For a series RLC circuit find the resonant frequency, quality factor and bandwidth.

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written 3.9 years ago by

teamques10 ★ 69k

modified 3.3 years ago by

pedsangini276 • 4.8k

For a series RLC circuit having R=10Ω, L=0.01 H and C=100?F, find the resonant frequency, quality factor and bandwidth.

basic electrical engineering

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written 3.9 years ago by

teamques10 ★ 69k

Answer:

$R= 10 \ \Omega$

$L=0.01 \mathrm H$

$C=100 \mu F$

At resonance, $X_L \ =\ X_c$

$f_r = \dfrac{1}{2\pi \sqrt{L\times C}}$

$f_r = \dfrac{1}{2\pi \sqrt{0.0110010^{-6}}}$

$f_r= 159.154\ \mathrm Hz$

The resonant frequency is 159.154Hz

$Q = \dfrac{X_L}{R}$

$Q = \dfrac{2\pi159.1540.01}{10}$ $\Big[\because X_L = 2\pi \times f_r \times …

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Answer:

R=10 ΩR= 10 \ \Omega

L=0.01HL=0.01 \mathrm H

C=100μFC=100 \mu F

At resonance, XL = XcX_L \ =\ X_c

fr=12π√L×Cf_r = \dfrac{1}{2\pi \sqrt{L\times C}}

fr=12π√0.01∗100∗10−6f_r = \dfrac{1}{2\pi \sqrt{0.01*100*10^{-6}}}

fr=159.154 Hz f_r= 159.154\ \mathrm Hz

Q=XLRQ = \dfrac{X_L}{R}

Q=2π∗159.154∗0.0110Q = \dfrac{2\pi*159.154*0.01}{10}$\Big[\because X_L = 2\pi \times f_r \times …