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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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