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Test: Sinusoidal Steady State Analysis


20 Questions MCQ Test RRB JE for Electrical Engineering | Test: Sinusoidal Steady State Analysis


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Attempt Test: Sinusoidal Steady State Analysis | 20 questions in 60 minutes | Mock test for Railways preparation | Free important questions MCQ to study RRB JE for Electrical Engineering for Railways Exam | Download free PDF with solutions
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

Solution:

R-Ccauses a positive phase shift in voltage

QUESTION: 2

vo (t) = ?

Solution:

10 sin (t + 30°) = 10 cos (t - 60°)


QUESTION: 3

Vo = ?

Solution:
QUESTION: 4

The circuit is as shown in fig.

i1( t) = ?
Solution:




QUESTION: 5

The circuit is as shown in fig

i2(t) = ?
Solution:

QUESTION: 6

Ix = ?
Solution:


QUESTION: 7

Vx = ?
Solution:

Let Vo be the voltage across current source

Vo(20 + j10) - (20 + j40) Vx = j600



QUESTION: 8

Determine the complex power for hte given values in question.

P = 269 W, Q = 150 VAR (capacitive)
Solution:

S = P-jQ = 269-j150 VA

QUESTION: 9

Determine the complex power for hte given valuesin question.

Q = 2000 VAR, pf =09. (leading)
Solution:

pf = cos θ = 0.9 ⇒ θ = 25.84°
Q = S sin θ ⇒ 

QUESTION: 10

Determine the complex power for hte given values in question.

S = 60 VA, Q = 45 VAR (inductive)
Solution:

Q = S sin θ ⇒ 

QUESTION: 11

Determine the complex power for hte given values in question.

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


 = 0.8264 or θ = 34.26°,

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
Solution:

S = Vrms I*rms = (21∠20°)(8.5∠50°)

= 61+j167.7VA

QUESTION: 13

Determine the complex power for hte given values in question.

Vrms = 120∠30°V, Z = 40 + j80Ω
Solution:

= 72 + j144 VA

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
Solution:
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
Solution:

Z= 30 + j(0.5)(2π)(50) = 30 + j157,

Apparent power  = 275.6 VA 

QUESTION: 16

In the circuit shown in fig. power factor is
Solution:


= 4 - j6 = 7.21∠ - 56.31°, pf = cos 56.31° = 0.555 leading

QUESTION: 17

The power factor seen by the voltage source is
Solution:

 

I1 = 1∠36.9°

 
pf = cos 36.9° = 0.8 leading

QUESTION: 18

The average power supplied by the dependent source is
Solution:

(2∠ - 90°)4.8 = -Ix (4.8 + j1.92) + 0.6Ix(8)

Ix = 5∠0°,   Va = 0.6 x 5 x 8 = 24∠0°, 

QUESTION: 19

In the circuit of fig. the maximum power absorbed by ZL is
Solution:


QUESTION: 20

The value of the load impedance, that would absorbs the maximum average power is
Solution:

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