GATE 2017 EE (EE2) – Question 58

The diode in the shown circuit is ideal. Find the input power factor to two decimal places.

Show answer and explanation

Correct answer: 0.70 to 0.71

Explanation

For the 100 V-peak sinusoidal source and 10 Ω load, peak current is 10 A. The diode passes a half-wave sine, whose rms current is Im/2=5 A.

Real input power equals resistor power, $I_{rms}^2R=25(10)=250$ W. Source rms voltage is 100/√2 V, so apparent power is $(100/\sqrt2)(5)=353.553$ VA.

Power factor is 250/353.553=1/√2=0.707107, or **0.71**. The non-sinusoidal current makes it less than one despite the resistive load.

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