Q: In L.C.R. series resonant circuit, an inductance of 10 mH , a resistance of 30 ohms and a capacitance of 0.1 mF are connected with an A.C. the e.m.f. of the A.C. source is given a E = 20 cosωt . What will be the amplitude of current at a frequency 20% less than the resonant frequency?

(a) 0.66 A

(b) 6.6 A

(c) 66 A

(d) 6.6 × 10^{-3} A

Ans: (a)

Sol: Here, L = 10 mH = 10^{-2} H ,R = 30 Ω,

C = 0.1 mF = 10^{-4} F

Resonance frequency $ \displaystyle \omega = \frac{1}{\sqrt{L C}}$

$ \displaystyle \omega = \frac{1}{\sqrt{10^{-2}\times 10^{-4}}}$

= 1000 rad/sec

Actual frequency, ω’ = 80%×1000 = 800 rad/s

X_{L} = ω’ L = 800×10^{-2} = 8 ohm

$ \displaystyle X_C = \frac{1}{\omega’ } = \frac{1}{800 \times 10^{-4}} $

= 12 .5 ohm

$ \displaystyle Z = \sqrt{R^2 + (X_L – X_C)^2} $

$ \displaystyle Z = \sqrt{30^2 + (4.5)^2} $

Z = 30.33 ohm

$ \displaystyle I_0 = \frac{E_0}{Z} = \frac{20}{30.33} $

= 0.66 A

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