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An inductor, a capacitor, and a resistor are all connected in series across an ac source. If the resistance, inductance, and capacitance are all doubled, by what factor does each of the following quantities change? Indicate whether they increase or decrease: (a) the resonance angular frequency; (b) the inductive reactance; (c) the capacitive reactance. (d) Does the impedance double?

Short Answer

Expert verified
(a) Decreases by \( \frac{1}{\sqrt{2}} \), (b) Doubles, (c) Halves, (d) No, impedance doesn't double.

Step by step solution

01

Understanding Resonance Angular Frequency

The resonance angular frequency \( \omega_0 \) for an RLC circuit is given by the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \). If both the inductance \( L \) and capacitance \( C \) are doubled, the new resonance frequency becomes \( \omega_0' = \frac{1}{\sqrt{2L \cdot 2C}} = \frac{1}{2} \cdot \frac{1}{\sqrt{LC}} = \frac{1}{2} \omega_0 \). Thus, the resonance angular frequency decreases by a factor of \( \frac{1}{\sqrt{2}} \).
02

Analyzing Inductive Reactance Change

Inductive reactance \( X_L \) is given by \( X_L = \omega L \). If the inductance \( L \) is doubled, then \( X_L' = \omega (2L) = 2X_L \). Hence, the inductive reactance doubles.
03

Analyzing Capacitive Reactance Change

Capacitive reactance \( X_C \) is given by \( X_C = \frac{1}{\omega C} \). If the capacitance \( C \) is doubled, the new reactance becomes \( X_C' = \frac{1}{\omega (2C)} = \frac{1}{2} X_C \). Thus, the capacitive reactance is halved.
04

Evaluating Impedance Change

The impedance \( Z \) in a series RLC circuit is \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). With doubled \( R \), \( L \), and \( C \), the resistance becomes \( 2R \), the inductive reactance becomes \( 2X_L \), and the capacitive reactance changes to \( \frac{1}{2}X_C \). To find out how the impedance changes, it is required to calculate the new \( Z' = \sqrt{(2R)^2 + ((2X_L) - (\frac{1}{2} X_C))^2} \). Although individual reactances and resistances change significantly, the overall formula for \( Z \) must be evaluated to conclude if it doubles. Generally, it does not simply double.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resonance Angular Frequency
In an RLC circuit, the resonance angular frequency is crucial as it determines when the circuit oscillates most efficiently. At resonance, the impedance is minimized, and the voltage across the circuit is at its peak. The formula for the resonance angular frequency, denoted as \( \omega_0 \), is given by:\[\omega_0 = \frac{1}{\sqrt{LC}}\]Here, \( L \) stands for inductance and \( C \) for capacitance. When both inductance and capacitance are doubled, the formula changes to:\[\omega_0' = \frac{1}{\sqrt{2L \cdot 2C}} = \frac{1}{\sqrt{4LC}} = \frac{1}{2} \cdot \frac{1}{\sqrt{LC}}\]This result shows that the new resonance angular frequency is half of its original, meaning it has decreased. This decrease implies that the circuit now naturally oscillates at a lower frequency.
Inductive Reactance
Inductive reactance describes how much an inductor resists the change in current. It depends on the angular frequency \( \omega \) and the inductance \( L \). The formula for inductive reactance, \( X_L \), is:\[ X_L = \omega L\]If the inductance is doubled, the new inductive reactance becomes:\[ X_L' = \omega (2L) = 2X_L\]This shows that when the inductance is doubled, the inductive reactance also doubles. Inductors store energy in a magnetic field, and by doubling the inductance, you effectively increase the opposition to current change, leading to higher reactance.
Capacitive Reactance
Capacitive reactance measures the opposition that a capacitor presents to the change of voltage. The formula for capacitive reactance \( X_C \) is as follows:\[X_C = \frac{1}{\omega C}\]Doubling the capacitance changes the reactance to:\[X_C' = \frac{1}{\omega (2C)} = \frac{1}{2} X_C\]This indicates that the capacitive reactance is halved when capacitance is doubled. Capacitors store energy in an electric field, and more capacitance means more storage capacity, which reduces the current's opposition, leading to decreased reactance.
Impedance in RLC Circuit
Impedance in an RLC circuit combines resistance, inductive reactance, and capacitive reactance, showing how much the total circuit resists the electric current. It is represented as \( Z \) and calculated by:\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]Upon doubling \( R \), \( L \), and \( C \), the resistance becomes \( 2R \), inductive reactance becomes \( 2X_L \), and capacitive reactance becomes \( \frac{1}{2}X_C \). The new impedance is:\[Z' = \sqrt{(2R)^2 + ((2X_L) - (\frac{1}{2} X_C))^2}\]Hence, the impedance doesn't just double straightforwardly but needs to be recalculated as a whole to determine the new effect. This highlights the complexity of impedance where changes in individual parameters can affect the overall impedance in non-linear ways.

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Most popular questions from this chapter

The \(L \cdot R-C\) Parallel Circuit. A resistor, inductor, and capacitor are connected in parallel to an ac source with voltage amplitude \(V\) and angular frequency \(\omega\) . Let the source voltage be given by \(v=V \cos \omega t .\) (a) Show that the instantaneous voltages \(v_{R}, v_{L},\) and \(v_{C}\) at any instant are each equal to \(v\) and that \(i=i_{R}+i_{L}+i_{C},\) where \(i\) is the current through the source and \(i_{R},\) \(i_{L},\) and \(i_{C}\) are the currents through the resistor, the inductor, and the capacitor, respectively. (b) What are the phases of \(i_{R}, i_{L},\) and \(i_{C}\) with respect to \(v ?\) Use current phasors to represent \(i, i_{R}, i_{L},\) and \(i_{C}\) In a phasor diagram, show the phases of these four currents with respect to \(v\) . (c) Use the phasor diagram of part (b) to show that the current amplitude \(I\) for the current \(i\) through the source is given by \(I=\sqrt{I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}}\) (d) Show that the result of part (c) can be written as \(I=V / Z,\) with \(1 / Z=\sqrt{1 / R^{2}+(\omega C-1 / \omega L)^{2}}\)

A Step-Up Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply \(13,000 \mathrm{V}(\mathrm{rms})\) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit; exceeds 8.50 \(\mathrm{mA}\) . (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 \(\mathrm{mA}\) ? (c) What current rating should the fuse in the primary circuit have?

Analyzing an \(L \cdot R-C\) Circuit. You have a \(200-\Omega\) resistor, a \(0.400-\mathrm{H}\) inductor, a \(5.00-\mu \mathrm{F}\) capacitor, and a variable-frequency ac source with an amplitude of 3.00 \(\mathrm{V} .\) You connect all four elements together to form a series circuit. (a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency? (b) What will be the current amplitude at an angular frequency of 400 \(\mathrm{rad} / \mathrm{s} ?\) At this frequency, will the source voltage lead or lag the current?

An \(L-R-C\) series circuit is connected to an ac source of constant voltage amplitude \(V\) and variable angular frequency \(\omega\) (a) Show that the current amplitude, as a function of \(\omega,\) is $$I=\frac{V}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}}$$ (b) Show that the average power dissipated in the resistor is $$P=\frac{V^{2} R / 2}{R^{2}+(\omega L-1 / \omega C)^{2}}$$ (c) Show that \(I\) and \(P\) are both maximum when \(\omega=1 / \sqrt{L C},\) the resonance frequency of the circuit. (d) Graph \(P\) as a function of \(\omega\) for \(V=100 \mathrm{V}, R=200 \Omega, L=2.0 \mathrm{H},\) and \(C=0.50 \mu \mathrm{F}\) . Compare to the light purple curve in Fig. \(31.19 .\) Discuss the behavior of \(I\) and \(P\) in the limits \(\omega=0\) and \(\omega \rightarrow \infty\) .

In an \(L-R-C\) series circuit, the components have the following values: \(L=20.0 \mathrm{mH}, C=140 \mathrm{nF},\) and \(R=350 \Omega .\) The generator has an rms voltage of 120 \(\mathrm{V}\) and a frequency of 1.25 \(\mathrm{kHz}\) . Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

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