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Q#1 Sinusoidal Steady State Analysis GATE EE 1994 MCQ +1 mark -0.33 marks

At resonance, the given parallel circuit constituted by an iron-coil and a capacitor behaves like

Z:\PY\EE\Redreaw figure\Network\updated\67-05.jpg

An open-circuit

A short-circuit

A pure resistor of value R

A pure resistor of value much higher than R

Explanation:

The phasor diagram for the given circuit is shown below,

21.jpg

At resonance, only in phase component of current remains so that power factor should be unity.

Hence, quadrature component of current must cancel out.  

Total Current,   

 is the impedance angle of ‘R - L’ branch.

The current in RL branch would be,

        

Since

Total impedance 

Since, the power factor is unity the impedance will be resistive.

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