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A sinusoidal voltage of peak value 283 V and frequency 50Hz is applied to a series LCR circuit in which R= 3 ohm L= 25.48mH C= 796 mu F. Suppose the frequency of the source can be varied. (a) What is thefrequency of the source at which resonance occurs? (b) Calculate the impedance, the current,and the power dissipated at the resonant condition. |
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Answer» Solution :a. The frequency at which the RESONANCE occurs is `omega_0 = (1)/(sqrtLC) = (1)/(sqrt(25.48 xx 10^(-3)xx 796 xx 10^(-6) )) = 222.1` rad/s `v_r = (omega_0)/(2pi) =(222.1)/(2 xx 3.14) Hz= 35.4 Hz` b. The impedance Z at resonance CONDITION is equal to the RESISTANCE `Z = R = 3 Omega` The rms current resonance is `= V/Z = V/R = ((283)/(sqrt2)) 1/3 = 66.7A` The POWER dissipated at resonance is `P = I^2 xx R = (66.7)^2 xx = 13.35 KW` Power dissipated at resonance is more than the power dissipated in the above example. |
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