Problem 18 - College Physics

Question

A series ac circuit contains a 250\(\Omega\) resistor, a 15 \(\mathrm{mH}\) inductor, a 3.5\(\mu\) 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 is 0.302. (b) Average power to circuit is 0.37 W. (c) Power to resistor is 0.37 W; to capacitor and inductor is 0 W.

Step 1: Calculate the inductive reactance

The inductive reactance \( X_L \) can be calculated using the formula \( X_L = \omega L \), where \( \omega = 360 \, \mathrm{rad/s} \) and \( L = 15 \, \mathrm{mH} = 0.015 \, \mathrm{H} \). So, \( X_L = 360 \times 0.015 = 5.4 \, \Omega \).

Step 2: Calculate the capacitive reactance

The capacitive reactance \( X_C \) is given by \( X_C = \frac{1}{\omega C} \), where \( C = 3.5 \, \mu\mathrm{F} = 3.5 \times 10^{-6} \, \mathrm{F} \). Thus, \( X_C = \frac{1}{360 \times 3.5 \times 10^{-6}} = 793.65 \, \Omega \).

Step 3: Calculate the impedance of the circuit

The total impedance \( Z \) is given by \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Substituting the values, \( Z = \sqrt{250^2 + (5.4 - 793.65)^2} = \sqrt{250^2 + (-788.25)^2} \). Calculating gives \( Z \approx 826.87 \, \Omega \).

Step 4: Calculate the power factor

The power factor \( \cos \phi \) is defined as the ratio \( \frac{R}{Z} \). Here, \( \cos \phi = \frac{250}{826.87} \approx 0.3023 \).

Step 5: Calculate the average power delivered to the circuit

The average power \( P \) is given by \( P = \frac{1}{2} V_m I_m \cos \phi \), where \( V_m = 45 \, \mathrm{V} \) and \( I_m = \frac{V_m}{Z} = \frac{45}{826.87} \approx 0.0544 \, \mathrm{A} \). Substitute these into the formula: \( P \approx \frac{1}{2} \times 45 \times 0.0544 \times 0.3023 \approx 0.37 \, \mathrm{W} \).

Step 6: Calculate the average power delivered to each component

The average power delivered to a purely resistive component is \( P_R = \frac{1}{2} I_m^2 R \) for the resistor, which simplifies to \( P_R \approx 0.37 \, \mathrm{W} \). For the inductor and capacitor, \( P_L = P_C = 0 \, \mathrm{W} \) because they do not dissipate power over a complete cycle in an AC circuit.

Key Concepts

Inductive Reactance

Inductive reactance is a measure of an inductor's opposition to the change of current in an AC circuit. It is represented by the symbol \( X_L \) and is calculated using the formula \( X_L = \omega L \), where \( \omega \) is the angular frequency in radians per second, and \( L \) is the inductance in henrys.

In the given exercise, the angular frequency \( \omega \) is 360 rad/s, and the inductance \( L \) is 15 mH (or 0.015 H).
Therefore, the inductive reactance \( X_L \) = 360 \( \times 0.015 = 5.4 \; \Omega \).

The larger the value of \( X_L \), the greater the inductor resists changes in current. Inductive reactance increases linearly with frequency, meaning at higher frequencies, inductors offer more opposition.

Capacitive Reactance

Capacitive reactance, represented as \( X_C \), quantifies how much a capacitor resists changes in voltage in an AC circuit.
It's calculated using \( X_C = \frac{1}{\omega C} \), where \( \omega \) is the angular frequency, and \( C \) is the capacitance.

In our exercise, \( C \) is given as 3.5 \( \mu \)F (or \( 3.5 \times 10^{-6} \) F).
Thus, \( X_C = \frac{1}{360 \times 3.5 \times 10^{-6}} \approx 793.65 \; \Omega \).

A higher \( X_C \) implies that the capacitor is more effective at blocking current flow. Interestingly, unlike inductors, capacitive reactance is inversely related to frequency, meaning capacitors become less resistive at higher frequencies.

Power Factor

The power factor of an AC circuit is the cosine of the phase angle \( \phi \) between the voltage and current waveforms. It gives insight into how effectively the circuit uses electric power, with a value ranging from 0 to 1.

It's calculated as \( \cos \phi = \frac{R}{Z} \), where \( R \) is the resistance and \( Z \) is the total impedance.

In our scenario, using \( R = 250 \; \Omega \) and \( Z = 826.87 \; \Omega \),
the power factor is \( \cos \phi = \frac{250}{826.87} \approx 0.3023 \).

This suggests that only 30.23% of the power is effectively used, typically indicating a lagging current (common in inductive circuits) or leading current (in capacitive circuits). A low power factor can lead to inefficient power use and higher costs.

Average Power

Average power in an AC circuit represents the real power consumed by the resistive components as opposed to reactive power, which flows back and forth between the source and reactive components.

The formula for average power \( P \) is given as \( P = \frac{1}{2} V_m I_m \cos \phi \), where \( V_m \) is the peak voltage, \( I_m \) is the peak current, and \( \cos \phi \) is the power factor. In our exercise:

• \( V_m = 45 \; V \)
• \( I_m = \frac{V_m}{Z} \approx 0.0544 \; A \)
• \( \cos \phi = 0.3023 \)

Plugging these values into the formula gives approximately \( P \approx 0.37 \; W \).

This reflects the power consumed by the resistor in the circuit, as only resistors dissipate real power in AC circuits. Inductors and capacitors only store and return energy to the system, resulting in no net energy loss.

Impedance

Impedance, denoted as \( Z \), is a comprehensive measurement of a circuit's resistance to AC and includes resistive (\( R \)) and reactive (both inductive and capacitive) components.

It's calculated using \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). In the exercise, \( R = 250 \; \Omega \), \( X_L = 5.4 \; \Omega \), and \( X_C = 793.65 \; \Omega \):

Thus, \( Z = \sqrt{250^2 + (5.4 - 793.65)^2} \approx 826.87 \; \Omega \).

\( Z \) gives a holistic idea of AC circuit resistance.

• An increase in \( X_L \) or \( X_C \) raises \( Z \).
• In an ideal situation, matching \( X_L \) and \( X_C \) can minimize \( Z \), reducing losses and improving power efficiency.

Understanding impedance helps in designing circuits for optimal performance.