A 240 resistor is in series with a 0.110 H inductor and a 0.600 F capacitor. a)C

Question

A 240 resistor is in series with a 0.110 H inductor and a 0.600 F capacitor. a)Compute the impedance of the circuit at a frequency of f1= 500 Hz and at a frequency of f2= 1000 Hz. Enter your answer as two numbers separated with a comma.
b)In each case, compute the phase angle of the source voltage with respect to the current. Enter your answer as two numbers separated with a comma. c)State whether the source voltage lags or leads the current at a frequency 500 Hz d)State whether the source voltage lags or leads the current at a frequency 1000 Hz A 240 resistor is in series with a 0.110 H inductor and a 0.600 F capacitor. a)Compute the impedance of the circuit at a frequency of f1= 500 Hz and at a frequency of f2= 1000 Hz. Enter your answer as two numbers separated with a comma.
b)In each case, compute the phase angle of the source voltage with respect to the current. Enter your answer as two numbers separated with a comma. c)State whether the source voltage lags or leads the current at a frequency 500 Hz d)State whether the source voltage lags or leads the current at a frequency 1000 Hz a)Compute the impedance of the circuit at a frequency of f1= 500 Hz and at a frequency of f2= 1000 Hz. Enter your answer as two numbers separated with a comma. b)In each case, compute the phase angle of the source voltage with respect to the current. Enter your answer as two numbers separated with a comma. c)State whether the source voltage lags or leads the current at a frequency 500 Hz d)State whether the source voltage lags or leads the current at a frequency 1000 Hz b)In each case, compute the phase angle of the source voltage with respect to the current. Enter your answer as two numbers separated with a comma. c)State whether the source voltage lags or leads the current at a frequency 500 Hz d)State whether the source voltage lags or leads the current at a frequency 1000 Hz

Explanation / Answer

at f1
impedance due to C = 1/2*500*.6*10^-6 = 530.51

impedane due to L = 2*500*.110 = 345.577

impedane due to C and L =184.94
total impedance = (240^2+184.94^2) = 302.9

at f2
impedance due to C = 1/2*1000*.6*10^-6 = 265.26

impedane due to L = 2*1000*.110 = 691.15

impedane due to C and L =425.894
total impedance = (240^2+425.894^2) = 488.86

a) 302.9,488.86

b) angle at f1 = arccos(240/302.9) = 37.59

angle at f2 = arccos(240/488.86) = 60.59

c) first case capacitance is higher so voltage lags current

d inductance impedance is higher so voltage leads current

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