In a series R-L-C circuit, the components have the following value, L= 60 mH, C=

Question

In a series R-L-C circuit, the components have the following value, L= 60 mH, C= 140 nF, and R= 60 ohms. The generator has an rms voltage of 120 V and a frequency of 1502 Hz.

In a series R-L-C circuit, the components have the following value, L 60 mH, C 140 nF, and R 60 2. The generator has an rms voltage of 120 V and a frequency of 1502 Hz (note: 21t 9. Find the impedance of the circuit (in 2) A) 50 E) 200 B) 60 C) 140 D) 120 10. Determine the phase angle, between the voltage and the current in degrees E) -79.5 D) -72.5 C) -50.0 A) 15.5 B) 22.5 11. If the frequency is tunable, at certain value, the impedance will reach the minimum at (in 2) C) 140 D) 120 E) 200 A) 50 (B) 60 12. AM0911 the minimum, the (resonant) frequency is (in Hz) B) 1502 C) 3004 D) 6008 en Z reaches E) 1737 13. The physics meaning of minus sign at (10) is, in the circuit, B) I leads V C) Lenz's Law D LV in phase E) resonance

Explanation / Answer

9) Impedence is given by

Z = sqrt [(Xl - Xc)^2 + R^2]

Xl = 2 pi f L = 2 x 3.14 x 1502 x 60 x 10^-3 = 565.95 Ohm

Xc = 1/2 pi f C = 1/2 x 3.14 x 1502 x 140 x 10^-9 = 757.26 Ohm

Xl - Xc = 565.95 - 757.26 = -191.31 Ohm

Z = sqrt [-191.31^2 + 60^2] = 200 Ohm

Hence, Z = 200 Ohm (E)200

10)phi = tan-1 [(Xl - Xc)/R]

phi = tan-1 (-191.31/60) = -72.5 deg

Hence, phi = -72.5 deg (D)-72.5 deg

12)Z will be min when

Xl = Xc

2 pi f L = 1/ 2 pi f C

f = 1/2 pi sqrt (1/LC)

f = 1/2 x 3.14 x sqrt (1/60 x 10^-3 x 140 x 10^-9) = 1737 Hz

Hence, f = 1737 Hz

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