GATE EE 2014 Set 3 — Question 42
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Electric Circuits → Transient & AC Steady-State → Series & Parallel Resonance
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Question
A series circuit is observed at two frequencies. At , we note that source voltage results in a current . At , the source voltage results in a current . The closest values for out of the following options are
Correct answer
(B) R = 50 Ω; L = 10 mH; C = 25 μF;
Solution
1.Analyze the circuit at :
The source voltage is and the current is . Since the current is in phase with the voltage, the circuit is at resonance at this frequency.At resonance, the impedance is purely resistive:
2.Resonance Condition:
At resonance, .3.Analyze the circuit at :
The current leads the voltage by . This means the circuit is capacitive at this frequency.The phase angle of the impedance is .
4.Solve for and :
Substitute into the equation:Now, calculate :
5.Conclusion:
The values are , which corresponds to option (B).Continue learning with Success Tracker
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