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 Vrms = 16.0 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 189 mA. When the switch is closed in position a, the rms current is 295 mA. When the switch is closed in position b, the rms current is 137 mA.

(a) Determine the possible values of R. (Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.)



(b) Determine the possible values of C. (Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.)
µF
µF

(c) Determine the possible values of L. (Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.)
mH
mH

Explanation / Answer

Xc = 1 / 2pi fC = 1/ (2 pi x 60 x C)

and xL = 2pi f L

when switch is open.

Impedance of circuit:

Z1 = R + j(XL - Xc)

|Z1| = sqrt(R^2 + (XL - Xc)^2)

|Z1| = Vrsm / Irms = 16 / 0.189 = 84.66 ohm

R^2 + (XL - XC)^2 = 7167.32 ........(i)

when switch is close at b:

now iductor is short circuited.

hence Z3 = R - jXc

sqrt(R^2 + Xc^2) = 16/0.137 = 116.79

R^2 + Xc^2 = 13639.5 .........(ii)

when switch is closed at a:

R' = R/2

sqrt((R/2)^2 + (XL - Xc)^2 ) = 16 / 0.295 = 54.24

R^2 / 4 + (XL - Xc)^2 = 2941.68 ........(iii)

a) (i) - (iii)

R^2 - R^2/4 = 7167.32 - 2941.68

R = 75.06 ohm

b) putting in (ii)

75.06^2 + Xc^2 = 13639.5

Xc = 89.5 ohm

1/(2 x pi x 60 x C) = 89.5

C = 29.65 x 10^-6 F = 29.65 uF

c) putting values in (i)

75.06^2 + (XL - 89.5)^2 = 7167.32

XL - 89.5 = +_39.16

XL = 128.66 ohm Or 50.34 ohm

XL = 2 x pi x 60 x L = 128.66 ohm Or 50.34 ohm

L = 0.1335 Or 0.3413 H

L = 133.5 mH Or 341.3 mH

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