Problem 26 - College Physics

Question

\(\bullet\) In an \(R-L-C\) series circuit, \(R=300 \Omega, L=0.400 \mathrm{H},\) and \(C=6.00 \times 10^{-8} \mathrm{F}\) . When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 \(\mathrm{A}\) . (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

Step-by-Step Solution

Answer

(a) 150 V, (b) Resistor: 150 V, Inductor: 150 V, Capacitor: 150 V, (c) 75 W

Step 1: Understanding Resonance in RLC Circuit

In an R-L-C series circuit at resonance, the impedance is at its minimum and is equal to the resistance, that is, 300 \( \Omega \). The resonance condition implies that the inductive reactance and capacitive reactance cancel out each other.

Step 2: Calculate the Voltage Amplitude of the Source

At resonance, the impedance \( Z \) is purely real and equal to \( R \). The voltage amplitude \( V_0 \) of the source can be calculated using Ohm's Law: \[ V_0 = I_0 \cdot R \] where the current amplitude \( I_0 = 0.500 \text{ A} \) and \( R = 300 \Omega \). Thus, \[ V_0 = 0.500 \times 300 = 150 \text{ V} \].

Step 3: Voltage Across the Resistor

Since the current amplitude through the resistor is 0.500 A and \( V_R = I_0 \cdot R \), we find \( V_R = 0.500 \times 300 = 150 \text{ V} \).

Step 4: Calculate Voltage Across the Inductor and Capacitor

At resonance, the voltage across the inductor (\( V_L \)) is equal in magnitude and opposite in phase to the voltage across the capacitor (\( V_C \)). Both can be calculated as \[ V_L = V_C = I_0 \cdot X_L = I_0 \cdot \omega L \] and \[ V_C = I_0 \cdot \frac{1}{\omega C} \]. At resonance, \( X_L = \omega L = \frac{1}{\omega C} = X_C \), thus \( V_L = V_C = 150 \text{ V} \).

Step 5: Average Power Supplied by the Source

The average power supplied by the source in a series RLC circuit at resonance is given by \[ P = I_0^2 \cdot R \]. Using \( I_0 = 0.500 \text{ A} \) and \( R = 300 \Omega \), calculate: \[ P = (0.500)^2 \times 300 = 0.250 \times 300 = 75 \text{ W} \].

Key Concepts

Resonance Frequency in RLC Circuit

An RLC circuit consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series. The resonance frequency is a critical concept in such circuits. It is the frequency at which the circuit naturally oscillates. During resonance, the inductive reactance (denoted by \( X_L \)) and the capacitive reactance (denoted by \( X_C \)) are equal and cancel each other out.

• This balance leads the circuit to exhibit minimum impedance.
• The impedance at resonance is solely resistive, meaning it equals the resistance \( R \).

Resonance occurs at angular frequency \( \omega_0 \), which can be calculated using the formula:\[ \omega_0 = \frac{1}{\sqrt{LC}} \]This results in the circuit being at its most efficient state, where it consumes only the energy it can dissipate as resistive power.

Voltage Amplitude

In the context of an RLC circuit at resonance, voltage amplitude is a vital measurement. It is the maximum voltage delivered by the ac source to the circuit. Calculating this is straightforward using the current amplitude (\( I_0 \)) and the resistance (\( R \)) since the impedance is purely resistive at resonance.

• By applying Ohm's Law, the voltage amplitude (\( V_0 \)) is given as:

\[ V_0 = I_0 \cdot R \]For our example with \( I_0 = 0.500 \text{ A} \) and \( R = 300 \Omega \), we find:\[ V_0 = 0.500 \times 300 = 150 \text{ V} \]This voltage amplitude helps in understanding the power handling and efficiency of the circuit at resonance.

Average Power in RLC Circuit

At resonance, the average power supplied to an RLC circuit is defined through energy dissipation across the resistor. Since only the resistor imparts resistance at resonance, power is determined by:

• The square of the current amplitude (\( I_0^2 \)).
• Multiply it by the resistance (\( R \)).

The formula to find average power (\( P \)) is:\[ P = I_0^2 \cdot R \]Substituting the values from our exercise, where \( I_0 = 0.500 \text{ A} \) and \( R = 300 \Omega \):\[ P = (0.500)^2 \times 300 = 75 \text{ Watts} \]Understanding power in this manner allows us to determine how efficiently energy is being used or dissipated in the system.

Inductive Reactance and Its Role

Inductive reactance, denoted by \( X_L \), gives insight into how an inductor behaves in an AC circuit. It is defined by the equation:\[ X_L = \omega L \] where \( \omega \) is the angular frequency and \( L \) is the inductance.

• It acts to oppose changes in current through an inductor.
• In RLC circuits at resonance, it balances with capacitive reactance.
• Both reactances ensure that their effects cancel each other out, leaving only the resistor to influence current and voltage.

By understanding inductive reactance, one gains insights into the energy storage capabilities and resonant behaviors of RLC circuits, which contribute to their unique responses at different frequencies.