Problem 25 - University Physics with Modern Physics

Question

A series ac circuit contains a \(250-\Omega\) resistor, a 15 -m \(\mathrm{H}\) inductor, a \(3.5-\mu \mathrm{F}\) capacitor, and an ac power source of voltage amplitude 45 \(\mathrm{V}\) operating at an angular frequency of 360 \(\mathrm{rad} / \mathrm{s}\) , (a) What is the power factor of this circuit? (b) Find the average power delivered to the entire circuit. (c) What is the average power delivered to the resistor, to the capacitor, and to the inductor?

Step-by-Step Solution

Answer

(a) Power factor: 0.302, (b) Average power: 0.367 W, (c) Resistor: 0.367 W, Capacitor: 0 W, Inductor: 0 W.

Step 1: Calculate the Inductive Reactance

The inductive reactance \( X_L \) can be calculated using the formula \( X_L = \omega \times L \), where \( \omega = 360 \, \mathrm{rad/s} \) and \( L = 15 \, \mathrm{mH} = 15 \times 10^{-3} \, \mathrm{H} \). Thus, \( X_L = 360 \times 15 \times 10^{-3} = 5.4 \, \Omega \).

Step 2: Calculate the Capacitive Reactance

The capacitive reactance \( X_C \) is found using the formula \( X_C = \frac{1}{\omega \times C} \), where \( C = 3.5 \times 10^{-6} \, \mathrm{F} \). Therefore, \( X_C = \frac{1}{360 \times 3.5 \times 10^{-6}} \approx 793.7 \, \Omega \).

Step 3: Calculate the Impedance of the Circuit

The total impedance \( Z \) is given by the formula \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Substituting the values, \( R = 250 \, \Omega \), \( X_L = 5.4 \, \Omega \), and \( X_C = 793.7 \, \Omega \), we find \( Z = \sqrt{250^2 + (5.4 - 793.7)^2} = \sqrt{250^2 + (-788.3)^2} \approx 827.5 \, \Omega \).

Step 4: Calculate the Power Factor

The power factor (\( \cos \phi \)) is given by \( \cos \phi = \frac{R}{Z} \). Therefore, \( \cos \phi = \frac{250}{827.5} \approx 0.302 \).

Step 5: Calculate the Average Power Delivered to the Circuit

The average power delivered to an AC circuit is found using \( P = V_{rms} I_{rms} \cos \phi \), where the rms voltage \( V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{45}{\sqrt{2}} \approx 31.8 \, \mathrm{V} \). First, calculate the rms current \( I_{rms} = \frac{V_{rms}}{Z} = \frac{31.8}{827.5} \approx 0.0384 \, \mathrm{A} \). Finally, \( P = 31.8 \times 0.0384 \times 0.302 \approx 0.367 \, \mathrm{W} \).

Step 6 : Calculate Average Power Delivered to Each Component

The average power delivered to a resistor \( P_R \) is \( I_{rms}^2 \times R \). Here, \( P_R = (0.0384)^2 \times 250 \approx 0.367 \, \mathrm{W} \). The average power to the capacitor and inductor is zero because their average power over a full cycle is zero due to their storing and releasing energy without dissipation. Therefore, \( P_C = 0 \, \mathrm{W} \), \( P_L = 0 \, \mathrm{W} \).

Key Concepts

Impedance Calculation

Impedance is a key concept when analyzing AC circuits. It represents the total opposition a circuit presents to the flow of alternating current. In a series circuit containing resistors, inductors, and capacitors, the impedance replaces resistance typically found in DC circuits.
To calculate impedance in an AC circuit, we use the formula:

• \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

Here, \( R \) represents the resistance, \( X_L \) is the inductive reactance, and \( X_C \) is the capacitive reactance. Inductive reactance is calculated using \( X_L = \omega \times L \), and capacitive reactance is \( X_C = \frac{1}{\omega \times C} \), where \( \omega \) is the angular frequency, \( L \) is inductance, and \( C \) is capacitance.
The key idea is that impedance not only considers resistance but also how inductors and capacitors affect current. Inductors store energy in a magnetic field and create a lag (positive phase angle) while capacitors store energy in an electric field and create a lead (negative phase angle). This interplay results in an overall phase difference affecting the flow of electric current in complex AC circuits.

Power Factor

The power factor in an AC circuit measures the efficiency with which electrical power is converted into useful work. It is defined as the cosine of the phase angle \( \phi \), which represents the phase difference between voltage and current.

• Mathematically, power factor is given by:\[ \text{Power Factor} (\cos \phi) = \frac{R}{Z} \]

This formula implies that the power factor is the ratio of the resistor's resistance \( R \) to the total impedance \( Z \) of the circuit.
A power factor of 1 (or 100%) indicates the most efficient condition where all the power is being used for useful work. When the circuit has inductive or capacitive components, it can result in a lower power factor. A power factor less than 1 means that some of the power is reactive, stored and then returned, not used for any productive work.
Improving power factor involves reducing the phase difference, which can often be achieved by balancing inductive and capacitive elements. This balance can help supply systems minimize losses and maximize efficiency.

Average Power

In AC circuits, average power is the real power consumed by the resistors in the circuit, often described in watts. This differs from reactive power, associated with the energy storage by capacitors or inductors, which doesn't dissipate energy as heat or drive a mechanical load.
The average power delivered to an AC circuit with given RMS voltage \( V_{rms} \) and RMS current \( I_{rms} \) can be calculated using:

• \[ P = V_{rms} \times I_{rms} \times \cos \phi \]

Where \( \cos \phi \) is the power factor.
RMS voltage \( V_{rms} \) is calculated from the peak voltage \( V_0 \) as \( \frac{V_0}{\sqrt{2}} \). RMS current follows similarly as \( I_{rms} = \frac{V_{rms}}{Z} \).
Average power in resistive components is not zero and is given by \( I_{rms}^2 \times R \). However, for inductors and capacitors, the average power over a complete cycle is zero due to their nature of storing and returning energy rather than dissipating it.
This understanding is crucial for designing efficient electrical systems and ensuring devices operate correctly under alternating current scenarios.