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A three phase bridge inverter is fed from a 500 V dc source. The inverter is operated in 180° conduction mode and it is supplying a purely resistive, star – connected load. The RMS value of the output (line) voltage is
- a)450 V
- b)259.80 V
- c)408 V
- d)235.56 V
Correct answer is option 'C'. Can you explain this answer?
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A three phase bridge inverter is fed from a 500 V dc source. The inver...
° conduction mode. The load is a balanced Y-connected resistive load with each phase having a resistance of 10 Ω. Determine the line voltage and line current of the load.
To determine the line voltage and line current, we first need to calculate the output voltage of the inverter.
In a three-phase bridge inverter, the output voltage is given by:
V_out = V_dc / √3
where V_dc is the DC input voltage.
Given that V_dc = 500 V, we can calculate the output voltage:
V_out = 500 V / √3
≈ 288.7 V
The line voltage is equal to the output voltage, so the line voltage of the load is 288.7 V.
Next, we can calculate the line current of the load.
In a balanced Y-connected load, the line current is equal to the phase current.
The phase current can be calculated using Ohm's Law:
I = V / R
where V is the voltage across the load and R is the resistance of the load.
Given that V = 288.7 V and R = 10 Ω, we can calculate the line current:
I = 288.7 V / 10 Ω
= 28.87 A
Therefore, the line current of the load is 28.87 A.
To determine the line voltage and line current, we first need to calculate the output voltage of the inverter.
In a three-phase bridge inverter, the output voltage is given by:
V_out = V_dc / √3
where V_dc is the DC input voltage.
Given that V_dc = 500 V, we can calculate the output voltage:
V_out = 500 V / √3
≈ 288.7 V
The line voltage is equal to the output voltage, so the line voltage of the load is 288.7 V.
Next, we can calculate the line current of the load.
In a balanced Y-connected load, the line current is equal to the phase current.
The phase current can be calculated using Ohm's Law:
I = V / R
where V is the voltage across the load and R is the resistance of the load.
Given that V = 288.7 V and R = 10 Ω, we can calculate the line current:
I = 288.7 V / 10 Ω
= 28.87 A
Therefore, the line current of the load is 28.87 A.
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A three phase bridge inverter is fed from a 500 V dc source. The inverter is operated in 180° conduction mode and it is supplying a purely resistive, star – connected load. The RMS value of the output (line) voltage isa)450 Vb)259.80 Vc)408 Vd)235.56 VCorrect answer is option 'C'. Can you explain this answer?
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