A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series L-C-R circuit in which R=3 Ω, L=25.48 mH and C=796 μ F. Calculate the impedance, the current and the power dissipated at the resonant condition.

Question

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series L-C-R circuit in which R=3 Ω, L=25.48 mH and C=796 μ F. Calculate the impedance, the current and the power dissipated at the resonant condition. (A) 4 Ω, 13.35 A , 60 W (B) 2 Ω, 65 A , 13 kW (C) 8 Ω, 66.7 A , 13.35 kW (D) 3 Ω, 66.7 A , 13.35 kW

Tardigrade Admin · Accepted Answer

The impedance Z at resonant condition is equal to the resistance Z=R=3 Ω The rms current at resonance is =(V/Z)=(V/R)=((283/√2)) (1/3)=66.7 A The power dissipated at resonance is P=I2 × R=(66.7)2 × 3=13.35 kW.

Q. A sinusoidal voltage of peak value $283 V$ and frequency $50 Hz$ is applied to a series $L - C - R$ circuit in which $R = 3 Ω, L = 25.48 m H$ and $C = 796 μ F$ .
Calculate the impedance, the current and the power dissipated at the resonant condition.

$4 Ω, 13.35 A, 60 W$

$2 Ω, 65 A, 13 kW$

$8 Ω, 66.7 A, 13.35 kW$

$3 Ω, 66.7 A, 13.35 kW$

The impedance $Z$ at resonant condition is equal to the resistance
$Z = R = 3 Ω$
The rms current at resonance is
$= \frac{V}{Z} = \frac{V}{R} = (\frac{283}{2}) \frac{1}{3} = 66.7 A$
The power dissipated at resonance is
$P = I^{2} \times R = (66.7)^{2} \times 3 = 13.35 kW$ .

Q. A sinusoidal voltage of peak value 283V and frequency 50Hz is applied to a series L−C−R circuit in which R=3Ω,L=25.48mH and C=796μF. Calculate the impedance, the current and the power dissipated at the resonant condition.

4Ω,13.35A,60W

2Ω,65A,13kW

8Ω,66.7A,13.35kW

3Ω,66.7A,13.35kW