The 60 Hz ac source of a series circuit has a voltage amplitude of 120 V. The ca

Question

The 60 Hz ac source of a series circuit has a voltage amplitude of 120 V. The capacitive and inductive reactances are 800 omega and 280 omega, respectively. The resistance is 490 omega. In the figure, the rms current in the circuit is closest to?

The 60 Hz ac source of a series circuit has a voltage amplitude of 120 V. The capacitive and inductive reactances are 800 omega and 270 omega, respectively. The resistance is 470 omega. In the figure, the resistance is changed from its original value so that the power factor is now 0.80. The original values of capacitance, inductance, and frequency remain unchanged. The new resistance is closest to?

A series ac circuit is shown. The inductor has a reactance of 50 ohms and an inductance of 230 mH. A 40 ohm resistor and a capacitor whose reactance is 100 ohms are also in the circuit. The rms current in the circuit is 1.2 A. In the figure, the capacitance of the capacitor is closest to?

A series ac circuit is shown. The inductor has a reactance of 80 ohms and an inductance of 210 mH. A 90 ohm resistor and a capacitor whose reactance is 110 ohms are also in the circuit. The rms current in the circuit is 1.4 A. In the figure, the peak magnetic energy in the inductor is closest to?

Explanation / Answer

1)

rms current = 120/(sqrt(490^2+(800-280)^2)*sqrt(2)) = 0.119 A

2)

0.8 = cos(atan((270-800)/R) in degrees)

So, R = 707 ohm <----------------answer

3)

f = 50/(2*3.14*0.23) = 34.6 Hz

So, C = 1/(2*3.14*34.6*100) = 4.6*10^-5 Hz <---------------answer

4)

peak magnetic energy= 0.5*0.21*1.4^2*2 = 0.412 J

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