- 3 For The Following Questions You May Use Any Resources You Wish To Answer Them You Must Write Your Solutions By Hand 1 (33.52 KiB) Viewed 22 times
- 3 For The Following Questions You May Use Any Resources You Wish To Answer Them You Must Write Your Solutions By Hand 2 (43.27 KiB) Viewed 22 times
- 3 For The Following Questions You May Use Any Resources You Wish To Answer Them You Must Write Your Solutions By Hand 3 (52.57 KiB) Viewed 22 times
x= 4
y=11.28
3. For the following questions, you may use any resources you wish to answer them. You must write your solutions by hand, cite all your references, and show all your calculations. [a] For Example 23.9 (Calculating Characteristic Time and Current) in Sec. 23.10 (RL Circuits) of the OpenStax College Physics textbook, replace the resistance of the resistor with X ohms and the initial current with Y amperes. Then solve the example, showing all your work. For Example 23.10 (Calculating Inductive Reactance) in Sec. 23.11 (Reactance, Inductive and Capacitive) of the textbook, replace the inductance with YmH and the applied mms voltage with X volts. Then solve the example, showing all your work.
T- R 3 EXAMPLE 23.9 Calculating Characteristic Time and Current in an RL Circuit (a) What is the characteristic time constant for a 7.50 mH inductor in series with a 3.00 S2 resistor? (b) Find the current 5.00 ms after the switch is moved to position 2 to disconnect the battery, if it is initially 10.0 A. Strategy for (a) The time constant for an RL circuit is defined by r = LIR. Solution for (a) Entering known values into the expression for given in t = UR yields L 7.50 mH = 2.50 ms. 23.48 3.00 Ω Discussion for (a) This is a small but definitely finite time. The coil will be very close to its full current in about ten time constants, or about 25 ms. Strategy for (b) We can find the current by using 1 = loete, or by considering the decline in steps. Since the time is twice the characteristic time, we consider the process in steps. Solution for (b) In the first 2.50 ms, the current declines to 0.368 of its initial value, which is 1 = = 0.36870 =0.368)(10.0 A) 1 3.68 A att = 2.50 ms. After another 2.50 ms, or a total of 5.00 ms, the current declines to 0.368 of the value just found. That is 1' = 0.3681 =(0.368)(3.68 A) = 1.35 A at r = 5.00 ms. Discussion for (b) 23.49 , 23.50 ha cu
23.54 EXAMPLE 23.10 Calculating Inductive Reactance and then Current (a) Calculate the inductive reactance of a 3.00 mH inductor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the mms current at each frequency if the applied mms voltage is 120 V? Strategy The inductive reactance is found directly from the expression XL = 2n/L. Once X, has been found at each frequency, Ohm's law as stated in the Equation I = V/X can be used to find the current at each frequency Solution for (a) Entering the frequency and inductance into Equation X2 = 2x/L gives X1 = 2fl = 6.28(60.0/)(3.00 mH)= 1.13 22 at 60 Hz 23.53 Similarly, at 10 kHz X = 21/l = 6.28(1.00 x 10/s)(3.00 mH)= 188 at 10 kHz. Solution for (b) The rms current is now found using the version of Ohm's law in Equation / = V/X., given the applied ims voltage is 120 V. For the first frequency, this yields 120 V X2 1.132 = 106 A at 60 Hz Similarly, at 10 kHz, V 120 V 1 0.637 A at 10 kHz 23:56 1882 X Discussion The inductor reacts very differently at the two different frequencies. At the higher frequency, its reactance is large and the current is small consistent with how an inductor impedes rapid change. Thus high frequencies are impeded the most Inductors can be used to filter out high frequencies for example, a large inductor can be put in series with a sound reproduction cystem or in series with your home computer to reduce high-frequency sound output from your speakers or high-frequency power spikes into your computer. V 23.55 Note that although the resistance in the circuit considered is negligible, the AC current is not extremely large because inductive reactance impedes its flow. With AC, there is no time for the current to become extremely large