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NCERT Class XII Chapter
Alternating Current
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Question : 21 of 26
Marks: +1, -0
Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Solution:  
vrv_r = 12πLC\frac{1}{2\pi\sqrt{LC}} = 12×3.14×3×27×106\frac{1}{2\times3.14\times\sqrt{3\times27\times10^{-6}}} = 10006.28×9\frac{1000}{6.28\times9} = 17.7
resonant frequency, ωr\omega_r = 2πvr2\pi v_r = 2 × 3.14 × 17.7 = 111 rad sec1\sec^{-1}
Quality factor in the given resonant circuit
Q = 1L/C\frac{1}{\text{₹}\sqrt{L/C}} = 17.4327×106\frac{1}{7.4}\sqrt{\frac{3}{27\times10^{-6}}} = 45
we want to improve the quality factor to twice, without changing resonant frequency (without changing L and C).
Q' = 2Q = 90 = 1RLC\frac{1}{R'}\sqrt{\frac{L}{C}}or R' = 190327×106\frac{1}{90}\sqrt{\frac{3}{27\times10^{-6}}} = 3.7 Ω
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