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An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by
- a)v= 622 sin (314t) Volts
i = 7.07 cos (314t-90°) Amps - b)v= 311 sin (314t) Volts
i = 14 .14 sin (314t - 90°) Amps - c)v= 311 sin (314t) Volts
i = 7.07 sin (314t - 90°) Amps - d)v=622 sin (314t) Volts
i = 14.14 cos (314 t - 90°) Amps
Correct answer is option 'C'. Can you explain this answer?
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An ac voltage of 220 V is applied to a pure inductance of 50 H. If the...
Max. value of current,
Assuming voltage as the reference phasor,
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An ac voltage of 220 V is applied to a pure inductance of 50 H. If the...
The instantaneous value of voltage and current in an inductor can be given by the following equations:
v = Vmax sin(ωt) Volts
i = Imax cos(ωt - φ) Amperes
Where:
Vmax is the maximum voltage amplitude
Imax is the maximum current amplitude
ω is the angular frequency
t is the time
φ is the phase angle between voltage and current
In this case, the given values are:
Vmax = 220 V
Imax = 5 A
ω = 2πf = 2π(50 Hz) = 314 rad/s
For the voltage:
v = Vmax sin(ωt) = 220 sin(314t) Volts
For the current:
i = Imax cos(ωt - φ) = 5 cos(314t - φ) Amperes
The phase angle φ can be determined using the relationship between voltage and current in an inductor:
φ = 90 degrees
Therefore, the instantaneous value of voltage and current will be:
v = 220 sin(314t) Volts
i = 5 cos(314t - 90) Amperes
v = Vmax sin(ωt) Volts
i = Imax cos(ωt - φ) Amperes
Where:
Vmax is the maximum voltage amplitude
Imax is the maximum current amplitude
ω is the angular frequency
t is the time
φ is the phase angle between voltage and current
In this case, the given values are:
Vmax = 220 V
Imax = 5 A
ω = 2πf = 2π(50 Hz) = 314 rad/s
For the voltage:
v = Vmax sin(ωt) = 220 sin(314t) Volts
For the current:
i = Imax cos(ωt - φ) = 5 cos(314t - φ) Amperes
The phase angle φ can be determined using the relationship between voltage and current in an inductor:
φ = 90 degrees
Therefore, the instantaneous value of voltage and current will be:
v = 220 sin(314t) Volts
i = 5 cos(314t - 90) Amperes
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An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer?
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An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer?.
An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer?.
Solutions for An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given bya)v= 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v= 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v= 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v=622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
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