Problem 31 - University Physics with Modern Physics

Question

In an \(L-R-C\) series circuit, \(R=150 \Omega, L=0.750 \mathrm{H},\) and \(C=0.0180 \mu \mathrm{F} .\) The source has voltage amplitude \(V=150 \mathrm{V}\) and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with \(C=\) 0.0360\(\mu \mathrm{F}\) and the source frequency is adjusted to the new resonance value. Then what is the average power delivered by the source?

Step-by-Step Solution

Answer

(a) Power factor is 1; (b) Average power is 150 W; (c) New average power is 150 W.

Step 1: Calculate the Resonant Angular Frequency

The resonant angular frequency \( \omega_0 \) for an L-R-C circuit is given by the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \). Substituting the provided values into the equation:\[ \omega_0 = \frac{1}{\sqrt{(0.750)(0.0180 \times 10^{-6})}} \approx 8.6 \times 10^3 \text{ rad/s} \]

Step 2: Determine the Power Factor at Resonance

At the resonant frequency, the power factor is 1 because the inductive reactance \(X_L\) equals the capacitive reactance \(X_C\), making the overall impedance purely resistive.

Step 3: Calculate the Average Power Delivered at Resonance

The average power \(P\) delivered by the source can be calculated using the formula \(P = \frac{V^2}{R} \), where \(V\) is the voltage amplitude, and \(R\) is the resistance.\[ P = \frac{(150)^2}{150} \approx 150 \text{ W} \]

Step 4: Update the Capacitance and Calculate the New Resonance

When the capacitance is replaced by \(C = 0.0360 \mu F\), the new resonant angular frequency \(\omega_0'\) is:\[ \omega_0' = \frac{1}{\sqrt{(0.750)(0.0360 \times 10^{-6})}} \approx 6.1 \times 10^3 \text{ rad/s} \]

Step 5: Calculate Average Power with New Capacitance

Since the circuit is at resonance again at the adjusted frequency, the power factor remains 1. Thus, the average power delivered by the source, with the new capacitance, is the same:\[ P = 150 \text{ W} \]

Key Concepts

Understanding Resonance Frequency in an L-R-C Circuit

To comprehend the resonance frequency in an L-R-C circuit, imagine it as the point where the circuit naturally oscillates at its maximum efficiency. This happens when the inductive reactance (\(X_L\)) and the capacitive reactance (\(X_C\)) are equal. At this precise point, they cancel each other out, leaving the circuit operating purely on resistance.

The formula to find the resonant angular frequency is given by:

• \[ \omega_0 = \frac{1}{\sqrt{LC}} \]

Here, \(\omega_0\) is the resonant angular frequency, \(L\) is the inductance, and \(C\) is the capacitance. When the circuit achieves this resonance frequency, it allows for optimal energy transfer, as there's no net reactance. This means the circuit will draw the minimum possible current needed to supply the voltage, enhancing efficiency.

In the exercise, calculating the resonant frequency involved substituting given values into this equation, revealing the natural frequency for both original and adjusted capacitor values.

Examining the Power Factor in L-R-C Circuits

The power factor in an L-R-C circuit is a measure of efficiency, indicating how effectively the circuit's input power is converted to useful output power. It is defined by the cosine of the phase angle (\(\phi\)) between voltage and current:

• \[ \text{Power Factor} = \cos(\phi) \]

At resonance, the power factor reaches its maximum value of 1. This occurs because the phase difference between voltage and current is zero; they are perfectly in sync. Since there's no phase difference, the circuit experiences no reactive power losses; it operates on pure resistance.

In practical terms, a power factor of 1 means all the energy supplied by the power source is doing useful work (i.e., none is being wasted). This is crucial in circuits as it signifies the optimal working condition, especially in applications that demand high efficiency. This is why, even when the capacitor value changes, as long as the system is at resonance, the power factor stays at 1, ensuring optimal performance.

Calculating Average Power Delivered in the Circuit

In electrical circuits, average power represents the actual power consumed over time. Specifically, for an L-R-C circuit at resonance, it is calculated using the formula:

• \[ P = \frac{V^2}{R} \]

where \(P\) is the average power, \(V\) is the voltage amplitude, and \(R\) is the resistance in the circuit. This equation highlights that at resonance, the average power depends solely on the resistance and the voltage amplitude, as reactances are canceled out.

In this exercise, the average power was calculated for both the initial and adjusted capacitance values. Regardless of the capacitance change, when the circuit is set to resonate with the adjusted frequency, the impedance becomes purely resistive again, stabilizing the power factor at 1, thus delivering the same average power of 150 W.

This consistency in power delivery demonstrates the stability and reliability of L-R-C circuits at resonance, reflecting on their predictable performance under changing conditions.