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Q.

An alternating voltage $E = 6 \sin 20 t + 8 \cos 20 t$ is applied to a series resonant circuit as shown. The correct statements are:

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a

The capacitance C is 12.5 mF.

b

The resonant rms current in the circuit is $\sqrt{2} A$

c

The quality factor of the current is 0.8

d

Average power dissipated in the circuit is 10W

answer is A, B, C, D.

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Detailed Solution

$E = 6 \sin 20 t + 8 \cos 20 t$

$= 10 \sin (20 t + ϕ_{0}) ϕ_{0} = \tan^{- 1} (\frac{8}{6}) = 53^{0} E_{0} = 10 v o l t; I_{0} = 2 A P = \frac{E_{0} I_{0}}{2} = 10 W Q = ω_{0} \frac{L}{R} = 20 (\frac{2}{10 (5)}) = 0.8 ω_{0} L = \frac{1}{ω_{0} C} \Rightarrow C = \frac{1}{ω_{0}^{2} L} = 12.5 m F$

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