- C For Example 23 12 Calculating Impedance And Current In Sec 23 12 Rlc Series Ac Circuits Of The Textbook Replac 1 (23.99 KiB) Viewed 22 times
- C For Example 23 12 Calculating Impedance And Current In Sec 23 12 Rlc Series Ac Circuits Of The Textbook Replac 2 (23.99 KiB) Viewed 22 times
- C For Example 23 12 Calculating Impedance And Current In Sec 23 12 Rlc Series Ac Circuits Of The Textbook Replac 3 (57.75 KiB) Viewed 22 times
- C For Example 23 12 Calculating Impedance And Current In Sec 23 12 Rlc Series Ac Circuits Of The Textbook Replac 4 (45.69 KiB) Viewed 22 times
EXAMPLE 23.12 Calculating Impedance and Current An RLC series circuit has a 40.0 22 resistor, a 3.00 mH inductor, and a 5.00 uF capacitor. (a) Find the circuit's impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for Land C are the same as in Example 23.10 and Example 23.11. (b) If the voltage source has Vms = 120 V, what is Irms at each frequency? Chapter 23 • Electromagnetic Induction, AC Circuits, and Electrical Technologies . Strategy For each frequency, we use Z = VR2+(X - Xc)2 to find the impedance and then Ohr'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 X = 1.13 and in Example 23.11 to be Xc = 53122. Entering these and the given 40.0 82 for resistance into Z = R2+(X -- Xc)2 yields Z = R2+(X2 - Xc) = V(40.02)2+(1.13 22-531 (2)2 23.67 = 531 2 at 60.0 Hz. Similarly, at 10.0 kHz, X_ = 188 S and Xc = 3.18 22, so that Z = V(40.02)2 +(188 22 - 3.18 2)2 = 1902 at 10.0 kHz. Discussion for (a) In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual 23.68
EXAMPLE 23.13 Calculating Resonant Frequency and Current For the same RLC series circuit having a 40.0 22 resistor, a 3.00 mH inductor, and a 5.00 pF capacitor:(a) Find the resonant frequency. (b) Calculate Ims at resonance if Vrms is 120 V. Strategy The resonant frequency is found by using the expression info 2XVEC . The current at that frequency is the same as if the resistor alone were in the circuit. Solution for (a) Entering the given values for L and C into the expression given for fo info = zavro yields to 2VLC = 1.30 kHz. 24V (500XTO HX5.00X10F) Discussion for (a) We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at this intermediate frequency Solution for (b) The current is given by Ohm's law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone. Thus, 23.73