5 (part1) A 57 ? resistor is connected in series with a 91 mH inductor and a 0.2

Question

5(part1) A 57 ? resistor is connected in series with a 91 mH inductor and a 0.2 ?F capac- itor. The applied voltage has the form E = 190 V sin(2 ? f t), where the frequency is f = 922 cycles/s.

Find the rms current. Answer in units of A

6.(part 2)Find the rms voltage across R. Answer in units of V

7.(part 3)Find the rms voltage across L. Answer in units of V

8.(part 4)Find the rms voltage across C. Answer in units of V

9.(part 1)A 0.353 H inductor is connected to a capacitor and a 23.5 ? resistor along with a 28.6 Hz, 14.1 V generator.

To what value would the capacitor have to be adjusted to produce resonance?

Answer in units of ?F

10.(part 2)Find the voltage drop across the resistor at resonance.

Answer in units of V

Explanation / Answer

inductive reactance=2*pi*922*0.091=527.17 ohms

capacitive reactance=1/(2*pi*922*0.2*10^(-6))=863.1 ohms

so net impedance=57+j(527.17-863.1)=57-j335.92 ohms

so maximum voltage=190/(magnitude of net impedance)=0.557 A

rms value=max voltage/sqrt(2)=0.394 A

6)voltage across R=0.394*57=22.458 volts

7)voltage across of L=0.394*527.17=207.7 volts

8)voltage across C=0.394*335.92=132.35 volts

9)resonant frequency=1/(2*pi*sqrt(L*C))

so C=87.72 uF

10) voltage=14.1 volts(as impedance gets cancelled between capacitor and inductor)

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