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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 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 |
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 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
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,
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
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
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