Problem 62 - University Physics with Modern Physics

Question

An \(L \cdot R-C\) series circuit consists of a \(2.50-\mu \mathrm{F}\) capacitor, a 5.00 -mH inductor, and a \(75.0-\Omega\) resistor connected across an ac source of voltage amplitude 15.0 \(\mathrm{V}\) having variable frequency. (a) Under what circumstances is the average power delivered to the circuit equal to \(\frac{1}{2} V_{\mathrm{Vms}} I_{\mathrm{rms}} ?\) (b) Under the conditions of part (a), what is the average power delivered to each circuit element and what is the maximum current through the capacitor?

Step-by-Step Solution

Answer

Resonance occurs when \(X_L = X_C\); power is realized as \(\frac{1}{2} V_{\mathrm{rms}} I_{\mathrm{rms}}\) at resonance. Resistor power is \(P_R\); capacitor max current \(I_{C,max} = \omega C V_{C_{max}}\).

Step 1: Identify Resonance Condition

The average power delivered equals \( \frac{1}{2} V_{\mathrm{rms}} I_{\mathrm{rms}} \) when the circuit is at resonance. The resonance condition occurs when the impedance of the inductor and capacitor cancel each other out, i.e., their reactances are equal. This results in the circuit having a purely resistive impedance.

Step 2: Calculate Reactances

The inductive reactance is given by \( X_L = 2\pi f L \), and the capacitive reactance is \( X_C = \frac{1}{2\pi f C} \). At resonance, \( X_L = X_C \).

Step 3: Solve for Resonant Frequency

At resonance, \( 2\pi f L = \frac{1}{2\pi f C} \). Solving for \( f \), we get the resonant frequency \[ f_0 = \frac{1}{2\pi \sqrt{LC}}. \] For \( L = 5.00 \, \text{mH} \) and \( C = 2.50 \, \mu \text{F} \), substitute to find \( f_0 \).

Step 4: Calculate RMS Current

Under resonance condition, the impedance is purely resistive (\( R \)). Thus, \( I_{\text{rms}} = \frac{V_{\text{rms}}}{R} \). Given \( V\_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}} \), calculate \( V_{\text{rms}} \) and subsequently \( I_{\text{rms}} \).

Step 5: Calculate Average Power Delivered

The average power delivered at resonance is \[ P = \frac{1}{2} V_{\mathrm{rms}} I_{\mathrm{rms}}. \] Substitute the previously found values for \( V_{\mathrm{rms}} \) and \( I_{\mathrm{rms}} \) to compute \( P \).

Step 6: Power in Each Element

At resonance, the reactance parts do not consume power, only the resistor does. The power across the resistor is \( P_R = I_{\mathrm{rms}}^2 R \). The power for inductor and capacitor is effectively zero.

Step 7: Calculate Maximum Current through Capacitor

The maximum current in the capacitor occurs when voltage across it is at its peak: \( I_{C,max} = \omega C V_{C_{max}}. \) Here, \( V_{C_{max}} = V_{\text{max}} \) at resonance. Therefore compute \( I_{C,max} \) using known values.

Key Concepts

Resonance Condition in RLC Circuits

In an RLC circuit, the resonance condition is reached when the impedance of the circuit is purely resistive. This means the inductive reactance and capacitive reactance cancel each other out completely. Here's a simple way to think about it:

• The inductive reactance (\( X_L = 2\pi f L \)) comes from the coil in the circuit. It tries to "oppose" changes in the current.
• The capacitive reactance (\( X_C = \frac{1}{2\pi f C} \)) comes from the capacitor, and it tries to "store and release" electrical energy.

At resonance, these two forces are equal and opposite. Mathematically, this can be shown as: \( X_L = X_C \).
The result is the cancellation of their effects, and the circuit behaves like it has only a resistor. The resonance frequency where this happens is important because that's when you get maximum efficient power transfer. In formula terms, the resonance frequency \( f_0 \) is given by: \[ f_0 = \frac{1}{2\pi \sqrt{LC}} \].This equation shows the relationship between frequency (\( f \)), inductance (\( L \)), and capacitance (\( C \)). Understanding this balance is crucial for analyzing RLC circuits, especially in applications like tuning radios or designing filters.

Reactance Calculation in RLC Circuits

Reactance in RLC circuits refers to the resistance opposing the change of current in inductors and capacitors. It acts differently than ordinary resistance.

• Inductive Reactance (\(X_L\)) - Given by the formula \( X_L = 2\pi f L \), where \( L \) is the inductance in henries and \( f \) is the frequency in hertz. This reactance increases with frequency, making it higher as the frequency goes up.
• Capacitive Reactance (\(X_C\)) - Calculated using the formula \( X_C = \frac{1}{2\pi f C} \), where \( C \) is the capacitance in farads. This reactance decreases as the frequency increases, meaning capacitors resist less as frequency rises.

When calculating reactance in circuits, these formulas help predict how a circuit will behave at different frequencies. By setting \( X_L = X_C \), you identify the resonance condition, where the effects cancel out, leading to efficient circuit operation. This understanding is fundamental for designing systems where specific frequencies need maximized efficiency or filtering.

Average Power in AC Circuits

The average power in AC circuits is a measure of the actual power consumed or delivered. Because AC voltage varies with time, it's not as simple as in DC circuits.
In AC, average power \( P \) is calculated using the formula: \[ P = \frac{1}{2} V_{\text{rms}} I_{\text{rms}} \],where:

• \( V_{\text{rms}} \) is the root mean square voltage, a measure of the effective voltage.
• \( I_{\text{rms}} \) is the root mean square current, representing the effective current flow.

At resonance, with equal reactances canceling out, the circuit's impedance is purely resistive, and maximum power is delivered to the circuit. Only the resistive elements, like a resistor, consume power, calculated as: \( P_R = I_{\text{rms}}^2 R \).
Meanwhile, inductors and capacitors only store and return energy, so their average power consumption is zero at resonance. This distinction helps in applications where specific components need to receive or deliver defined power quantities.