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UPRVUNL AE EE 2014 Official Paper

Option 1 : 10.9 W

Concept:

For a full controlled thyristor bridge rectifier, the average output voltage is given by,

\({{\rm{V}}_{\rm{o}}} = \frac{{2{{\rm{V}}_{\rm{m}}}}}{{\rm{\pi }}}\cos {\rm{\alpha }}\)

The average current \({{\rm{I}}_0} = \frac{{{{\rm{V}}_{\rm{o}}}}}{{\rm{R}}}\)

For constant load current, Irms = I0

Load power \({{\rm{P}}_0} = {\left( {{{\rm{I}}_{{\rm{rms}}}}} \right)^2}{\rm{R}}\)

Calculation:

Given that, V = 110 V, f = 50 Hz, R_{L }= 60 ohm

Firing angle, α = 75°

\({{\rm{V}}_{\rm{o}}} = \frac{{2 \times 110 \times \sqrt 2 }}{{\rm{\pi }}}(\cos 75^\circ ) = 25.63{\rm{\;V}}\)

Therefore, \({{\rm{I}}_0} = \frac{{{{\rm{V}}_0}}}{{\rm{R}}} = \frac{{25.63}}{{50}} = 0.427{\rm{\;A}}\)

Irms = 0.427 A

P = (0.427)2 × 50 = 10.9 W