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
XL = 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