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All questions of Sinusoidal Steady State Analysis for Electrical Engineering (EE) Exam
In the bridge shown in fig. P1.7.13, Z1 = 300Ω, Z2= ( 300 - 600) Ω, Z3 = (200 + 100)Ω. The Z4 at balance is
- a)400 + j 300Ω
- b)400 - j 300 Ω
- c)j 100 Ω
- d)-j 900 Ω
Correct answer is option 'B'. Can you explain this answer?
In the bridge shown in fig. P1.7.13, Z1 = 300Ω, Z2= ( 300 - 600) Ω, Z3 = (200 + 100)Ω. The Z4 at balance is
a)
400 + j 300Ω
b)
400 - j 300 Ω
c)
j 100 Ω
d)
-j 900 Ω
 | Yash Patel answered |
Z1Z4 = Z3Z2
300Z4 = (300-j600)(200+j100)
⇒ Z4 = 400 - j300
300Z4 = (300-j600)(200+j100)
⇒ Z4 = 400 - j300
In a balanced Y–connected three phase generator Vab = 400 Vrms
Q. If phase sequence is acb then phase voltage are- a)231∠00 , 231∠1200, 231∠2400
- b)231∠-300 , 231∠1500, 231∠2400
- c)231∠300 , 231∠1500, 231∠900
- d)231∠600 , 231∠1800, 231∠2400
Correct answer is option 'C'. Can you explain this answer?
In a balanced Y–connected three phase generator Vab = 400 Vrms
Q. If phase sequence is acb then phase voltage are
Q. If phase sequence is acb then phase voltage are
a)
231∠00 , 231∠1200, 231∠2400
b)
231∠-300 , 231∠1500, 231∠2400
c)
231∠300 , 231∠1500, 231∠900
d)
231∠600 , 231∠1800, 231∠2400
 | Ritika Sarkar answered |
For the acb sequence
Determine the complex power for the given values in question.
Q. P = 269 W, Q = 150 VAR (capacitive)- a)150 - j269 VA
- b)150 + j269 VA
- c)269 - j150 VA
- d)269 + j150 VA
Correct answer is option 'C'. Can you explain this answer?
Determine the complex power for the given values in question.
Q. P = 269 W, Q = 150 VAR (capacitive)
Q. P = 269 W, Q = 150 VAR (capacitive)
a)
150 - j269 VA
b)
150 + j269 VA
c)
269 - j150 VA
d)
269 + j150 VA
 | Harsh Kulkarni answered |
Complex Power Calculation:
Complex power is a combination of real power (P) and reactive power (Q) in an AC circuit. It is represented in the form S = P + jQ, where j is the imaginary unit.
Given Values:
- Real Power (P) = 269 W
- Reactive Power (Q) = 150 VAR (capacitive)
Calculation:
- The complex power (S) can be calculated as S = P + jQ
- Substituting the given values, S = 269 + j150 VAR
- Therefore, the complex power is 269 + j150 VAR
Final Answer:
- The correct representation of the complex power for the given values is 269 - j150 VAR (Option C).
A three-phase circuit has two parallel balanced Δ loads, one of the 6Ω resistor and one of 12 Ω resistors. The magnitude of the total line current, when the line-to-line voltage is 480 Vrms ,is- a)120 A rms
- b)360 A rms
- c)208 A rms
- d)470 A rms
Correct answer is option 'C'. Can you explain this answer?
A three-phase circuit has two parallel balanced Δ loads, one of the 6Ω resistor and one of 12 Ω resistors. The magnitude of the total line current, when the line-to-line voltage is 480 Vrms ,is
a)
120 A rms
b)
360 A rms
c)
208 A rms
d)
470 A rms
 | Saranya Mishra answered |
Z A = 6 ||12 = 4,
In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66 sin (103t) V, i(t) = 2.3 sin (103t + 68.3.0) A. The nature of the elements would be- a)R - C
- b)L - C
- c)R - L
- d)R - R
Correct answer is option 'A'. Can you explain this answer?
In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66 sin (103t) V, i(t) = 2.3 sin (103t + 68.3.0) A. The nature of the elements would be
a)
R - C
b)
L - C
c)
R - L
d)
R - R
 | Mahi Bose answered |
R - C causes a positive phase shift in voltage
In a balanced Y–connected three phase generator Vab = 400 Vrms
Q. If phase sequence is abc then phase voltage Va, Vb,and Vc are respectively- a)231∠00 , 231∠1200, 231∠2400
- b)231∠-300 , 231∠1500, 231∠900
- c)231∠300 , 231∠1500, 231∠-900
- d)231∠600 , 231∠1800, 231∠-600
Correct answer is option 'B'. Can you explain this answer?
In a balanced Y–connected three phase generator Vab = 400 Vrms
Q. If phase sequence is abc then phase voltage Va, Vb,and Vc are respectively
Q. If phase sequence is abc then phase voltage Va, Vb,and Vc are respectively
a)
231∠00 , 231∠1200, 231∠2400
b)
231∠-300 , 231∠1500, 231∠900
c)
231∠300 , 231∠1500, 231∠-900
d)
231∠600 , 231∠1800, 231∠-600
 | Kunal Sharma answered |
Vx = ?
- a)29.11∠1660 V
- b)29.11∠-1660V
- c)43.24∠1240 V
- d)43.24∠-1240 V
Correct answer is option 'B'. Can you explain this answer?
Vx = ?
a)
29.11∠1660 V
b)
29.11∠-1660V
c)
43.24∠1240 V
d)
43.24∠-1240 V
 | Anirban Gupta answered |
Let Vo be the voltage across current source
i(t) = ?
- a)20 cos ( 300 t + 68.20 ) A
- b)20 cos( 300 t - 68.20) A
- c)2.48 cos( 300 t + 68.20) A
- d)2.48 cos( 300 t - 68.20) A
Correct answer is option 'D'. Can you explain this answer?
i(t) = ?
a)
20 cos ( 300 t + 68.20 ) A
b)
20 cos( 300 t - 68.20) A
c)
2.48 cos( 300 t + 68.20) A
d)
2.48 cos( 300 t - 68.20) A
 | Divya Nair answered |
Z = 3 + j(25m)( 300) = 3 + j7.5 Ω = 8.08 68.20
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