A single phase half bridge inverter has a resistive load of 5 - and the center tap dc input voltage is 94 V- The fundamental power consumed by the load is W-

Given, V_dc2 = 94 V_dc = 188 V nbsp; Rms value of the fundamental voltage in the output , V_01 = 2V_dc√(2)π = 2 × 188√(2)π = 84.629 V Fundamental power ...

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A single phase half bridge inverter has a resistive load of $5 Ω$ and the center tap dc input voltage is 94 V. The fundamental power consumed by the load is _____W.
1432.41

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Solution

The correct option is A 1432.41
Given, $\frac{V_{d c}}{2} = 94$

$V_{d c} = 188 V$

Rms value of the fundamental voltage in the output ,

$V_{01} = \frac{2 V_{d c}}{\sqrt{2} π} = \frac{2 \times 188}{\sqrt{2} π} = 84.629 V$

Fundamental power in the output

$= \frac{(V_{01})^{2}}{R} = \frac{(84.629)^{2}}{5} = 1432.41 W$

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A single phase half bridge inverter has a resistive load of 5 Ω and the center tap dc input voltage is 94 V. The fundamental power consumed by the load is _____W.1432.41

The correct option is A 1432.41Given, Vdc2=94 Vdc=188 V Rms value of the fundamental voltage in the output , V01=2Vdc√2π=2×188√2π=84.629 V Fundamental power in the output =(V01)2R=(84.629)25=1432.41 W

A single phase half bridge inverter has a resistive load of $5 Ω$ and the center tap dc input voltage is 94 V. The fundamental power consumed by the load is _____W.
1432.41