For the following series AC circuit, the angle of the circuit impedance is 0 deg

Question

For the following series AC circuit, the angle of the circuit impedance is 0 degrees +90 degrees +45 degrees -45 degrees For the following series AC circuit the phase angle of the circuit current is 0 degrees +90 degrees -45 degrees +45 degrees For the following series AC circuit, the total circuit impedance is: 100 - j 100 Ohm 100 + j 100 Ohm 100 - j 200 Ohm 100 + j 300 Ohm For the parallel AC circuit shown below, find the circuit current I S. Note that v in AC is mislabeled, it should be 10V, 10 sin (omega t ). 0.1 + j 0.1 A 0.1 - j 0.1 A 0.1 A j 0.1 A

Explanation / Answer

q7.

IMPEDANCE=100+100J

ANGLE=+45 DEGREES

Q8.

IMPEDANCE=100

ANGLE=0 DEGREES

Q9.
capacitive reactance=-j/(2500*2*10^(-6))=-j200

inductive reactance=j0.04*2500=j100

IMPEDANCE=100-j100

Q10.

100*j100/(100+j100)

=j100/(100+j100)

=j100*(100-j100)/(100^2+100^2)

=50+j50

current=voltage/impedance=10/(50+j50)=0.1-j0.1

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