Problem 33 - University Physics with Modern Physics

Question

A series circuit consists of an ac source of variable frequency, a \(115-\Omega\) resistor, a \(1.25-\mu F\) 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; (c) half the resonance angular frequency.

Step-by-Step Solution

Answer

(a) 115 Ω; (b) 287 Ω; (c) 1085.5 Ω.

Step 1: Define the Impedance Formula

The impedance \( Z \) of a series RLC circuit is given by \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where:- \( R \) is the resistance,- \( X_L = \omega L \) is the inductive reactance,- \( X_C = \frac{1}{\omega C} \) is the capacitive reactance, - \( \omega \) is the angular frequency.To find \( Z \), we need to calculate \( X_L \) and \( X_C \) based on the given \( \omega \).

Step 2: Find the Resonance Angular Frequency

The resonance angular frequency \( \omega_0 \) is calculated using \( \omega_0 = \frac{1}{\sqrt{LC}} \). Here, \( L = 4.50 \times 10^{-3} \mathrm{H} \) and \( C = 1.25 \times 10^{-6} \mathrm{F} \).Plug these values into the formula:\[ \omega_0 = \frac{1}{\sqrt{4.50 \times 10^{-3} \times 1.25 \times 10^{-6}}} \approx 1333.33 \text{ rad/s} \]

Step 3: Calculate Impedance at Resonance Frequency

At resonance, \( \omega = \omega_0 \). The impedance \( Z \) simplifies to just \( R \) as the inductive and capacitive reactances are equal:\[ X_L = \omega_0 L = X_C = \frac{1}{\omega_0 C} \]Thus, \( Z = R = 115 \Omega \).

Step 4: Calculate Impedance at Twice Resonance Frequency

Set \( \omega = 2\omega_0 = 2 \times 1333.33 = 2666.66 \text{ rad/s} \).Calculate \( X_L = \omega L = 2666.66 \times 4.50 \times 10^{-3} \approx 12.0 \Omega \).Calculate \( X_C = \frac{1}{\omega C} = \frac{1}{2666.66 \times 1.25 \times 10^{-6}} \approx 300 \Omega \).Calculate the total impedance:\[ Z = \sqrt{115^2 + (12.0 - 300)^2} \approx 287 \Omega \]

Step 5: Calculate Impedance at Half Resonance Frequency

Set \( \omega = \frac{1}{2} \omega_0 = 0.5 \times 1333.33 = 666.66 \text{ rad/s} \).Calculate \( X_L = \omega L = 666.66 \times 4.50 \times 10^{-3} \approx 3.0 \Omega \).Calculate \( X_C = \frac{1}{\omega C} = \frac{1}{666.66 \times 1.25 \times 10^{-6}} \approx 1200 \Omega \).Calculate the total impedance:\[ Z = \sqrt{115^2 + (3.0 - 1200)^2} \approx 1085.5 \Omega \]

Key Concepts

Resonance Frequency

In an RLC circuit, the resonance frequency is a special frequency at which the circuit's impedance is minimized, effectively resulting in the circuit behaving purely resistively. At this frequency, the inductive reactance (\(X_L\)) and capacitive reactance (\(X_C\)) are equal in magnitude but opposite in phase, causing them to cancel each other out. This is because the

• reactance of the inductor and the capacitor are completely out of phase by 180 degrees,
• making the net reactance zero, and
• leaving only the resistance \(R\) to contribute to the impedance.

At the resonance frequency \(\omega_0\), given by the formula \(\omega_0 = \frac{1}{\sqrt{LC}}\), the circuit reaches its maximum energy transfer, with the impedance \(Z\) reducing to just the resistance \(R\). For our exercise example, this resonance angular frequency is approximately 1333.33 rad/s.

Inductive Reactance

Inductive reactance \(X_L\) is the opposition that an inductor presents to the change in current flow in an AC circuit. It is frequency-dependent and is given by the formula \(X_L = \omega L\), where \(\omega\) is the angular frequency and \(L\) is the inductance of the inductor. As frequency increases:

• The inductive reactance increases because the opposing effect of the inductor on the AC changes becomes stronger,
• This means that at higher frequencies, the inductor resists the flow of the alternating current more effectively.

In our step-by-step solution, calculating \(X_L\) at different frequencies helps determine the total impedance of the circuit by understanding how strongly the inductor opposes the AC current.

Capacitive Reactance

Capacitive reactance \(X_C\), in contrast to inductive reactance, is the opposition presented by a capacitor to the change in current flow in an AC circuit. It is calculated using the formula \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the angular frequency and \(C\) is the capacitance of the capacitor. As frequency increases:

• The capacitive reactance decreases because the capacitor allows higher frequencies to pass more freely,
• This effect explains why capacitors are used in applications to allow AC signals to pass while blocking DC.

The balance between \(X_L\) and \(X_C\) at different frequencies dictates the overall impedance of the RLC circuit.

Angular Frequency

Angular frequency \(\omega\) is a measure of how quickly the AC voltage or current oscillates in the circuit and is measured in radians per second (rad/s). It provides a direct link between the time domain and the behavior of the circuit elements:

• A larger \(\omega\) implies a higher frequency of oscillation, which influences the reactance values of both inductive and capacitive components.
• The relationship \(\omega = 2\pi f\) connects angular frequency to the standard frequency \(f\), measured in Hertz (Hz).

By adjusting \(\omega\), one can determine how different frequencies affect the impedance of the entire RLC circuit, as seen in the examples where impedance changes significantly at resonance, twice the resonance, and half the resonance frequencies.