A capacitor, a coil, and two resistors of equal resistance are arranged in an AC

Question

A capacitor, a coil, and two resistors of equal resistance are arranged in an AC circuit as shown in the figure below. An AC source provides an emf of AVrms = 14.5 V at a frequency of 60.0 Hz. When the double-throw switch S is open as shown in the figure, the rms current is 190 mA. When the switch is closed in position a, the rms current is 294 mA. When the switch is closed in position b, the rms current is 137 mA. C Determine the possible values of R. (Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.) Determine the possible values of C. (Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.) Determine the possible values of L. (Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.)

Explanation / Answer

Notice the different combinations that result from the different switch positions.In the open position, the resistor is ignored, leaving the basic series RLC circuit. In position a, there is a parallel resistor combination to the RLC circuit .

In position b, the inductor can be ignored (due to the short) leaving a RC circuit (L = 0 ===> XL = 0)

Then Z2 = (Vrms/Irms)2 = R2 + (XL - XC)2

Considering each case:

1) R2 + (XL - XC)2 = (14.5/190*10^-3)2 = 5824.1

2) Req = [1/R + 1/R]^-1 = R/2

Req2 + (XL - XC)2 = (Vrms/Irms)2

R2/4 + (XL - XC)2 = (14.5/294*10^-3)2 = 2432.44

3) XL = 0 =====> R2 + XC2 = (14.5/137*10^-3)2 = 11201.98

Solving for R, XC, and XL, we get,

from 1) and 2) =====> R2 - R2/4 = 3391.66 =====> R = 67.25 ohm

From 3) ====> XC2 = 11201.98 - 4522.21 ====> XC = 81.73 ohm

From 1) XL = XC +- sqrt[5824.1 - R2] =====> XL = 117.8, 45.65 ohm

Now XC = 1/wC =====> C = 1/wXC= 1/2pi*f*XC = 1/(2pi*60*81.73) = 3.25*10^-5 F

XL = wL ====> L = XL /w = XL /2pi*f = 0.312 H and 0.121 H

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