Problem 25 - College Physics

Question

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) 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 0.600 A. What is the resistance \(R\) of the resistor? (c) At the resonance frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Step-by-Step Solution

Answer

(a) 250 rad/s. (b) 400 Ω. (c) 30 V across inductor and capacitor, 240 V across resistor.

Step 1: Identify Variable and Formula for Resonance Angular Frequency

In a series R-L-C circuit, the resonance angular frequency \( \omega_0 \) is given by \( \omega_0 = \frac{1}{\sqrt{LC}} \). Here we have \( L = 0.200 \text{ H} \) and \( C = 80.0 \underline{\phantom{xxx}}\mu\text{F} = 80.0 \times 10^{-6} \text{ F} \).

Step 2: Calculate Resonance Angular Frequency

Calculate \( \omega_0 \) using the formula. Substitute \( L = 0.200 \text{ H} \) and \( C = 80.0 \times 10^{-6} \text{ F} \) into \( \omega_0 = \frac{1}{\sqrt{LC}} \):\[ \omega_0 = \frac{1}{\sqrt{0.200 \times 80.0 \times 10^{-6}}} \approx 250 \text{ rad/s}. \]

Step 3: Relate Current Amplitude to Resistance

At resonance, the impedance of the circuit is purely resistive, so the current amplitude is given by \( I_0 = \frac{V_0}{R} \). Given \( I_0 = 0.600 \text{ A} \) and \( V_0 = 240 \text{ V} \), use this formula to solve for \( R \).

Step 4: Calculate the Resistance of the Resistor

Substitute \( I_0 = 0.600 \text{ A} \) and \( V_0 = 240 \text{ V} \) into the equation \( I_0 = \frac{V_0}{R} \):\[ R = \frac{V_0}{I_0} = \frac{240}{0.600} = 400 \underline{\phantom{xxx}} \Omega. \]

Step 5: Determine Peak Voltages across Components

At resonance, the peak voltages across the inductor \( V_L \) and capacitor \( V_C \) are calculated as:- \( V_L = I_0 \cdot \omega_0 \cdot L \)- \( V_C = \frac{I_0}{\omega_0 \cdot C} \)For the resistor \( R \), it is simply \( V_R = I_0 \cdot R \).

Step 6: Compute Peak Voltage across the Inductor

Substitute \( I_0 = 0.600 \text{ A} \), \( \omega_0 = 250 \text{ rad/s} \), and \( L = 0.200 \text{ H} \) into \( V_L = I_0 \cdot \omega_0 \cdot L \):\[ V_L = 0.600 \times 250 \times 0.200 = 30 \text{ V}. \]

Step 7: Compute Peak Voltage across the Capacitor

Substitute \( I_0 = 0.600 \text{ A} \), \( \omega_0 = 250 \text{ rad/s} \), and \( C = 80.0 \times 10^{-6} \text{ F} \) into \( V_C = \frac{I_0}{\omega_0 \cdot C} \):\[ V_C = \frac{0.600}{250 \times 80.0 \times 10^{-6}} = 30 \text{ V}. \]

Step 8: Compute Peak Voltage across the Resistor

Substitute \( I_0 = 0.600 \text{ A} \) and \( R = 400 \underline{\phantom{xxx}} \Omega \) into \( V_R = I_0 \cdot R \):\[ V_R = 0.600 \times 400 = 240 \text{ V}. \]

Key Concepts

Resonance Angular Frequency

In an RLC circuit, the resonance angular frequency is the frequency at which the circuit naturally oscillates. This point is notable because, at resonance, the impedance is minimized and the current reaches its maximum for a given voltage. The formula to find the resonance angular frequency \( \omega_0 \) is:\[ \omega_0 = \frac{1}{\sqrt{LC}} \]Here, \( L \) is the inductance and \( C \) is the capacitance of the circuit. For example, if you have \( L = 0.200 \text{ H} \) and \( C = 80.0 \underline{\phantom{xxx}} \mu\text{F} = 80.0 \times 10^{-6} \text{ F} \), you would substitute these values into the formula to find:\[ \omega_0 = \frac{1}{\sqrt{0.200 \times 80.0 \times 10^{-6}}} \approx 250 \text{ rad/s} \]This frequency is crucial, as at \( 250 \text{ rad/s} \), the circuit behaves optimally with respect to the impedance and current flow.

Impedance and Current Amplitude

At resonance, the impedance of a series RLC circuit is purely resistive, meaning that the reactance components cancel each other out. Therefore, the impedance \( Z \) is equal to just the resistance \( R \). This simplifies the relationship between voltage and current amplitude, making calculations straightforward. The current amplitude \( I_0 \) can be found using the formula:\[ I_0 = \frac{V_0}{R} \]Where \( V_0 \) is the voltage amplitude. For instance, if \( V_0 = 240 \text{ V} \) and the desired current amplitude \( I_0 = 0.600 \text{ A} \), the resistance \( R \) can be calculated as:\[ R = \frac{240}{0.600} = 400 \underline{\phantom{xxx}} \Omega \]Understanding this relationship is key in designing circuits that require specific current outputs based on known voltages.

Peak Voltages in RLC Circuit

In a series RLC circuit at resonance, the voltage across the components can be significantly different despite the presence of the same current through each component. Calculating these peak voltages is important for assessing component voltage tolerances and potential dangers, such as over-voltages.- **Voltage across the Inductor (\( V_L \))**: This is given by the formula \[ V_L = I_0 \cdot \omega_0 \cdot L \]Suppose \( I_0 = 0.600 \text{ A} \), \( \omega_0 = 250 \text{ rad/s} \), and \( L = 0.200 \text{ H} \); then \[ V_L = 0.600 \times 250 \times 0.200 = 30 \text{ V} \]- **Voltage across the Capacitor (\( V_C \))**: Can be calculated using \[ V_C = \frac{I_0}{\omega_0 \cdot C} \]For \( I_0 = 0.600 \text{ A} \) and \( C = 80.0 \times 10^{-6} \text{ F} \), \[ V_C = \frac{0.600}{250 \times 80.0 \times 10^{-6}} = 30 \text{ V} \]- **Voltage across the Resistor (\( V_R \))**: Purely resistive and directly proportional to \( R \) and \( I_0 \):\[ V_R = I_0 \cdot R = 0.600 \times 400 = 240 \text{ V} \]These calculations show the different peak voltages each component experiences even if maintained at resonance conditions, conveying that each element requires proper rating to avoid failures.