Concept explainers
A lossless LC circuit can be used to provide controlled oscillations to generate a controlled frequency for wireless communications. (a) Design an LC circuit with amplitude of 5 V and frequency of 400 kHz, where the largest possible inductor available is 400 nH. Now suppose that you have an undesired resistance of 0.2 mΩ in series with the LC oscillator. (b) Determine if, and how much, the frequency changes as a result of the resistance. (c) Determine the maximum time that the oscillator can run before the voltage amplitude decays to 4.8 V, and (d) determine the energy dissipation during this time period (it will be very useful to use software such as MATLAB for calculating the energy dissipation!).
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Chapter 9 Solutions
Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf
- Assume that the voltage drop across the resistor, ER, is 78 V; the voltage drop across the capacitor, EC, is 104 V; and the circuit has a total impedance, Z, of 20 . The frequency of the AC voltage is 60 Hz. Find the missing values. ET ER78V EC104V IT IR IC Z20 R XC VA P VARSC PF Carrow_forwardAn AC circuit contains a 24 resistor, a 15.9-mH inductor, and a 13.3F capacitor connected in parallel. The circuit is connected to a 240-V, 400-Hz power supply. Find the following values. XL=XC=IR=AIL=AIC=AP=WVARsL=VARsC=IT=AVA=PF=%=arrow_forwardDetermine if each statement is True or False; if false, please explain whya) A forced oscillator is when a system is being yelled at to perform a specific motion.b) Impedance is a measure of the total resistance a RLC series circuit has towards current.c) The phase angle tells us how “out-of-phase” the charge on the capacitor iswith the driving voltage.arrow_forward
- One steradian represents .... .... ... (rad)^2 O degree O 2rad (rad)3 Oarrow_forwardA carrier signal with frequency of 1 MHz and peak-to-peak amplitude of 10√2v is added to a single tone message with frequency of 1 kHz and peak-to-peak amplitude of 4√2v. Then passed through a diode having the following c/cs: i(t) = e(t) + 0.5e²(t), find: 1- Modulation depth of DSB signal. 2- Frequency of each component in the diode current.arrow_forwardFind the voltage across and the current through the capacitor both at =20μSec Please try some part in typing format...arrow_forward
- An AC circuit is composed of a serial connection of: a resistor with resistance 50 0, a coil with inductance 0.3 H and a capacitor with capacitance 15 µF. The circuit is connected to an AC voltage source with amplitude 25 V and frequency 50 Hz. Determine the amplitude of the electric current in the circuit and a phase difference between the voltage and the current. R=50 2 C=15 µF omH L=0.3 H U.-25 V f=50 Hz Hi guys! Here is something you can practice with. Simplify the circuit above. The final answer is RLC circuit the amplitude of the current is approximate: /m = 0.2 A. %3D The phase difference between the voltage and the current is about: p = -67°. %3Darrow_forwardExplain the working of 3 stage RC phase shift Oscillator. Design a 5 stage RC phase shift oscillator to generate a 300Hz sinusoid. Assume the capacitance used is 3pFarrow_forwardA series R-L-C circuit is connected to a 0.2 V supply and the current is at its maximum value of 4 mA when the supply frequency is adjusted to 3 kHz. The Q-factor of the circuit under these conditions is 100. Determine the value of (a) the circuit resistance, (b) the circuit inductance, (c) the circuit capacitance and (d) the voltage across the capacitor. ANS: (a) 50 2; (b) 0.265 H; (c) 10.61 nF; (d) 20 V.arrow_forward
- Discussion 1. Comment on your results. 2. Compare between the practicl and theoretical results. 3. Find Va, Ve on the figure below: 15A32 8202 R, R. 2.2KQarrow_forward4) The function sin x (which you may remember from Calculus 1 limits) is important in signal processing and electrical engineering, and is known as the "sinc" function, often abbreviated as just sinc(x). Find § sinc(x) dx. Then, approximate ſ sinc(x) dx using the first five terms of the appropriate series.arrow_forward2_53528980697... -> For the circuit shown below, detemine the voltage of the -j10 2 capacitive reactance. -j5 0 ILOA(T) ==j102 jlon is Q 102 O 0.5L-90 52 A.arrow_forward
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning