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Salsk061 [2.6K]
3 years ago
10

A parallel RLC resonant circuit has a resistance of 200 Ω. If it is known that the bandwidth is 80 rad/s and the lower half-powe

r frequency is 800 rad/s, find the values of the parameters L and C.
Physics
1 answer:
d1i1m1o1n [39]3 years ago
5 0

Answer:

L= 3.6mH

C =9.9 microfarad

Explanation:

Resonant frequency fr

fr = fl + 1/2 BW

fr = 800+ 1/2×80

=800+40

=840 rad/s

Bandwidth (BW)

BW = fr/Q

Q = quality factor

Q= fr/ BW

Q = 840/80

Q= 10.5

Quality factor = R/Xl

Xl = inductive reactance

Xl = R/Q

Xl = 200/10.5

Xl = 19.05 ohms

Xl =2πfL

L= Inductance

L = Xl /2πf

L =19.05/5278.56

L= 3.6mH

Capacitor C

1= fr^2 × 4π^2LC

C = 1/fr^24π^2L

C = 1/100307.5

C= 9.9microfarad

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Answer:

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Explanation:

Given;

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tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

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195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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Answer:

Explanation:

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