Use the information given in the figure for the series RCL circuit to determine

Question

Use the information given in the figure for the series RCL circuit to determine its total impedance Z. (L = 0.035 H, R = 150 Ohm, C = 1.75 mu F, f = 60 H_x) A) 300 Ohm B) 500 Ohm C) 1500 Ohm D) 1700 Ohm E) 1900 Ohm When a resistor is connected by itself to an ac generator, the average power delivered to it is P_ave Gen = 1.000 W (1 W). When a capacitor is added in series with the resistor, the power delivered is _ave.C = 0.500 W90.5W. When an inductor is added in series with the resistor only, (without the capacitor), the power delivered is P_ave.L = 0.250 W. Find the power delivered when all the three elements, the capacitor, inductor and the resistor are connected, in series, to the same generator.

Explanation / Answer

when capacitor is connected in series with Resistor

P2 = Erms*Irms*R/sqrt(R^2+Xc^2)

P2 = P1*R/sqrt(R^2+Xc^2)

0.5 = 1*R/(sqrt(R^2+Xc)

4*R^2 = R^2 + Xc^2

Xc^2 = 3*R^2

Xc = R*sqrt3

when Inductor is connected in series with Resistor

P3 = Erms*Irms*R/sqrt(R^2+XL^2)

0.25 = 1*R/sqrt(R^2+XL^2)

sqrt(R^2+XL^2) = 4R

R^2 + XL^2 = 16R^2

XL^2 = 15*R^2

XL = R*sqrt15

for RLC circuit

P = Erms*Irms*R/sqrt(R^2+(XL-Xc)^2)

P = 1*R/sqrt(R^2 + (R^2*(sqrt15-sqrt3)^2))

P = 1/sqrt(1+((sqrt15-sqrt3)^2)

P = 0.423 W <<<-----answer

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