A series RLC circuit driven by a source with an amplitude of 120.0 V and a frequ

Question

A series RLC circuit driven by a source with an amplitude of 120.0 V and a frequency of 50.0 Hz has an inductance of 777 mH, a resistance of 212 2, and a capacitance of 45.0 (a) What are the maximum current and the phase angle between the current and the source emf in this circuit? Imax =1 (b) What are the maximum potential difference across the inductor and the phase angle between this potential difference and the current in the circuit? VL, max = (c) What are the maximum potential difference across the resistor and the phase angle between this potential difference and the current in this circuit? Vo, max = (d) What are the maximum potential difference across the capacitor and the phase angle between this potential difference and the current in this circuit? Vc, max

Explanation / Answer

A) Inductive reactance is XL = 2*pi*f*L = 2*3.142*50*0.777 = 244.10 ohm

Capacitive inductance is XC = 1/(2*pi*f*C) = 1/(2*3.142*50*45*10^-6) = 70.74 ohm

R = 212 ohm

then Impedence is Z = sqrt((XL-XC)^2+R^2) = sqrt((244.10-70.74)^2+212^2) = 273.86 ohm

maximum current is Imax = Vmax/Z = 120/273.86 = 0.438 A

phase angle is phi = tan^(-1)((XL-XC)/R) = tan^(-1)((244.10-70.74)/212) = 39.27 degrees

b) VL,max= Imax*XL = 0.438*244.10 = 106.92 V

current I leads Voltage V by a phase angle 90 degrees

C) VR,max = Imax*R = 0.438*212 = 92.856 V

Current I and Voltage V are in same phase

D) VC,max = Imax*XC = 0.438*70.74 = 30.98 V

Current I lags behind the Volatge V by a phase difference of 90 degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.