Question

#3. A 250-ohm resistor, a 0.450 H inductor, and a 6.45 uF capacitor are connected in series
across an emf with a 36.0 volt amplitude and an angular frequency of 260 rad/s.
a. What is the impedance?
b. What is the current amplitude?
c. What is the phase angle between the voltage and current? Does the voltage lag or lead?
d. What are the voltage amplitudes across the resistor, inductor and capacitor individually?
e. What is the power?
f. What is the resonant frequency of the circuit?
g. What is the impedance at the resonant frequency?
h. What is the power at resonant frequency?

Answer 1

R=2500, L=0 450H c=6.45 HF Vo= 36 v W =260 randl a) Impedance, za JR²+ (%-X.)² XL = LW= Xe=1/W 0.450*260 = 117.12 1 - 5962 6.usxio x260 291941 =540.32 2=1250²+ (117-596)² 162500 + 229441 = b) Current amplitude, Io = Vo Io = Vo = 36 c) Phase Tan (XL-Xc 0.066 A ㅈ 540.3 Tan 117 596 asoo 1,5596) - Tah'(-479 = Tan (1.916) = -62.44 e is ove Voltage: Current leads so voltage lags behind behind Current d) Voltage amplitude de across resistor, URO = IR=0.066x250 = 16.5 V Voltage amplitude alross indutor, Vo= IoXL = 0.066x117-7.72 v Voltage amplitele Veo=Io xe =0.066x596=39.34 U Capacitor, e) lower, Pau= Vems IRMs Cuse Vems = Vo Pav= 25.46 0.046x Cost 62.44) IRM = 16 aurau 36 Jz 25.46U =o066 -0.046 A = 0.54 W Jz f) Resonant frequeny, fal a The WISE 2x3.14 0 450x XE.USXICE 3 3 - 0.0936 X10 2x3 14 x 170 = 93.6 Hz 9) Imped ame at at resonanie - resistame - 250 R h) Power at resoname, Pav=URMs Irms = 25.46x 0.046 = 1.17 W