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A parallel resonant circuit has a midband admittance of 25 x 10-3 S, quality factor of 80 and a resonant frequency of 200 krad/s.The value of R is
- a)40Ω
- b)56.57Ω
- c)80Ω
- d)28.28Ω
Correct answer is option 'A'. Can you explain this answer?
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A parallel resonant circuit has a midband admittance of 25 x 10-3S, qu...
Ay mid-band frequency Z = R, Y = 1/R
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A parallel resonant circuit has a midband admittance of 25 x 10-3S, qu...
The quality factor Q of a parallel resonant circuit is given by the formula:
Q = ω0L/R
Where Q is the quality factor, ω0 is the resonant frequency, L is the inductance, and R is the resistance.
We are given that Q = 80 and ω0 = 200 krad/s.
Substituting these values into the formula, we can solve for R:
80 = (200 krad/s)L/R
Rearranging the formula, we get:
R = (200 krad/s)L/80
We are also given that the midband admittance (Y) is 25 x 10^-3 S.
The admittance Y of a parallel resonant circuit is given by the formula:
Y = 1/R + jω0C
Where Y is the admittance, R is the resistance, ω0 is the resonant frequency, and C is the capacitance.
Substituting the given values and rearranging the formula, we get:
25 x 10^-3 S = 1/R + j(200 krad/s)C
Since the circuit is purely resistive at resonance, the imaginary part of the admittance should be zero. Therefore, we can disregard the j(200 krad/s)C term.
Simplifying the equation, we get:
25 x 10^-3 S = 1/R
R = 1/(25 x 10^-3 S)
R = 40
Therefore, the value of R is 40.
Q = ω0L/R
Where Q is the quality factor, ω0 is the resonant frequency, L is the inductance, and R is the resistance.
We are given that Q = 80 and ω0 = 200 krad/s.
Substituting these values into the formula, we can solve for R:
80 = (200 krad/s)L/R
Rearranging the formula, we get:
R = (200 krad/s)L/80
We are also given that the midband admittance (Y) is 25 x 10^-3 S.
The admittance Y of a parallel resonant circuit is given by the formula:
Y = 1/R + jω0C
Where Y is the admittance, R is the resistance, ω0 is the resonant frequency, and C is the capacitance.
Substituting the given values and rearranging the formula, we get:
25 x 10^-3 S = 1/R + j(200 krad/s)C
Since the circuit is purely resistive at resonance, the imaginary part of the admittance should be zero. Therefore, we can disregard the j(200 krad/s)C term.
Simplifying the equation, we get:
25 x 10^-3 S = 1/R
R = 1/(25 x 10^-3 S)
R = 40
Therefore, the value of R is 40.
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A parallel resonant circuit has a midband admittance of 25 x 10-3S, quality factor of 80 and a resonant frequency of 200 krad/s.The value of R isa)40Ωb)56.57Ωc)80Ωd)28.28ΩCorrect answer is option 'A'. Can you explain this answer?
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A parallel resonant circuit has a midband admittance of 25 x 10-3S, quality factor of 80 and a resonant frequency of 200 krad/s.The value of R isa)40Ωb)56.57Ωc)80Ωd)28.28ΩCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A parallel resonant circuit has a midband admittance of 25 x 10-3S, quality factor of 80 and a resonant frequency of 200 krad/s.The value of R isa)40Ωb)56.57Ωc)80Ωd)28.28ΩCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel resonant circuit has a midband admittance of 25 x 10-3S, quality factor of 80 and a resonant frequency of 200 krad/s.The value of R isa)40Ωb)56.57Ωc)80Ωd)28.28ΩCorrect answer is option 'A'. Can you explain this answer?.
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