Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A 230 V, 50Hz , one pulse SCR controlled con...
Start Learning for Free
A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage is
- a)8.33μsec,84.84 V
- b)8.44m sec,83.38 V
- c)8.33 m sec 84.48 V
- d)8.33 sec, 84.48 V
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a ...
Given parameters:
- Voltage (V) = 230 V
- Frequency (f) = 50 Hz
- Firing angle (α) = 40°
- Extinction angle (β) = 210°
- Load resistance (R) = 5 Ω
- Load inductance (L) = 2 mH
To find:
- Circuit-turn off time
- Average output voltage
Solution:
1. Calculation of circuit-turn off time:
- Circuit-turn off time (tq) can be calculated using the following formula:
tq = (π - β + α) / (2πf)
- Substituting the given values in the formula, we get:
tq = (π - 210° + 40°) / (2π x 50 Hz)
tq = 8.33 ms
- Therefore, the circuit-turn off time is 8.33 ms.
2. Calculation of average output voltage:
- Average output voltage (Vo) can be calculated using the following formula:
Vo = (2V / π) x (cos α - cos β) - (2IR)
- Substituting the given values in the formula, we get:
Vo = (2 x 230 V / π) x (cos 40° - cos 210°) - (2 x 5 Ω x I)
Vo = 84.48 V - 10 I
- To find the value of I, we need to use the fact that the load current extinguishes at an angle of β = 210°. At this point, the load current is given by:
iL = Im sin (ωt + φ)
where Im = V / √(R^2 + ω^2L^2) = 45.67 A (rms value)
ω = 2πf = 314.16 rad/s
φ = tan^-1 (ωL / R) = 0.79°
- At β = 210°, we have:
iL = Im sin (ωβ + φ) = -36.64 A (rms value, negative sign indicates that the current is flowing in the opposite direction)
- Therefore, the value of I can be calculated as:
I = iL / √2 = -25.92 A (peak value, negative sign indicates that the current is flowing in the opposite direction)
- Substituting the value of I in the equation for Vo, we get:
Vo = 84.48 V - 10 x (-25.92 A)
Vo = 83.38 V
- Therefore, the average output voltage is 83.38 V.
Answer:
The circuit-turn off time is 8.33 ms and the average output voltage is 84.48 V. Therefore, the correct answer is option (c) 8.33 ms, 84.48 V.
- Voltage (V) = 230 V
- Frequency (f) = 50 Hz
- Firing angle (α) = 40°
- Extinction angle (β) = 210°
- Load resistance (R) = 5 Ω
- Load inductance (L) = 2 mH
To find:
- Circuit-turn off time
- Average output voltage
Solution:
1. Calculation of circuit-turn off time:
- Circuit-turn off time (tq) can be calculated using the following formula:
tq = (π - β + α) / (2πf)
- Substituting the given values in the formula, we get:
tq = (π - 210° + 40°) / (2π x 50 Hz)
tq = 8.33 ms
- Therefore, the circuit-turn off time is 8.33 ms.
2. Calculation of average output voltage:
- Average output voltage (Vo) can be calculated using the following formula:
Vo = (2V / π) x (cos α - cos β) - (2IR)
- Substituting the given values in the formula, we get:
Vo = (2 x 230 V / π) x (cos 40° - cos 210°) - (2 x 5 Ω x I)
Vo = 84.48 V - 10 I
- To find the value of I, we need to use the fact that the load current extinguishes at an angle of β = 210°. At this point, the load current is given by:
iL = Im sin (ωt + φ)
where Im = V / √(R^2 + ω^2L^2) = 45.67 A (rms value)
ω = 2πf = 314.16 rad/s
φ = tan^-1 (ωL / R) = 0.79°
- At β = 210°, we have:
iL = Im sin (ωβ + φ) = -36.64 A (rms value, negative sign indicates that the current is flowing in the opposite direction)
- Therefore, the value of I can be calculated as:
I = iL / √2 = -25.92 A (peak value, negative sign indicates that the current is flowing in the opposite direction)
- Substituting the value of I in the equation for Vo, we get:
Vo = 84.48 V - 10 x (-25.92 A)
Vo = 83.38 V
- Therefore, the average output voltage is 83.38 V.
Answer:
The circuit-turn off time is 8.33 ms and the average output voltage is 84.48 V. Therefore, the correct answer is option (c) 8.33 ms, 84.48 V.
Free Test
FREE
| Start Free Test |
Community Answer
A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a ...
Given: Vs=230V, ?=400 , β=2100, R=5Ω , L=2 mH
Circuit turn off time is time duration for which thyristor
is reverse bias.
For 1 pulse SCR controlled converter,
Circuit turn on time, 
Average output voltage
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer?
Question Description
A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect 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 A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect 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 A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer?.
A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect 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 A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect 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 A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 230 V, 50Hz , one pulse SCR controlled converter is triggered at a firing angle of 400 and the load current extinguishes at an angle of 2100 . For a load of R=5Ω and L=2mH. The circuit-turn off time and average output voltage isa)8.33μsec,84.84 Vb)8.44m sec,83.38 Vc)8.33 m sec 84.48 Vd)8.33 sec, 84.48 VCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup