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A single phase full bridge inverter feeds power to a load of R = 12 Ω and L = 0.04 H from a 400 V dc source. If the inverter operates the frequency of 50 Hz, determine the power delivered to load (in watt) for square wave operation. (Consider first and third harmonic).
- a)5372 W
- b)5782 W
- c)4272 W
- d)5272 W
Correct answer is option 'D'. Can you explain this answer?
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A single phase full bridge inverter feeds power to a load of R = 12 &O...
RMS value of fundamental voltage is
Load impedance at fundamental frequency
RMS value of resultant.
Load impedance at fundamental frequency
RMS value of resultant.
Most Upvoted Answer
A single phase full bridge inverter feeds power to a load of R = 12 &O...
Ω and L = 0.1 H. The input DC voltage is 24 V.
To find the output voltage of the inverter, we first need to determine the switching frequency of the inverter. Let's assume a switching frequency of 50 kHz.
Next, we need to calculate the duty cycle of the inverter. The duty cycle is the ratio of the on-time to the total switching period. For a full bridge inverter, the on-time is half of the switching period.
On-time = T/2 = 1/2f = 1/2 x 50,000 = 10 μs
The total switching period is the sum of the on-time and the off-time. The off-time is also 10 μs for a full bridge inverter.
Total switching period = T = 2 x 10 μs = 20 μs
The duty cycle is therefore:
D = On-time/Total switching period = 10 μs/20 μs = 0.5
The output voltage of the inverter can be calculated using the following formula:
Vout = Vdc x D
Where Vdc is the input DC voltage and D is the duty cycle.
Vout = 24 V x 0.5 = 12 V
Next, we need to calculate the rms value of the output voltage. For a square wave voltage with a duty cycle of 0.5, the rms value can be calculated using the following formula:
Vrms = Vmax/√2
Where Vmax is the peak value of the voltage.
Vmax = Vout = 12 V
Vrms = 12 V/√2 = 8.48 V
Finally, we need to calculate the current flowing through the load. The current can be calculated using Ohm's law:
I = V/R
Where V is the rms value of the output voltage and R is the resistance of the load.
I = 8.48 V/12 Ω = 0.71 A
Therefore, the output voltage of the inverter is 12 V rms and the current flowing through the load is 0.71 A.
To find the output voltage of the inverter, we first need to determine the switching frequency of the inverter. Let's assume a switching frequency of 50 kHz.
Next, we need to calculate the duty cycle of the inverter. The duty cycle is the ratio of the on-time to the total switching period. For a full bridge inverter, the on-time is half of the switching period.
On-time = T/2 = 1/2f = 1/2 x 50,000 = 10 μs
The total switching period is the sum of the on-time and the off-time. The off-time is also 10 μs for a full bridge inverter.
Total switching period = T = 2 x 10 μs = 20 μs
The duty cycle is therefore:
D = On-time/Total switching period = 10 μs/20 μs = 0.5
The output voltage of the inverter can be calculated using the following formula:
Vout = Vdc x D
Where Vdc is the input DC voltage and D is the duty cycle.
Vout = 24 V x 0.5 = 12 V
Next, we need to calculate the rms value of the output voltage. For a square wave voltage with a duty cycle of 0.5, the rms value can be calculated using the following formula:
Vrms = Vmax/√2
Where Vmax is the peak value of the voltage.
Vmax = Vout = 12 V
Vrms = 12 V/√2 = 8.48 V
Finally, we need to calculate the current flowing through the load. The current can be calculated using Ohm's law:
I = V/R
Where V is the rms value of the output voltage and R is the resistance of the load.
I = 8.48 V/12 Ω = 0.71 A
Therefore, the output voltage of the inverter is 12 V rms and the current flowing through the load is 0.71 A.
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A single phase full bridge inverter feeds power to a load of R = 12 Ω and L = 0.04 H from a 400 V dc source. If the inverter operates the frequency of 50 Hz, determine the power delivered to load (in watt) for square wave operation. (Consider first and third harmonic).a)5372 Wb)5782 Wc)4272 Wd)5272 WCorrect answer is option 'D'. Can you explain this answer?
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A single phase full bridge inverter feeds power to a load of R = 12 Ω and L = 0.04 H from a 400 V dc source. If the inverter operates the frequency of 50 Hz, determine the power delivered to load (in watt) for square wave operation. (Consider first and third harmonic).a)5372 Wb)5782 Wc)4272 Wd)5272 WCorrect answer is option 'D'. 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 single phase full bridge inverter feeds power to a load of R = 12 Ω and L = 0.04 H from a 400 V dc source. If the inverter operates the frequency of 50 Hz, determine the power delivered to load (in watt) for square wave operation. (Consider first and third harmonic).a)5372 Wb)5782 Wc)4272 Wd)5272 WCorrect answer is option 'D'. 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 single phase full bridge inverter feeds power to a load of R = 12 Ω and L = 0.04 H from a 400 V dc source. If the inverter operates the frequency of 50 Hz, determine the power delivered to load (in watt) for square wave operation. (Consider first and third harmonic).a)5372 Wb)5782 Wc)4272 Wd)5272 WCorrect answer is option 'D'. Can you explain this answer?.
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