quest.cns.utexas.edu A 0.646 H inductor is connected in series with fluorescent

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quest.cns.utexas.edu A 0.646 H inductor is connected in series with fluorescent lamp to limit the current drawn 2 of 2 y the lamp If the combination is connected to a 43.3 Hz, 130 V line, and if the voltage across the lamp is to be 35 V, what is the current in the circuit? (The lamp is a pure resistive load.) Answer in units of A 013 10.0 points A 15 ? resistor, 11 mH inductor, and 47 ?F capacitor are connected in series to a 77 V (rms) source having variable frequency Find the heat dissipated in the circuit dur- ing one period if the operating frequency is twice the resonance frequency Answer in units of J 014 (part 1 of 2) 10.0 points In a series RLC ac circuit, the resistance is 17.2 ?, the inductance is 24.5 mH, and the capacitance is 26.5 F. The maximum poten- tial is 94.9 V, and the angular frequency is 636.62 rad/s. Calculate the maximum current in the cir- cuit Answer in units of A 015 (part 2 of 2) 10.0 points What is the power factor for the circuit?

Explanation / Answer

0.12: XL = 2 pi f L

xL = 2 x pix 43.3 x 0.646 = 175.75 ohm

Z = sqrt[ R^2 + XL^2]

V_lamp = I R = V R / sqrt[R^2 + XL^2]

35 = (130 x R) / sqrt(R^2 + 175.75^2)

0.0725 R^2 + 2238.93 = R^2

R = 49.13 hm

so I = 130 / sqrt[49.13^2 + 175.75^2]

I = 0.712 A

013: w0 = 1/sqrt(LC) = 1391 rad/s

w = 2w0 = 2781.5 rad/s

xL = w L = 30.6 ohm

Xc = 1/wC = 7.65 ohm

Z = sqrt[ R^2 + (XL - XC)^2] = 27.42 ohm

Irms= V /Z = 77/27.42 = 2.808A

Pavg = Irms^2 R = 118.3 W

and t = 2 pi / w = 2.26 x 10^-3 s

Heat dissipated = Pavg t = 0.267 J

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