Problem 34 - University Physics with Modern Physics

Question

In an \(L-R-C\) series circuit, \(L=0.280 \mathrm{H}\) and \(C=\) 4.00\(\mu \mathrm{F}\) . The voltage amplitude of the source is 120 \(\mathrm{V}\) . (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 A. What is the resistance \(R\) of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Step-by-Step Solution

Answer

(a) \(9.49 \times 10^3 \) rad/s, (b) \(70.59\) ohms, (c) Peak voltages: Resistor \(120\) V, Inductor \(4.52 \times 10^3\) V, Capacitor \(4.52 \times 10^3\) V.

Step 1: Calculate the Resonance Angular Frequency

In an L-C-R series circuit, the resonance angular frequency \( \omega_0 \) is given by the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \). Substitute the given values \( L = 0.280 \text{ H} \) and \( C = 4.00 \times 10^{-6} \text{ F} \) into this formula.\[ \omega_0 = \frac{1}{\sqrt{0.280 \times 4.00 \times 10^{-6}}} \]Calculating this gives:\[ \omega_0 \approx 9.49 \times 10^3 \text{ rad/s} \]

Step 2: Find the Resistance R

At resonance, the impedance of the circuit is purely resistive, meaning the resistance \( R \) is the same as the total impedance. Using Ohm's law \( V = I \times R \), where the voltage \( V = 120 \text{ V} \) and the current \( I = 1.70 \text{ A} \), solve for \( R \):\[ R = \frac{V}{I} = \frac{120}{1.70} \]Calculating this gives:\[ R \approx 70.59 \text{ ohms} \]

Step 3: Calculate Peak Voltages Across Components at Resonance

At resonance, the peak voltage across the inductor and capacitor are equal, and the peak voltage across the resistor \( V_R \) is given by \( V_R = I \times R \). Substituting the known values:For the resistor:\[ V_R = 1.70 \times 70.59 \approx 120 \text{ V} \]For the inductor and capacitor:Use \( V_L = I \times X_L \) and \( V_C = I \times X_C \), with \( X_L = L\omega_0 \) and \( X_C = \frac{1}{C\omega_0} \).\[ V_L = 1.70 \times (0.280 \times 9.49 \times 10^3) \approx 4.52 \times 10^3 \text{ V} \]\[ V_C = 1.70 \times \left(\frac{1}{4.00 \times 10^{-6} \times 9.49 \times 10^3}\right) \approx 4.52 \times 10^3 \text{ V} \]

Key Concepts

Resonance Angular Frequency

In an L-R-C series circuit, understanding resonance angular frequency is key. The concept of resonance in electrical circuits involves the condition where reactance from components like inductors and capacitors cancel each other out. This happens at a specific frequency, which is termed as resonance frequency. At this point, the impedance is minimized, allowing maximum voltage across the circuit.
To calculate the resonance angular frequency, use the formula:

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

This formula arises because at resonance, the inductive reactance \( \omega L \) equals the capacitive reactance \( \frac{1}{\omega C} \), simplifying to the above expression. Plug in values for the inductance \( L = 0.280 \text{ H} \) and capacitance \( C = 4.00 \times 10^{-6} \text{ F} \) to find:

• \( \omega_0 \approx 9490 \text{ rad/s} \)

Thus, this frequency marks the equilibrium where maximum energy transfer occurs with negligible impedance from reactance.

Ohm's Law

Ohm's Law is fundamental in understanding electrical circuits. It's a simple equation expressing the relationship between voltage, current, and resistance as:

• \( V = IR \)

In circuit analysis, it's used to find any of the three variables if the other two are known. For our L-R-C circuit at resonance, the impedance becomes purely resistive, which simplifies our circuit greatly since inductive and capacitive reactances cancel each other.
Using the known values, where the current amplitude \( I = 1.70 \text{ A} \) and the voltage amplitude \( V = 120 \text{ V} \), we calculate the resistance \( R \) as:

• \( R = \frac{V}{I} = \frac{120}{1.70} \approx 70.59 \Omega \)

This shows how Ohm's Law assists in solving for unknown values in a circuit, helping us to understand the condition of 'pure resistive' behavior at resonance.

Impedance

Impedance is a pivotal concept in AC electrical circuits, including L-R-C series circuits. It's like resistance but in the realm of alternating current, containing both resistive and reactive components.

• The total impedance in the circuit is denoted by \( Z \).

At resonance, the impedance is purely resistive, meaning the effects of inductance \( L \) and capacitance \( C \) are neutralized. This causes impedance to simplify to the resistance \( R \), making calculations easier.

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

Where \( X_L = L\omega \) is the inductive reactance and \( X_C = \frac{1}{C\omega} \) is the capacitive reactance. At the resonance angular frequency, \( X_L \) equals \( X_C \), minimization makes \( Z = R \).
The peculiar nature of impedance at resonance helps in maximizing energy transfer across the circuit, evident through big swings of voltages across the inductor and capacitor, yet they're effectively canceling each other due to their opposite nature.