for hte given values in question.

Vrms = 120∠30°V, Z = 40 + j80Ω

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 13  = 72 + j144 VA

Test: Sinusoidal Steady State Analysis - Question 14

In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66 sin (1000t) V, i(t) = 2.3sin (1000t + 68.3) 3 A. The nature of the elements would be

Test: Sinusoidal Steady State Analysis - Question 15

A relay coil is connected to a 210 V, 50 Hz supply. If it has resistance of 30Ω and an inductance of 0.5 H, the apparent power is

Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 15

Z= 30 + j(0.5)(2π)(50) = 30 + j157, Apparent power = 275.6 VA

Test: Sinusoidal Steady State Analysis - Question 16

In the circuit shown in fig. power factor is Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 16 = 4 - j6 = 7.21∠ - 56.31°, pf = cos 56.31° = 0.555 leading

Test: Sinusoidal Steady State Analysis - Question 17

The power factor seen by the voltage source is Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 17   I1 = 1∠36.9°  pf = cos 36.9° = 0.8 leading

Test: Sinusoidal Steady State Analysis - Question 18

The average power supplied by the dependent source is Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 18

(2∠ - 90°)4.8 = -Ix (4.8 + j1.92) + 0.6Ix(8) Ix = 5∠0°,   Va = 0.6 x 5 x 8 = 24∠0°, Test: Sinusoidal Steady State Analysis - Question 19

In the circuit of fig. the maximum power absorbed by ZL is Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 19  Test: Sinusoidal Steady State Analysis - Question 20

The value of the load impedance, that would absorbs the maximum average power is Detailed Solution for Test: Sinusoidal Steady State Analysis - Question 20  ## GATE Electrical Engineering (EE) 2024 Mock Test Series

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