A phase controlled single phase rectifier- supplied by an AC source feeds power to an RLE load as shown in the figure-The rectified output voltage has an average value given by V0-2Vm-cos - where Vm-252 V and - is the firing angle- If the power delivered to the lossless battery is 1600 W- then firing angle - in degrees is

The power delivered to the lossless battery is, E_bI_0=1600 I_0=160080=20 A The rectifier output voltage, V_0=E_b+I_0R 3V_mπcos α=80+(20× 2) 2× 252π cos α=1 ...

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1-Phase Semiconverter RL, RLE Load

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Question

A phase controlled single phase rectifier, supplied by an AC source feeds power to an RLE load as shown in the figure,

The rectified output voltage has an average value given by $V_{0} = \frac{2 V_{m}}{π} c o s α,$ where $V_{m} = 252 V$ and $α$ is the firing angle. If the power delivered to the lossless battery is $1600 W$ , then firing angle $α$ (in degrees) is ______

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Solution

The power delivered to the lossless battery is,
$E_{b} I_{0} = 1600$

$I_{0} = \frac{1600}{80} = 20 A$

The rectifier output voltage,
$V_{0} = E_{b} + I_{0} R$

$\frac{3 V_{m}}{π} c o s α = 80 + (20 \times 2)$

$\frac{2 \times 252}{π} c o s α = 120$

$160.42 c o s α = 120$

Firing angle $α = {41.58}^{o}$

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The power delivered to the lossless battery is, EbI0=1600 I0=160080=20A The rectifier output voltage, V0=Eb+I0R 3Vmπcosα=80+(20×2) 2×252πcosα=120 160.42cosα=120 Firing angle α=41.58o