Question

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.

Part A

What is the impedance of the circuit?

Part B

What is the current amplitude?

Part C

What is the phase angle of the source voltage with respect to the current?

Part E

What is the voltage amplitude across the resistor?

Answer 1

Part A:

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

Part B

Imax = Vmax/Z

Imax = 26.0/561 = 0.0463 Amp

Part C

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.

Part D.

Voltage drop across resistor will be

V_R = Imax*R = 0.0463*250 = 11.6 V

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