You have a resistor of resistance 250 Ω , an inductor of inductance 0.370 H , a capacitor of capacitance 5.90 μF and a voltage source that has a voltage amplitude of 26.0 V and an angular frequency of 280 rad/s . The resistor, inductor, capacitor, and voltage source are connected to form an L-R-C series circuit.
What is the impedance of the circuit?
What is the current amplitude?
What is the phase angle of the source voltage with respect to the current?
What is the voltage amplitude across the resistor?
Z = sqrt (R^2 + (XL - Xc)^2)
R = 250 Ohm
w = 280 rad/sec
XL = w*L = 280*0.370 = 103.6 ohm
Xc = 1/(w*C) = 1/(280*5.90*10^-6) = 605.33 Ohm
So impedance will be
Z = sqrt (250^2 + (103.6 - 605.33)^2)
Z = 560.56 = 561 Ohm
Imax = Vmax/Z
Imax = 26.0/561 = 0.0463 Amp
phase Angle is given by:
phi = arctan ((XL - Xc)/R)
phi = arctan ((103.6 - 605.33)/250)
phi = -63.5 deg, and since phase angle is negative, So voltage will lag.
Voltage drop across resistor will be
VR = Imax*R = 0.0463*250 = 11.6 V
Get Answers For Free
Most questions answered within 1 hours.