Problem 23
Question
A series circuit consists of an ac source of variable frequency, a 115\(\Omega\) resistor, a 1.25\(\mu\) capacitor, and a 4.50 \(\mathrm{mH}\) inductor. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to (a) the resonance angular frequency, (b) twice the resonance angular frequency, and (c) half the resonance angular frequency.
Step-by-Step Solution
Verified Answer
(a) 115Ω, (b) 154.59Ω, (c) 155.65Ω.
1Step 1: Determine the resonance angular frequency
For a series RLC circuit, the resonance angular frequency \( \omega_0 \) is given by:\[ \omega_0 = \frac{1}{\sqrt{LC}} \]Where \( L \) is the inductance and \( C \) is the capacitance. Plug in the given values:\[ L = 4.50 \times 10^{-3} \text{ H}, \quad C = 1.25 \times 10^{-6} \text{ F} \] Calculate \( \omega_0 \):\[ \omega_0 = \frac{1}{\sqrt{(4.50 \times 10^{-3})(1.25 \times 10^{-6})}} \approx 13,413.84 \text{ rad/s} \].
2Step 2: Calculate impedance at resonance angular frequency
At the resonance frequency, the inductive reactance \( X_L \) and capacitive reactance \( X_C \) cancel each other, and the impedance \( Z \) is purely resistive:\[ Z = R \]Given \( R = 115 \Omega \), the impedance \( Z \) at resonance is:\[ Z = 115 \Omega \].
3Step 3: Calculate impedance at twice the resonance angular frequency
When the angular frequency is twice the resonance frequency (\( 2\omega_0 \)), the impedance \( Z \) is calculated using:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]Where \( X_L = 2\omega_0 L \) and\( X_C = \frac{1}{2\omega_0 C} \). Calculate each:\[ X_L = 2(13,413.84)(4.50 \times 10^{-3}) \approx 120.73 \Omega \]\[ X_C = \frac{1}{2(13,413.84)(1.25 \times 10^{-6})} \approx 29.17 \Omega \]Substitute back into the impedance formula:\[ Z = \sqrt{115^2 + (120.73 - 29.17)^2} \approx 154.59 \Omega \].
4Step 4: Calculate impedance at half the resonance angular frequency
When the angular frequency is half the resonance frequency (\( \frac{\omega_0}{2} \)), the impedance \( Z \) is also calculated using:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]Where \( X_L = \frac{\omega_0}{2} L \) and\( X_C = \frac{1}{\frac{\omega_0}{2} C} \). Calculate each:\[ X_L = \frac{1}{2}(13,413.84)(4.50 \times 10^{-3}) \approx 30.18 \Omega \]\[ X_C = \frac{1}{\frac{1}{2}(13,413.84)(1.25 \times 10^{-6})} \approx 116.68 \Omega \]Substitute back into the impedance formula:\[ Z = \sqrt{115^2 + (30.18 - 116.68)^2} \approx 155.65 \Omega \].
Key Concepts
Resonance Angular FrequencyInductive ReactanceCapacitive ReactanceSeries CircuitImpedance Calculation
Resonance Angular Frequency
In a series RLC circuit, the resonance angular frequency, denoted as \( \omega_0 \), is a crucial parameter. It defines the frequency at which the circuit's reactive impedance components—inductive and capacitive reactances—cancel each other out. This results in the circuit's impedance being purely resistive. The formula to calculate this frequency is given by:\[\omega_0 = \frac{1}{\sqrt{LC}} \]where \( L \) is the inductance and \( C \) is the capacitance.
- This relationship indicates that while the resonance angular frequency depends on both the inductor and capacitor, it does not depend on the resistor.
- The unit for resonance angular frequency is radians per second (rad/s).
Inductive Reactance
Inductive reactance, denoted as \( X_L \), refers to the opposition that an inductor offers to a change in current. This opposition is frequency-dependent, meaning it varies with the changes in the frequency of the AC source.
- The formula for calculating inductive reactance is:\[ X_L = \omega L \] where \( \omega \) is the angular frequency and \( L \) is the inductance in henrys (H).
- Inductive reactance increases linearly with frequency, meaning higher frequencies lead to larger \( X_L \).
- The unit for \( X_L \) is ohms (\( \Omega \)).
Capacitive Reactance
Capacitive reactance, represented as \( X_C \), is the measure of a capacitor's opposition to changes in voltage in an AC circuit. Like inductive reactance, capacitive reactance also depends on the frequency of the signal.
- The formula used to calculate capacitive reactance is:\[ X_C = \frac{1}{\omega C} \] where \( \omega \) is the angular frequency and \( C \) is the capacitance in farads (F).
- Capacitive reactance inversely relates to frequency, meaning that \( X_C \) decreases as the frequency increases.
- Just like \( X_L \), \( X_C \) is also measured in ohms (\( \Omega \)).
Series Circuit
In a series circuit, components are connected end-to-end, forming a single path for current to flow. This is one of the most basic types of electrical circuits.
- The current that flows through each component is the same because there is only one path available.
- In a series RLC circuit, consisting of an inductor (L), a capacitor (C), and a resistor (R), the individual voltages across each component can differ even though the current remains constant.
- This circuit type simplifies analysis and makes it easier to predict circuit behavior, especially concerning resonance.
Impedance Calculation
Calculating impedance in a series RLC circuit involves understanding how both resistive and reactive components interact.
- Impedance \( Z \) is the total opposition a circuit presents to current, combining resistance \( R \) with reactance \( X \), and is given by:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
- At resonance, \( X_L \) and \( X_C \) are equal, cancel each other, and thus \( Z = R \), purely resistive.
- When operating off-resonance, the difference \( (X_L - X_C) \) contributes to an increase in \( Z \), indicating that the circuit behaves like a more complex resistor, with increased ohmic opposition to AC.
Other exercises in this chapter
Problem 20
\(\bullet\) (a) At what angular frequency will a 5.00\(\mu\) F capacitor have the same reactance as a 10.0 \(\mathrm{mH}\) inductor? (b) If the capacitor and in
View solution Problem 21
\(\bullet\) In an \(R-L-C\) series circuit, \(R=150 \Omega, L=0.750 \mathrm{H},\) and \(C=0.0180 \mu \mathrm{F}\) . The source has voltage amplitude \(V=150 \ma
View solution Problem 24
\(\bullet\) In a series \(R-L-C\) circuit, \(R=400 \Omega, L=0.350 \mathrm{H},\) and \(C=0.0120 \mu \mathrm{F}\) (a) What is the resonance angular frequency of
View solution Problem 25
\(\bullet\) In a series \(R-L-C\) circuit, \(L=0.200 \mathrm{H}, C=80.0 \mu \mathrm{F},\) and the voltage amplitude of the source is 240 \(\mathrm{V}\) . (a) Wh
View solution