Question

The input voltage to a converter is

\({V_i} = 100\sqrt 2 \sin \left( {100\pi t} \right)V\)

The current drawn by the converter is

\({i_1} = 10\sqrt 2 \sin \left( {100\pi t - \frac{\pi }{3}} \right) + 5\sqrt 2 \sin \left( {300\pi + \times \frac{\pi }{4}} \right) + 2\sqrt 2 \sin \left( {500\pi t - \pi /6} \right)A.\)

The input power factor of the converter
1. 0.31
2. 0.44
3. 0.5
4. 0.71

Answer 1

Correct Answer - Option 2 : 0.44

Input power factor

\(\begin{array}{l} = \frac{{{V_s}{I_{s1}}\cos {\phi _1}}}{{{V_s}{I_s}}}\\ = \frac{{{I_{{s_1}}}}}{{{I_s}}}cos{\phi _1}\\ {I_s} = \sqrt {{{10}^2} + {5^2} + {2^2}} = 11.35A\\ pf = \frac{{10}}{{11.35}} \times \cos 60 = 0.44 \end{array}\)