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In a series LCR circuit containing an AC voltage source of frequency ω, current and voltage V. Then for the given resistance, the correct statement is:
  • a)
    Voltage across R is maximum at ω2 = 1/LC 
  • b)
    Current is maximum at ω2 = 1/LC  
  • c)
    Voltage across R is minimum at ω2 = 1/LC 
  • d)
    Current is minimum at ω2 = 1/LC 
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
In a seriesLCRcircuit containing anACvoltage source of frequency&omega...
The correct answer is: Current is maximum at ω2 = 1/LC 
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In a seriesLCRcircuit containing anACvoltage source of frequencyω, currentIand voltageV. Then for the given resistance, the correct statement is:a)Voltage acrossRis maximum atω2= 1/LCb)Current is maximum atω2= 1/LCc)Voltage acrossRis minimum atω2= 1/LCd)Current is minimum atω2= 1/LCCorrect answer is option 'B'. Can you explain this answer?
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In a seriesLCRcircuit containing anACvoltage source of frequencyω, currentIand voltageV. Then for the given resistance, the correct statement is:a)Voltage acrossRis maximum atω2= 1/LCb)Current is maximum atω2= 1/LCc)Voltage acrossRis minimum atω2= 1/LCd)Current is minimum atω2= 1/LCCorrect answer is option 'B'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about In a seriesLCRcircuit containing anACvoltage source of frequencyω, currentIand voltageV. Then for the given resistance, the correct statement is:a)Voltage acrossRis maximum atω2= 1/LCb)Current is maximum atω2= 1/LCc)Voltage acrossRis minimum atω2= 1/LCd)Current is minimum atω2= 1/LCCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a seriesLCRcircuit containing anACvoltage source of frequencyω, currentIand voltageV. Then for the given resistance, the correct statement is:a)Voltage acrossRis maximum atω2= 1/LCb)Current is maximum atω2= 1/LCc)Voltage acrossRis minimum atω2= 1/LCd)Current is minimum atω2= 1/LCCorrect answer is option 'B'. Can you explain this answer?.
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