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 772 mH, a resistance of 289 ohm, and a capacitance of 46.4 mu F. (a) What are the maximum current and the phase angle between the current and the source emf in this circuit? I_max = A phi = degree (b) What are the maximum potential difference across the inductor and the phase angle between this potential difference and the current in the circuit? V_L, max = V phi = degree (c) What are the maximum potential difference across the resistor and the phase angle between this potential difference and the current in this circuit? V_R, max = V phi = degree (d) What are the maximum potential difference across the capacitor and the phase angle between this potential difference and the current in this circuit? V_C, max = V phi = degree

Explanation / Answer

a)

Maximum current,

Imax = V/sqrt(R^2 + (XL - Xc)^2)

XL = inductive reactance = 2*pi*f*L

Xc = capacitive reactance = 1/(2*pi*f*C)

So, Imax = 120/sqrt(289^2 + (2*pi*50*0.772 - 1/(2*pi*50*46.4*10^-6))^2)

= 0.356 A

phase angle = atan((XL - Xc)/R) = atan(((2*pi*50*0.772 - 1/(2*pi*50*46.4*10^-6))/289)

= 31 deg

b)

VL,max = Imax*XL

= 0.356*(2*pi*50*0.772)

= 86.3 V

phase angle between this potential difference and current = 90 deg

c)

Maximum potential difference across resistor = Imax*R = 0.356*289 = 102.9 V

Phase = 0 deg

d)

Maximum potential difference across capacitor = Imax*Xc = 0.356/(2*pi*50*46.4*10^-6)

= 24.4 V

Phase angle = -90 deg

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