A 120 resistor is in series with a 0.140 H inductor and a 0.440 F capacitor. Com

Question

A 120 resistor is in series with a 0.140 H inductor and a 0.440 F capacitor.

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.

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.

Part 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.

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Part 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.

Part C

State whether the source voltage lags or leads the current at a frequency 500 Hz .

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Part D

State whether the source voltage lags or leads the current at a frequency 1000 Hz .

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Part 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.

Z1, Z2 =

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Part 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.

1, 2 =

Part C

State whether the source voltage lags or leads the current at a frequency 500 Hz .

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Part D

State whether the source voltage lags or leads the current at a frequency 1000 Hz .

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Explanation / Answer

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

inductive reactance=Xl=j*2*pi*f*L

net impedance=R+Xl+Xc

where j=square root of -1

f=frequency

C=capacitance

L=inductance

part A:

for f1=500 Hz,

Xl=j 439.82 ohms

Xc=-j723.43 ohms

then impedance=Z1=120 -j 283.61 ohms

for f2=1000 Hz

Xl=j 879.65 ohms

xc=-j 361.72 ohms

then impedance=120 + j 517.93 ohms

part B:

voltage=current*impedance

hence angle of voltage with repsect to current is phase angle of impedance.

for f=500 Hz, angle =-67.066 degrees

for f=1000 Hz, angle=76.955 degrees

part C:

if the angle is negative then voltage is lagging and if the angle is positive then voltage is leading the current

hence for f=500 Hz, voltage is lagging

for f=1000 Hz, voltage is leading.

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