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Chapter 21: Problem 55

An RLC series circuit is driven by a sinusoidal emf at the circuit's resonant frequency. (a) What is the phase difference between the voltages across the capacitor and inductor? [Hint: since they are in series, the same current \(i(t) \text { flows through them. }]\) (b) At resonance, the rms current in the circuit is \(120 \mathrm{mA}\). The resistance in the circuit is \(20 \Omega .\) What is the rms value of the applied emf? (c) If the frequency of the emf is changed without changing its rms value, what happens to the rms current? (W) tutorial: resonance)

Short Answer

Expert verified
Answer: When the frequency of the emf is changed without changing its rms value, the rms current in the circuit will decrease.

Step by step solution

01

a) Phase Difference Between Capacitor and Inductor Voltages

At the resonant frequency, the impedance of the inductor (\(j\omega L\)) and the impedance of the capacitor (\(-\frac{j}{\omega C}\)) cancel each other out. Since the imaginary parts of these impedances are equal and opposite, the voltages across the capacitor and inductor are 180° out of phase. Therefore, the phase difference between the voltages across the capacitor and inductor is 180° or π radians.
02

b) RMS Value of the Applied EMF

At the resonant frequency, the total impedance of the circuit is purely resistive, and its magnitude is equal to the resistance in the circuit. The rms value of the applied emf can be found using Ohm's Law: \(E_{rms} = I_{rms} \times R\) where \(E_{rms}\) is the rms value of the applied emf, \(I_{rms}\) is the rms current, and \(R\) is the resistance. Given the rms current (\(I_{rms}\)) is 120 mA or 0.12 A, and the resistance (\(R\)) is 20 Ohms, we can calculate the rms value of the applied emf: \(E_{rms} = 0.12 \text{ A} \times 20 \Omega = 2.4 \text{ V}\)
03

c) Change in RMS Current with EMF Frequency Change

If the frequency of the emf is changed without changing its rms value, the impedance of the inductor and capacitor will no longer perfectly cancel each other out. This means that the total impedance of the circuit will not be purely resistive, and its magnitude will be greater than the resistance in the circuit. As the impedance increases, the rms current in the circuit will decrease, according to Ohm's Law: \(I_{rms} = \frac{E_{rms}}{Z}\) where \(Z\) is the total impedance of the circuit. So, if the frequency of the emf is changed without changing its rms value, the rms current in the circuit will decrease.

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Most popular questions from this chapter

A certain circuit has a \(25-\Omega\) resistor and one other component in series with a \(12-\mathrm{V}\) (rms) sinusoidal ac source. The rms current in the circuit is 0.317 A when the frequency is \(150 \mathrm{Hz}\) and increases by \(25.0 \%\) when the frequency increases to \(250 \mathrm{Hz}\). (a) What is the second component in the circuit? (b) What is the current at $250 \mathrm{Hz} ?$ (c) What is the numerical value of the second component?
A series combination of a \(22.0-\mathrm{mH}\) inductor and a \(145.0-\Omega\) resistor are connected across the output terminals of an ac generator with peak voltage \(1.20 \mathrm{kV},\) (a) \(\mathrm{At}\) \(f=1250 \mathrm{Hz},\) what are the voltage amplitudes across the inductor and across the resistor? (b) Do the voltage amplitudes add to give the source voltage (i.e., does \(\left.V_{R}+V_{L}=1.20 \mathrm{kV}\right) ?\) Explain. (c) Draw a phasor diagram to show the addition of the voltages.
A 25.0 -mH inductor, with internal resistance of \(25.0 \Omega\) is connected to a \(110-\mathrm{V}\) ms source. If the average power dissipated in the circuit is \(50.0 \mathrm{W},\) what is the frequency? (Model the inductor as an ideal inductor in series with a resistor.)
Show that over one complete cycle, the average value of a sine function squared is \(\frac{1}{2} .\) [Hint: Use the following trigonometric identities: \(\sin ^{2} a+\cos ^{2} a=1 ; \cos 2 a=\) \(\left.\cos ^{2} a-\sin ^{2} a .\right]\)
A variable inductor can be placed in series with a lightbulb to act as a dimmer. (a) What inductance would reduce the current through a \(100-\) W lightbulb to \(75 \%\) of its maximum value? Assume a \(120-\mathrm{V}\) rms, \(60-\mathrm{Hz}\) source. (b) Could a variable resistor be used in place of the variable inductor to reduce the current? Why is the inductor a much better choice for a dimmer?
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