Problem 18 - College Physics

Question

A series ac circuit contains a \(250 \Omega\) resistor, a \(15 \mathrm{mH}\) 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) 0.302, (b) 0.371 W, (c) Resistor: 0.371 W, Capacitor & Inductor: 0 W

Step 1: Calculate the inductive reactance

The inductive reactance \( X_L \) can be calculated using the formula \( X_L = \omega L \), where \( \omega \) is the angular frequency and \( L \) is the inductance. Substitute the given values: \( \omega = 360 \, \text{rad/s} \) and \( L = 15 \, \text{mH} = 15 \times 10^{-3} \, \text{H} \). So, \( X_L = 360 \times 15 \times 10^{-3} = 5.4 \, \Omega \).

Step 2: Calculate the capacitive reactance

The capacitive reactance \( X_C \) is calculated using \( X_C = \frac{1}{\omega C} \), where \( C \) is the capacitance. Given \( C = 3.5 \, \mu\text{F} = 3.5 \times 10^{-6} \, \text{F} \), calculate \( X_C = \frac{1}{360 \times 3.5 \times 10^{-6}} = 793.65 \, \Omega \).

Step 3: Calculate the impedance of the circuit

The impedance \( Z \) of the circuit is calculated using the formula: \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Substitute the given resistance \( R = 250 \Omega \), calculated \( X_L = 5.4 \, \Omega \), and \( X_C = 793.65 \, \Omega \): \( Z = \sqrt{250^2 + (5.4 - 793.65)^2} = \sqrt{250^2 + (-788.25)^2} \).

Step 4: Solve for the impedance value

Calculate \( Z = \sqrt{250^2 + (-788.25)^2} = \sqrt{62500 + 621356.5625} = \sqrt{683856.5625} \approx 826.85 \, \Omega \).

Step 5: Compute the power factor

The power factor \( \text{pf} \) is given by \( \text{pf} = \frac{R}{Z} \). Using \( R = 250 \Omega \) and \( Z = 826.85 \Omega \), calculate \( \text{pf} = \frac{250}{826.85} \approx 0.302 \).

Step 6: Calculate the average power delivered to the circuit

The average power \( P \) delivered to the circuit is given by \( P = V_{rms} I_{rms} \cos \phi \), where \( V_{rms} = \frac{V_m}{\sqrt{2}} \) and \( I_{rms} = \frac{V_{rms}}{Z} \). Substituting \( V_m = 45 \, V \), \( V_{rms} = \frac{45}{\sqrt{2}} \approx 31.82 \, V \), and \( I_{rms} = \frac{31.82}{826.85} \approx 0.0385 \, A \). Then, \( P = 31.82 \times 0.0385 \times 0.302 \approx 0.371 \, W \).

Step 7: Find the average power delivered to the resistor

The average power delivered to the resistor is \( P_R = I_{rms}^2 R \). Substitute \( I_{rms} = 0.0385 \, A \) and \( R = 250 \Omega \), giving \( P_R = (0.0385)^2 \times 250 = 0.371 \, W \).

Step 8: Calculate the average power delivered to the capacitor and the inductor

Since reactive components (the capacitor and inductor) do not consume average power, the average power delivered to both the capacitor and the inductor is \( 0 \, W \).

Key Concepts

Inductive Reactance

Inductive reactance is all about understanding how inductors behave in an AC circuit. An inductor opposes changes in current, and this opposition is quantified by what's called inductive reactance, denoted as \( X_L \). To calculate \( X_L \), we use the formula \( X_L = \omega L \), where \( \omega \) is the angular frequency of the AC source, and \( L \) is the inductance (measured in Henries). This parameter tells us how much opposition an inductor provides to the AC current.
The higher the frequency or the larger the inductance, the greater the inductive reactance. In our example, with an angular frequency of \( 360 \text{ rad/s} \) and an inductance of \( 15 \text{ mH} \), \( X_L = 5.4 \underline{\phantom{xxx}} \Omega \). This means that the inductor opposes the AC current with 5.4 ohms of resistance.

Capacitive Reactance

Capacitive reactance relates to how capacitors work in AC circuits. Unlike inductors, capacitors store energy in an electric field and also oppose changes in the voltage. Capacitive reactance, represented as \( X_C \), is calculated by \( X_C = \frac{1}{\omega C} \). Here, \( C \) is the capacitance, given in Farads, and \( \omega \) is the angular frequency.
A high-frequency signal will face lower capacitive reactance, while a low-frequency signal will face higher resistance from the capacitor. For our circuit, with a capacitance of \( 3.5 \mu \text{F} \), \( X_C = 793.65 \underline{\phantom{xxx}} \Omega \). This reveals that the capacitor significantly opposes the AC current, contributing an additional 793.65 ohms of resistance.

Impedance Calculation

The term impedance in AC circuit analysis combines resistance and reactance into a single term that reflects how much a circuit resists the flow of AC electricity. Impedance, symbolized as \( Z \), is calculated using the formula \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Here, \( R \) is the pure resistance, while \( X_L \) and \( X_C \) represent the inductive and capacitive reactances, respectively.
In the example problem, we have \( R = 250 \underline{\phantom{xxx}} \Omega \), \( X_L = 5.4 \underline{\phantom{xxx}} \Omega \), and \( X_C = 793.65 \underline{\phantom{xxx}} \Omega \). When we put these values into our formula, it results in an impedance \( Z \approx 826.85 \underline{\phantom{xxx}} \Omega \). This tells us the overall opposition the circuit has towards the AC current.

Power Factor

The power factor is a measure of how effectively electrical power is being transformed into useful work. It is defined as the cosine of the phase angle between the voltage and current, represented as \( \cos \phi \). Mathematically, it's given by \( \text{pf} = \frac{R}{Z} \).
A power factor of 1 indicates that all the power is being used effectively, while a power factor less than 1 indicates inefficiency. In our AC circuit, with \( R = 250 \underline{\phantom{xxx}} \Omega \) and \( Z = 826.85 \underline{\phantom{xxx}} \Omega \), we calculated a power factor of approximately 0.302. This means that less than a third of the total power is being utilized for useful work, i.e., the real power, while the rest is lost to the reactance in the circuit.

Average Power Calculation

Average power in an AC circuit is the actual power consumed by the resistive components of the circuit. It's calculated using the formula \( P = V_{\text{rms}} I_{\text{rms}} \cos \phi \), where \( V_{\text{rms}} \) is the root mean square voltage, and \( I_{\text{rms}} \) is the root mean square current.
In our problem, \( V_{\text{rms}} = \frac{45}{\sqrt{2}} \underline{\phantom{xxx}} V \), and \( I_{\text{rms}} = \frac{31.82}{826.85} \underline{\phantom{xxx}} A \). Multiplying these with the power factor, the circuit delivers \( P \approx 0.371 \underline{\phantom{xxx}} W \) of average power. Interestingly, only resistors consume real power, while capacitors and inductors, being reactive components, do not contribute to the average power. Their energy storage but non-consuming nature means the power associated with them averages out to zero over a complete cycle.