For an RLC series circuit, R = 150N, L = 150 mH, and C = 0.35µF. (a) If an ac source of variable frequency is connected to the circuit, at what frequency is maximum power dissipated in the resistor? (b) What is the quality factor of the circuit? Hint a. At the driving frequency Hz, maximum power is dissipated in the resistor. Additional hints for (a). b. The quality factor is Additional hints for (b)

For an RLC series circuit, R = 150N, L = 150 mH, and C = 0.35µF. (a) If an ac source of variable frequency is connected to the circuit, at what frequency is maximum power dissipated in the resistor? (b) What is the quality factor of the circuit? Hint a. At the driving frequency Hz, maximum power is dissipated in the resistor. Additional hints for (a). b. The quality factor is Additional hints for (b)

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter15: Alternating-current Circuits

Section: Chapter Questions

Problem 33P: For an RLC series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz,...

See similar textbooks

Similar questions

Calculate the rms currents for an ac source is given by v(t)=v0sint , where V0=100V and =200rad/s when connected across (a) a 20F capacitor, (b) a 20-mH inductor, and (c) a 50 resistor.
An ac source of voltage amplitude 10 V delivers electric energy at a rate of 0.80 W when its current output is 2.5 A. What is the phase angle between the emf and the current?
For an RLC series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively; R=500 ; and L = 0.20H . Find the average power dissipated in the resistor for the following values for the capacitance: (a) C=2.0F and (b) C=2.0F .
An ac source of voltage amplitude 100 V and variable frequency f drives an RLC series circuit with R=10 , L = 2.0 mH, and C=25F (a) Plot the canal through the resistor as a function of the frequency f. (b) Use the plot to determine the resonant frequency of the circuit.
At 1000 Hz, the reactance of a 5.0-mH inductor is equal to the reactance of a particular capacitor. What is the capacitance of the capacitor?
A 20-mH inductor is connected across an AC source with a variable frequency and a constant-voltage amplitude of 9.0 V. (a) Determine the reactance of the circuit and the maximum current through the inductor when the frequency Is set at 20 kHz. (b) Do the same calculations for a frequency of 60 Hz.
A 700-pF capacitor is connected across an ac source with a voltage amplitude of 160 V and a frequency of 20 kHz. (a) Determine the capacitive reactance of the capacitor and the amplitude of the output current of the source. (b) If the frequency is changed to 60 Hz while keeping the voltage amplitude at 160 V, what are the capacitive reactance and the current amplitude?
The emf of an ac source is given by v(t)=v0sint. where V0=100V and at =200rad/s . Calculate the average power output of the source if it is connected across (a) a 20F capacitor, (b) a 20-mH inductor, and (c) a 50 resistor.
A 1.5k resistor and 30-mH inductor are connected in series, as below, across a120-V(rms)ac power source oscillating at 60-Hz frequency. (a) Find the current in the circuit. (b) Find the voltage drops across the resistor and inductor. (C) Find the impedance of the circuit. (d) Find the power dissipated in the resistor. (e) Find the power dissipated in the inductor. (1) Find the power produced by the source.
An LC circuit in an AM tuner (in a car stereo) uses a coil with an inductance of 2.5 mH and a variable capacitor. If the natural frequency of the circuit is to be adjustable over the range 540 to 1600 kHz (the AM broadcast band), that range of capacitance is required?
What is the impedance of an RLC series circuit at the resonant frequency?
Check Your Understanding Show that the tins voltages across a resistor, a capacitor, and an inductor in an ac circuit where the tins current is Irmsare given by IrmsR,IrmsXC , and IrmsXL, respectively. Determine these values for the components of the RLC circuit of Equation 15.12. Pave=12t0v0cos (15.12)