Problem 72 - University Physics with Modern Physics

Question

An \(L-R-C\) series circuit draws 220 \(\mathrm{W}\) from a \(120-\mathrm{V}\) (rms), 50.0 -Hz ac line. The power factor is 0.560 , and the source voltage leads the current. (a) What is the net resistance \(R\) of the circuit? (b) Find the capacitance of the series capacistor that will result in a power factor of unity when it is added to the original circuit. (c) What power will then be drawn from the supply line?

Step-by-Step Solution

Answer

(a) R ≈ 20.5 Ω, (b) C ≈ 106 µF, (c) P ≈ 392.4 W

Step 1: Identify Given Values and Relevant Equations

We know the real power \( P = 220 \, \mathrm{W} \), the rms voltage \( V_{rms} = 120 \, \mathrm{V} \), frequency \( f = 50.0 \, \mathrm{Hz} \), and power factor \( \text{pf} = 0.560 \). Use the formula for power in an AC circuit: \( P = V_{rms} \cdot I_{rms} \cdot \cos(\theta) \) where \( \cos(\theta) \) is the power factor and \( I_{rms} \) is the rms current.

Step 2: Calculate RMS Current

Rearrange the power formula to find \( I_{rms} \): \[ I_{rms} = \frac{P}{V_{rms} \cdot \cos(\theta)} = \frac{220}{120 \times 0.560} \approx 3.27 \, \mathrm{A} \]

Step 3: Determine Net Resistance R

Use Ohm's law for AC circuits \( V_{rms} = I_{rms} \cdot Z \), and solve it for impedance \( Z \): \( Z = \frac{V_{rms}}{I_{rms}} \). Then, the resistance \( R \) can be found using \( R = Z \cdot \cos(\theta) \). Calculate, \[ Z = \frac{120}{3.27} \approx 36.7 \, \Omega \] and \[ R = 36.7 \times 0.560 \approx 20.5 \, \Omega \]

Step 4: Calculate Required Capacitance for Unity Power Factor

To achieve a unity power factor, the impedance must be purely resistive, meaning the inductive reactance \( X_L \) and capacitive reactance \( X_C \) must cancel out. Calculate total reactance \( X = \sqrt{Z^2 - R^2} \), then use \( X_C = \frac{1}{2 \pi f C} = X \).\[ X = \sqrt{36.7^2 - 20.5^2} \approx 30.1 \]\[ \frac{1}{2 \pi \times 50 \times C} = 30.1 \Rightarrow C \approx 106 \times 10^{-6} \mathrm{F} \]

Step 5: Calculate New Power with Unity Power Factor

The new power is calculated using the same formula as before but with a power factor of unity: \( P = V_{rms} \cdot I_{rms} \cdot 1 \). Hence the new power is \[ P = 120 \times 3.27 \approx 392.4 \, \mathrm{W} \]

Key Concepts

Power Factor Correction

Power factor correction is a crucial aspect in enhancing the efficiency of AC circuits. It involves adjusting the power factor of the electrical system to make it as close to 1 as possible. A power factor of 1, or unity, means that the circuit is entirely resistive, and all the power supplied by the source is effectively used for work.
In the case of an L-R-C circuit, the power factor is influenced by the presence of reactive components like inductors and capacitors. These elements do not consume power but create a phase difference between the voltage and current. The power factor correction aims to eliminate this phase difference, often by adding capacitors or inductors to balance out the effect of the circuit's reactance.

• With a low power factor, more current is required to deliver the same amount of real power as compared to a circuit with a higher power factor.
• This can lead to higher losses in the electrical system and increased load on generators and transformers.
• Correcting the power factor can reduce energy costs and increase the electrical system's overall stability and efficiency.

Impedance in AC Circuits

Impedance is a term used in AC circuits to describe the total opposition a circuit presents to the flow of alternating current. It is a complex quantity, involving both resistance, which dissipates real power, and reactance, which stores and releases energy.
In our L-R-C circuit, impedance is crucial because it combines all elements of resistance and reactance to define the circuit's behavior. While resistance (R) is associated with energy dissipation, inductive reactance (\(X_L = 2\pi fL \)) relates to energy storage in a magnetic field, and capacitive reactance (\(X_C = \frac{1}{2\pi fC} \)) is tied to energy storage in an electric field.

• Impedance in an AC circuit can vary with frequency, meaning as the frequency changes, the circuit's response to AC signals can alter significantly.
• This property makes devices like filters and tuners possible.
• Along with phase angle (\(\theta \)), impedance determines how much current flows through the circuit for a given voltage.

Calculation of Reactance

Reactance is the measure of opposition that inductors and capacitors present to the change in current in AC circuits. Like resistance, reactance affects how currents behave in the circuit, but unlike resistance, it does not dissipate energy.
The total reactance in our circuit can be calculated using the difference between inductive reactance (\(X_L = 2\pi fL \)) and capacitive reactance (\(X_C = \frac{1}{2\pi fC} \)). These values affect the total impedance (\(Z\)) of the circuit, where \[X = X_L - X_C\] represents the net reactive factor.

• The goal is to manage these two kinds of reactance such that they cancel each other out, leading to a purely resistive impedance.
• This cancellation is crucial for achieving a power factor of \(1\), optimizing efficiency.
• The calculated values for reactance are key to designing circuits that respond predictably to changes in frequency.