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A 3 phase bridge inverter delivers power to a resistive load 400 V DC source. For a start connected load of 10 Ω/phase. Determine load power (in watt) 1200 conduction mode.
Correct answer is between '7990,8010'. Can you explain this answer?
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A 3 phase bridge inverter delivers power to a resistive load 400 V DC ...
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A 3 phase bridge inverter delivers power to a resistive load 400 V DC ...
Ohms, determine:
a) The RMS value of the output voltage
b) The peak value of the output voltage
c) The average power delivered to the load
a) The RMS value of the output voltage can be calculated using the formula:
Vrms = Vp / √2
where Vp is the peak value of the output voltage. Since the load is resistive, the output voltage waveform will be a square wave. Therefore, the peak value can be calculated as:
Vp = Vdc / √3
where Vdc is the DC voltage source. Substituting the values, we get:
Vp = 400 / √3 = 230.94 V
Therefore, the RMS value of the output voltage is:
Vrms = 230.94 / √2 = 163.38 V
b) The peak value of the output voltage is already calculated in part (a) as 230.94 V.
c) The average power delivered to the load can be calculated using the formula:
Pavg = Vrms^2 / R
where R is the load resistance. Substituting the values, we get:
Pavg = 163.38^2 / 10 = 2679.53 W
Therefore, the average power delivered to the load is 2679.53 W.
a) The RMS value of the output voltage
b) The peak value of the output voltage
c) The average power delivered to the load
a) The RMS value of the output voltage can be calculated using the formula:
Vrms = Vp / √2
where Vp is the peak value of the output voltage. Since the load is resistive, the output voltage waveform will be a square wave. Therefore, the peak value can be calculated as:
Vp = Vdc / √3
where Vdc is the DC voltage source. Substituting the values, we get:
Vp = 400 / √3 = 230.94 V
Therefore, the RMS value of the output voltage is:
Vrms = 230.94 / √2 = 163.38 V
b) The peak value of the output voltage is already calculated in part (a) as 230.94 V.
c) The average power delivered to the load can be calculated using the formula:
Pavg = Vrms^2 / R
where R is the load resistance. Substituting the values, we get:
Pavg = 163.38^2 / 10 = 2679.53 W
Therefore, the average power delivered to the load is 2679.53 W.
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A 3 phase bridge inverter delivers power to a resistive load 400 V DC ...
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A 3 phase bridge inverter delivers power to a resistive load 400 V DC source. For a start connected load of 10 Ω/phase. Determine load power (in watt) 1200 conduction mode.Correct answer is between '7990,8010'. Can you explain this answer?
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A 3 phase bridge inverter delivers power to a resistive load 400 V DC source. For a start connected load of 10 Ω/phase. Determine load power (in watt) 1200 conduction mode.Correct answer is between '7990,8010'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 3 phase bridge inverter delivers power to a resistive load 400 V DC source. For a start connected load of 10 Ω/phase. Determine load power (in watt) 1200 conduction mode.Correct answer is between '7990,8010'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 3 phase bridge inverter delivers power to a resistive load 400 V DC source. For a start connected load of 10 Ω/phase. Determine load power (in watt) 1200 conduction mode.Correct answer is between '7990,8010'. Can you explain this answer?.
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