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An RLC series circuit has a 40.0 2 resistor, a 3.0o mH inductor, and a 5.00 µF capacitor. (a) Find the circuit's impedance at 60.0 Hz and 10.o kHz, noting that these frequencies and the values for L and C are the same as in Example 23.10 and Example 23.11. (b) If the voltage source has Vrms = 120 V, what is Ims at each frequency? Strategy For each frequency, we use Z = vR²+(XL – Xc)² to find the impedance and then Ohm's law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again. Solution for (a) At 60.0 Hz, the values of the reactances were found in Example 23.10 to be X1, = 1.13 Q and in Example 23.11 to be Xc = 531 Q. Entering these and the given 40.0 2 for resistance into Z = VR2+(XL – Xc)² yields Z = V R? +(X, — Хc)? (40.0N)²+(1.13 Q– 531 N)² 23.67 531 Q at 60.0 Hz. Similarly, at 10.o kHz, X1, = 188 Q and Xc = 3.18 Q, so that V(40.0 2)2+(188 Q – 3.18 N)² 23.68 190 Ω at 10.0 kHz.

An RLC series circuit has a 40.0 2 resistor, a 3.0o mH inductor, and a 5.00 µF capacitor. (a) Find the circuit's impedance at 60.0 Hz and 10.o kHz, noting that these frequencies and the values for L and C are the same as in Example 23.10 and Example 23.11. (b) If the voltage source has Vrms = 120 V, what is Ims at each frequency? Strategy For each frequency, we use Z = vR²+(XL – Xc)² to find the impedance and then Ohm's law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again. Solution for (a) At 60.0 Hz, the values of the reactances were found in Example 23.10 to be X1, = 1.13 Q and in Example 23.11 to be Xc = 531 Q. Entering these and the given 40.0 2 for resistance into Z = VR2+(XL – Xc)² yields Z = V R? +(X, — Хc)? (40.0N)²+(1.13 Q– 531 N)² 23.67 531 Q at 60.0 Hz. Similarly, at 10.o kHz, X1, = 188 Q and Xc = 3.18 Q, so that V(40.0 2)2+(188 Q – 3.18 N)² 23.68 190 Ω at 10.0 kHz.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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AnLC circuit consists of a 3.00 mH inductor and a 5.00 μF capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z with those found in Example 23.12 in which there was also a resistor.

An RLC series circuit has a 40.0 2 resistor, a 3.0o mH inductor, and a 5.00 µF capacitor. (a) Find the circuit's impedance at 60.0 Hz and 10.o kHz, noting that these frequencies and the values for L and C are the same as in Example 23.10 and Example 23.11. (b) If the voltage source has Vrms = 120 V, what is Ims at each frequency? Strategy For each frequency, we use Z = vR²+(XL – Xc)² to find the impedance and then Ohm's law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again. Solution for (a) At 60.0 Hz, the values of the reactances were found in Example 23.10 to be X1, = 1.13 Q and in Example 23.11 to be Xc = 531 Q. Entering these and the given 40.0 2 for resistance into Z = VR2+(XL – Xc)² yields Z = V R? +(X, — Хc)? (40.0N)²+(1.13 Q– 531 N)² 23.67 531 Q at 60.0 Hz. Similarly, at 10.o kHz, X1, = 188 Q and Xc = 3.18 Q, so that V(40.0 2)2+(188 Q – 3.18 N)² 23.68 190 Ω at 10.0 kHz.
Transcribed Image Text:An RLC series circuit has a 40.0 2 resistor, a 3.0o mH inductor, and a 5.00 µF capacitor. (a) Find the circuit's impedance at 60.0 Hz and 10.o kHz, noting that these frequencies and the values for L and C are the same as in Example 23.10 and Example 23.11. (b) If the voltage source has Vrms = 120 V, what is Ims at each frequency? Strategy For each frequency, we use Z = vR²+(XL – Xc)² to find the impedance and then Ohm's law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again. Solution for (a) At 60.0 Hz, the values of the reactances were found in Example 23.10 to be X1, = 1.13 Q and in Example 23.11 to be Xc = 531 Q. Entering these and the given 40.0 2 for resistance into Z = VR2+(XL – Xc)² yields Z = V R? +(X, — Хc)? (40.0N)²+(1.13 Q– 531 N)² 23.67 531 Q at 60.0 Hz. Similarly, at 10.o kHz, X1, = 188 Q and Xc = 3.18 Q, so that V(40.0 2)2+(188 Q – 3.18 N)² 23.68 190 Ω at 10.0 kHz.
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