A single phase half bridge inverter has a resistive load of 10 - and the center tap dc input voltage is 96 V- The fundamental power consumed by the load is

Given, V_dc2=96 V_dc=192 V Rms value of the fundamental voltage in the output, V_01=2V_f√(2π)=2× 192√(2π)=86.43 V Fundamental power in the output =(V_01)^2R=(86 ...

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Standard XII

Physics

Power Factor

A single phas...

Question

A single phase half bridge inverter has a resistive load of $10 Ω$ and the center tap dc input voltage is 96 V. The fundamental power consumed by the load is

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Solution

Given,
$\frac{V_{d c}}{2} = 96$
$V_{d c} = 192 V$
Rms value of the fundamental voltage in the output,
$V_{01} = \frac{2 V_{f}}{\sqrt{2 π}} = \frac{2 \times 192}{\sqrt{2 π}} = 86.43 V$
Fundamental power in the output
$= \frac{(V_{01})^{2}}{R} = \frac{(86.43)^{2}}{10} = 747 W$

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A single phase half bridge inverter has a resistive load of 10 Ω and the center tap dc input voltage is 96 V. The fundamental power consumed by the load is

Given, Vdc2=96 Vdc=192V Rms value of the fundamental voltage in the output, V01=2Vf√2π=2×192√2π=86.43V Fundamental power in the output =(V01)2R=(86.43)210=747W

A single phase half bridge inverter has a resistive load of $10 Ω$ and the center tap dc input voltage is 96 V. The fundamental power consumed by the load is

Given,
$\frac{V_{d c}}{2} = 96$
$V_{d c} = 192 V$
Rms value of the fundamental voltage in the output,
$V_{01} = \frac{2 V_{f}}{\sqrt{2 π}} = \frac{2 \times 192}{\sqrt{2 π}} = 86.43 V$
Fundamental power in the output
$= \frac{(V_{01})^{2}}{R} = \frac{(86.43)^{2}}{10} = 747 W$