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Chapter 31: Problem 46

At a frequency \(\omega_1\) the reactance of a certain capacitor equals that of a certain inductor. (a) If the frequency is changed to \(\omega_2\) = \(2\omega_1\), what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (b) If the frequency is changed to \(\omega3\)= \(\omega_1\)/3 , what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (c) If the capacitor and inductor are placed in series with a resistor of resistance R to form an \(L-R-C\) series circuit, what will be the resonance angular frequency of the circuit?

Short Answer

Expert verified
(a) Ratio: 4, Inductor has larger reactance; (b) Ratio: 1/9, Capacitor has larger reactance; (c) Resonance frequency is \( \omega_1 \).

Step by step solution

01

Determine initial condition at frequency \( \omega_1 \)

The reactance of a capacitor is given by \( X_C = \frac{1}{\omega C} \), and for an inductor it is \( X_L = \omega L \). At \( \omega_1 \), the reactance of both the capacitor and inductor are equal, so: \[ X_C = X_L \Rightarrow \frac{1}{\omega_1 C} = \omega_1 L. \] This gives the initial condition, \( \omega_1^2 = \frac{1}{LC} \).
02

Analyze reactance at frequency \( \omega_2 = 2\omega_1 \)

When the frequency is changed to \( \omega_2 = 2\omega_1 \), the reactance of the inductor is \( X_L' = \omega_2 L = 2\omega_1 L \), whereas the reactance of the capacitor is \( X_C' = \frac{1}{\omega_2 C} = \frac{1}{2\omega_1 C} \). The ratio of the inductive to capacitive reactance is: \[ \frac{X_L'}{X_C'} = \frac{2\omega_1 L}{\frac{1}{2\omega_1 C}} = 4 \cdot \omega_1^2 LC. \] Given \( \omega_1^2 = \frac{1}{LC} \), the ratio is \( 4 \). The inductive reactance is larger.
03

Analyze reactance at frequency \( \omega_3 = \frac{\omega_1}{3} \)

When the frequency is changed to \( \omega_3 = \frac{\omega_1}{3} \), the reactance of the inductor is \( X_L'' = \omega_3 L = \frac{\omega_1}{3} L \), and the reactance of the capacitor is \( X_C'' = \frac{1}{\omega_3 C} = 3\frac{1}{\omega_1 C} \). The ratio of the inductive to capacitive reactance is: \[ \frac{X_L''}{X_C''} = \frac{\frac{\omega_1}{3} L}{3\frac{1}{\omega_1 C}} = \frac{1}{9} \cdot \omega_1^2 LC. \] Again, using \( \omega_1^2 = \frac{1}{LC} \), the ratio is \( \frac{1}{9} \). The capacitive reactance is larger.
04

Calculate the resonance frequency of the L-R-C circuit

For an L-R-C series circuit, the resonance angular frequency is defined as \( \omega = \frac{1}{\sqrt{LC}} \). From the initial condition, we found that \( \omega_1^2 = \frac{1}{LC} \), which shows that at \( \omega = \omega_1 \), the circuit will be at resonance.

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Key Concepts

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

Reactance in L-R-C Circuits
Reactance is a key concept in understanding how circuits containing capacitors and inductors behave with changing frequencies. It is important to note that reactance is different from resistance. While resistance is a measure of the opposition to the flow of direct current, reactance deals with alternating currents and revolves around how capacitors and inductors store and release energy.
Reactance is frequency-dependent, which can be illustrated through the formulas for capacitive and inductive reactance:
  • Capacitive reactance, given by the formula \( X_C = \frac{1}{\omega C} \), shows that as the frequency \( \omega \) increases, the capacitive reactance decreases.
  • Inductive reactance, given by \( X_L = \omega L \), increases with frequency, meaning that as frequency goes up, the inductive reactance also goes up.
Understanding these relationships helps us analyze changes in reactance with changes in frequency, which is essential in step-by-step problems like calculating the ratio of reactance at different frequencies.
Resonance Frequency in L-R-C Circuits
In an L-R-C circuit, resonance is a special condition where the capacitive and inductive reactances are equal in magnitude, but opposite in phase. This means they cancel each other out, resulting in minimal impedance from the reactance, and the circuit behaves like a simple resistor circuit.
Resonance is identified by the resonance angular frequency \( \omega_0 \), which is given by:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]
This equation shows that the resonance frequency is determined by the values of the inductance \( L \) and capacitance \( C \). At this frequency, the voltage and current in the circuit are in phase, enhancing the circuit's efficiency.
Many practical applications, such as tuning radios and filtering signals, rely on resonance. Understanding the condition for resonance allows engineers and physicists to design systems that maximize the desired outcomes.
The Roles of Capacitors and Inductors
Both capacitors and inductors play crucial roles in L-R-C circuits, each affecting the circuit in unique ways.
  • Capacitors: Store energy in an electric field and resist changes in voltage. Their opposition to current flow is measured by capacitive reactance \( X_C = \frac{1}{\omega C} \).
  • Inductors: Store energy in a magnetic field and resist changes in current. Inductive reactance \( X_L = \omega L \) quantifies their opposition to current flow.
These components determine how the circuit's reactance changes with frequency. By combining capacitors and inductors, designers can create circuits that selectively filter frequencies, control the timing of signals, or match impedances.
So, understanding the individual characteristics and behavior of capacitors and inductors is critical to mastering complex electrical circuits.

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Most popular questions from this chapter

A resistor, an inductor, and a 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 each of the instantaneous voltages \(v_R\) , \(v_L\), and \(v_C\) at any instant is equal to \(v\) and that \(i\) = \(i_R\) + \(i_L\) + \(i_C\), where i is the current through the source and iR , iL, and iC are the currents through the resistor, inductor, and 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, iR , iL, and iC. 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 \(I =\sqrt{I_{R ^2} + (I_C - I_L)^2}\). (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 transformer connected to a 120-V (rms) ac line is to supply 13,000 V (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 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 mA? (c) What current rating should the fuse in the primary circuit have?

(a) Show that for an \(L-R-C\) series circuit the power factor is equal to R/Z. (b) An \(L-R-C\) series circuit has phase angle -31.5\(^\circ\). The voltage amplitude of the source is 90.0 V. What is the voltage amplitude across the resistor?

A large electromagnetic coil is connected to a 120-Hz ac source. The coil has resistance 400 \(\Omega\), and at this source frequency the coil has inductive reactance 250 \(\Omega\). (a) What is the inductance of the coil? (b) What must the rms voltage of the source be if the coil is to consume an average electrical power of 450 W?

A toroidal solenoid has 2900 closely wound turns, cross-sectional area 0.450 cm\(^2\), mean radius 9.00 cm, and resistance \(R\) = 2.80 \(\Omega\). Ignore the variation of the magnetic field across the cross section of the solenoid. What is the amplitude of the current in the solenoid if it is connected to an ac source that has voltage amplitude 24.0 V and frequency 495 Hz?
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