Problem 23 - College Physics

Question

A series circuit consists of an ac source of variable frequency, a \(115 \Omega\) resistor, a \(1.25 \mu \mathrm{F}\) 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

Answer

The impedance is 115 Ω at resonance, ≈ 309.72 Ω at 2ω0, and ≈ 1206.78 Ω at 0.5ω0.

Step 1: Define Impedance in a Series Circuit

Impedance (\(Z\)) in a series circuit with a resistor, capacitor, and inductor is given by the formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where \( R \) is the resistance, \( X_L = \omega L \) is the inductive reactance, and \( X_C = \frac{1}{\omega C} \) is the capacitive reactance. Here \( \omega \) is the angular frequency, \( L \) is the inductance, and \( C \) is the capacitance.

Step 2: Determine Resonance Angular Frequency

The resonance angular frequency (\( \omega_0 \)) is found using the formula: \[ \omega_0 = \frac{1}{\sqrt{LC}} \] Plugging in the given values, \( L = 4.50 \times 10^{-3} \) H and \( C = 1.25 \times 10^{-6} \) F, we get \[ \omega_0 = \frac{1}{\sqrt{4.50 \times 10^{-3} \times 1.25 \times 10^{-6}}} \approx 1333 \text{ rad/s} \]

Step 3: Calculate Impedance at Resonance Frequency

At resonance (\( \omega = \omega_0 \)), \( X_L = X_C \) which makes \( Z = R \). Therefore: \[ Z = 115 \Omega \] as given in the resistance value.

Step 4: Calculate Impedance at Twice the Resonance Frequency

For twice the resonance frequency (\( \omega = 2\omega_0 = 2 \times 1333 \text{ rad/s} = 2666 \text{ rad/s} \)), calculate \( X_L \) and \( X_C \): \[ X_L = \omega L = 2666 \times 4.50 \times 10^{-3} \approx 11.997 \Omega \] \[ X_C = \frac{1}{\omega C} = \frac{1}{2666 \times 1.25 \times 10^{-6}} \approx 301.204 \Omega \] Now find \( Z \): \[ Z = \sqrt{115^2 + (11.997 - 301.204)^2} \approx \sqrt{115^2 + (-289.207)^2} \approx 309.72 \Omega \]

Step 5: Calculate Impedance at Half the Resonance Frequency

For half the resonance frequency (\( \omega = 0.5\omega_0 = 0.5 \times 1333 \text{ rad/s} = 666.5 \text{ rad/s} \)), calculate \( X_L \) and \( X_C \): \[ X_L = \omega L = 666.5 \times 4.50 \times 10^{-3} \approx 2.999 \Omega \] \[ X_C = \frac{1}{\omega C} = \frac{1}{666.5 \times 1.25 \times 10^{-6}} \approx 1204.818 \Omega \] Now find \( Z \):\[ Z = \sqrt{115^2 + (2.999 - 1204.818)^2} \approx \sqrt{115^2 + (-1201.819)^2} \approx 1206.78 \Omega \]

Key Concepts

Understanding Impedance

Impedance is a core concept in AC circuit analysis that represents the opposition a circuit presents to the flow of alternating current. In a series circuit, impedance is not just a simple resistance. It is a combination of resistance and reactance (both inductive and capacitive) which causes the current to lag or lead the voltage. The total impedance \( Z \) in a series circuit consisting of a resistor \( R \), an inductor \( L \), and a capacitor \( C \) is calculated using the equation:

• \( Z = \sqrt{R^2 + (X_L - X_C)^2} \)

Here, \( X_L = \omega L \) is the inductive reactance and \( X_C = \frac{1}{\omega C} \) is the capacitive reactance, with \( \omega \) being the angular frequency. The magnitude of these reactances depends on the frequency and thus the total impedance changes with frequency adjustments.

Resonance Frequency

The resonance frequency in an AC series circuit is achieved when the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \). At this frequency, the circuit's reactive components cancel each other out, leaving only the resistive component to determine the impedance. Calculating the resonance angular frequency \( \omega_0 \) is done using:

• \( \omega_0 = \frac{1}{\sqrt{LC}} \)

At resonance, the impedance \( Z \) of the series circuit lowers to the value of the resistance \( R \) since the reactance components neutralize each other:

• \( Z = R \)

Thus, the circuit has zero reactive power at resonance, maximizing the current flow as dictated purely by the resistive element.

Grasping Reactance

Reactance is the component of impedance that arises from the presence of inductors and capacitors in a circuit. It plays a significant role in determining how the total impedance changes with frequency. Reactance itself is broken into two types:- **Inductive Reactance (\(X_L\))**: - Calculated by \( X_L = \omega L \), where \( \omega \) is the angular frequency and \( L \) is inductance. - It increases with frequency, signifying greater opposition to current.- **Capacitive Reactance (\(X_C\))**: - Given by \( X_C = \frac{1}{\omega C} \), where \( C \) is the capacitance. - It decreases as frequency rises, providing lesser opposition with increasing frequency.The interplay between these reactances can significantly affect the behavior of the circuit, especially at frequencies other than the resonant frequency.

Exploring Series Circuits

A series circuit is a fundamental type of electrical circuit in which components are connected end-to-end so that the same current flows through each component. In the case of an AC series circuit consisting of a resistor, an inductor, and a capacitor, the total impedance is a main factor analyzed to determine how the circuit will behave with different frequencies. Key points about series circuits include: - All components share the same current through the circuit. - The total voltage across the circuit is the sum of the voltages across each component. - Impedance in a series circuit varies with frequency, which impacts the overall current flow as demonstrated by changes in reactance. In AC series circuit analysis, understanding the way impedance is affected by each component helps predict system behavior at different frequencies. This ensures efficient electrical energy use and effective circuit design.