A series L-R-C Circuit has values of (pictures of questions listed below) A seri

Question

A series L-R-C Circuit has values of (pictures of questions listed below)

A series L R-C circuit has the following values: Z. = 185 mH C = 66.8 mu F(R = 7.00 Ohm Suppose the capacitor is initially charged up. as shown at right, and the switch is closed. What will be the natural frequency f theta, of the current oscillations in the circuit? (Assume that the value of f0, is unaffected by the resistor's light damping.) Show your work. Show your work. (9 pis.) Suppose instead that the capacitor starts uncharged, and that the circuit is continuously driven by a 10.0-V-rms alternating voltage source, as shown here. Find the rms value of the current in the circuit when it is driven at each of the following frequencies: Show your work completely. Frequency f0, frequency 1/10 f0, frequency 10f0

Explanation / Answer

a)

fo = w/2pi

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

= (1/(0.185*66.8e-6))y0.5/(2 * 3.1416)

= 45.274

= 45.3 Hz

b)

(i) fo

Z = sqrt(R^2 + (2 pi f L - 1/(2 pi f C))^2)

Z = sqrt(7*7 + (2*3.1416*45.274*0.185 - 1/(2*3.1416*45.274*66.8e-6))^2)

Z = 7 ohms

i_rms = (V_rms)/Z = 10/7 = 1.43 A

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(ii) fo/10 = 4.5274

Z = sqrt(R^2 + (2 pi f L - 1/(2 pi f C))^2)

Z = sqrt(7*7 + (2*3.1416*4.5274*0.185 - 1/(2*3.1416*4.5274*66.8e-6))^2)

Z = 521.04 ohms

i_rms = V/Z = 10/521.04 = 0.0192 A

------------------------------------------------------------------------

(ii) 10 fo = 452.74

Z = sqrt(R^2 + (2 pi f L - 1/(2 pi f C))^2)

Z = sqrt(7*7 + (2*3.1416*452.74*0.185 - 1/(2*3.1416*452.74*66.8e-6))^2)

Z = 521.04 ohms

i_rms = V/Z = 10/521.04 = 0.0192 A

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