Please answer Q1 and Q2 A series L - R - C circuit consists of a 209 ohm resisto

Question

Please answer Q1 and Q2

A series L - R - C circuit consists of a 209 ohm resistor, a 7.8 mH inductor, a 10.61 muF capacitor, and an AC source of amplitude 41 V and variable frequency. If the voltage source is set to 2.6 times the resonance frequency, the impedance of this circuit is: Express the answer as a whole number. An AC source whose rms voltage is 75 V is in series with a 93 ohm resistor and a capacitor whose reactance Xc is 282 ohm at the frequency of the source. In the figure, the rms voltage (as a whole number) across the capacitor is:

Explanation / Answer

W = 1/sqrt(LC)

W = 1/sqrt(0.078 * 10.61uF)

W =3476.121 rad/s

w= 2pif = 3476.121

W new = 2.6 *3476.121 = 9037.91 rad/s

so inductive reacatance XL = wL = 9037.91 *0.078 = 70.49 ohms

capacitive reactantcne Xc = 1/wC =1/9037.91*10.61 = 10.42 ohms

now Impedence Z^2 = R^2+(Xl-Xc)^2

Z^2 = 209^2 +(70.49-10.42)^2

Z = 217.45 ohms

b. impedence Z^2 = 93^2 + 282^2

Z= 296.93 ohms

i = v/Z = 75/296.93 = 0.252 A

so Vacross C = 0.252 * 282 = 71.22 volts

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