A single-phase full bridge diode rectifier is supplied from 230 V- 50 Hz sinusoidal source- The load comprises of R - 20 - and large inductance such that load current is constant- The value of input supply power factor will be

Average value of output voltage, V_0 =2V_m/π =2 ×√(2)× 230/π = 207.07V ∴ Average load current, nbsp; nbsp; nbsp; nbsp; I_0 =V_0/R =207.07/20 = 1 ...

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A single-phase full bridge diode rectifier is supplied from 230 V, 50 Hz sinusoidal source. The load comprises of $R = 20 Ω$ and large inductance such that load current is constant. The value of input supply power factor will be

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Solution

Average value of output voltage,
$V_{0} = \frac{2 V_{m}}{π} = \frac{2 \times \sqrt{2} \times 230}{π} = 207.07 V$

$∴ A v e r a g e l o a d c u r r e n t,$

$I_{0} = \frac{V_{0}}{R} = \frac{207.07}{20} = 10.35 A$

$I n p u t s o u r c e c u r r e n t, I_{s} = I_{0} \sqrt{\frac{π}{π}} = I_{0} = 10.35 A$

Input source rms voltage,
$V_{s} = 230 V$

$I n p u t p o w e r f a c t o r = \frac{V_{0} I_{0}}{V_{s} I_{s}} = \frac{207.07 \times 10.35}{230 \times 10.35} = 0.9 l a g g i n g$

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A single-phase full bridge diode rectifier is supplied from 230 V, 50 Hz sinusoidal source. The load comprises of R=20 Ω and large inductance such that load current is constant. The value of input supply power factor will be

Average value of output voltage, V0=2Vmπ=2×√2×230π=207.07V ∴ Average load current, I0=V0R=207.0720=10.35A Input source current, Is=I0√ππ=I0=10.35A Input source rms voltage, Vs=230V Input power factor =V0I0VsIs=207.07×10.35230×10.35=0.9 lagging

A single-phase full bridge diode rectifier is supplied from 230 V, 50 Hz sinusoidal source. The load comprises of $R = 20 Ω$ and large inductance such that load current is constant. The value of input supply power factor will be