Question

A series ac circuit contains a 250-$\mathrm{\Omega }$ resistor, a 15-mH inductor, a 3.5-$\mu$F capacitor, and an ac power source of voltage amplitude 45 V operating at an angular frequency of 360 rad/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?

Short answer

(a) Power factor is 0.9598. (b) Average power delivered is 3.726 W. (c) Resistor receives 3.726 W, capacitor and inductor receive 0 W.

Step-by-step solution

Calculate the Impedance of the Inductor

To find the impedance of the inductor, use the formula $Z_L=j\omega L$, where $\omega =360\text{rad/s}$ is the angular frequency and $L=15\text{mH}=0.015\text{H}$. Thus, $Z_L=j\times 360\times 0.015=j5.4\mathrm{\Omega }$.

Calculate the Impedance of the Capacitor

For the impedance of the capacitor, use the formula $Z_C=\frac{-j}{\omega C}$, where $C=3.5\mu \text{F}=3.5\times {10}^{-6}\text{F}$. Thus, $Z_C=\frac{-j}{360\times 3.5\times {10}^{-6}}\approx -j79.37\mathrm{\Omega }$.

Calculate the Total Impedance of the Circuit

The total impedance $Z$ of the circuit is a combination of resistance and reactances, given by $Z=R+Z_L+Z_C$. Substitute the given values: $Z=250+j5.4-j79.37\equiv 250-j73.97$.

Calculate the Magnitude of the Total Impedance

The magnitude of $Z$ is $|Z|=\sqrt{R^2+(X_L-X_C{)}^2}$. Therefore, $|Z|=\sqrt{{250}^2+(5.4-79.37{)}^2}\approx \sqrt{{250}^2+(-73.97{)}^2}=260.50\mathrm{\Omega }$.

Determine the Power Factor of the Circuit

The power factor (pf) is given by $\cos \theta =\frac{R}{|Z|}$, where $|Z|=260.50$. Therefore, $\text{pf}=\frac{250}{260.50}\approx 0.9598$.

Find the Average Power Delivered to the Entire Circuit

The average power $P$ delivered is $P=V_{rms}{I}_{rms}\cos \theta$. First, find $I_{rms}=\frac{V_{rms}}{|Z|}$, where $V_{rms}=\frac{45}{\sqrt{2}}$. Thus, $I_{rms}=\frac{45/\sqrt{2}}{260.50}\approx 0.1224\text{A}$. Now, $P=0.1224\times 31.82\times 0.9598\approx 3.726\text{ W}$.

Calculate the Average Power Delivered to the Resistor

The average power delivered to the resistor $P_R$ is $I_{rms}^{2}R$. Therefore, $P_R=(0.1224{)}^2\times 250=3.726\text{ W}$.

Calculate the Average Power Delivered to the Capacitor and Inductor

Capacitors and inductors in an AC circuit do not consume power over a full cycle, so the average power delivered to both components is zero: $P_C=P_L=0\text{W}$.

Key concepts

Impedance Calculation

In AC circuit analysis, understanding impedance is crucial. Full impedance ($Z$) in a circuit incorporates resistance and reactance. Impedance is a vector quantity typically expressed as a complex number, allowing both magnitude and phase to be considered.
To calculate the impedance of an inductor, use $Z_L=j\omega L$, where $j$ denotes the imaginary unit, $\omega$ is the angular frequency, and $L$ is the inductance. For a capacitor, the formula is $Z_C=\frac{-j}{\omega C}$, where $C$ is the capacitance.
Once you find these individual impedances, total impedance in a series circuit combines resistance and reactance: $Z=R+Z_L+Z_C$. Here, resistance $R$ is a purely real number, while inductive and capacitive reactance are imaginary. This mix requires converting into polar form for magnitude: $|Z|=\sqrt{R^2+(X_L-X_C{)}^2}$.
This magnitude represents the "total opposition" to alternating current, factoring in both resistive and reactive components, essential for complete analysis.

Power Factor

The power factor is a crucial concept when analyzing AC circuits, as it indicates how effectively electrical power is being converted into useful work. It is calculated using the formula: $\cos \theta =\frac{R}{|Z|}$. The term $\theta$ is the phase angle between voltage and current.
A power factor of 1 (or 100%) means that all electrical power is effectively used for work. In practical circuits, values less than 1 suggest inefficiencies due to reactive components like inductors and capacitors. These components cause phase shifts: inductors delay current by 90°, while capacitors lead voltage by 90°.
An efficient circuit aims to have a power factor as close to 1 as possible. A lower power factor indicates more energy is stored rather than used, leading to higher energy costs and less efficient power use. Addressing these inefficiencies often involves reactive power compensation or power factor correction techniques.

Average Power

Average power in an AC circuit is the actual power consumed or used by the circuit. This power is calculated using the formula $P=V_{rms}{I}_{rms}\cos \theta$, where $V_{rms}$ is the root mean square voltage, $I_{rms}$ is the root mean square current, and $\cos \theta$ is the power factor.
In resistive components, average power reflects real, consumable power, calculated as $P_R=I_{rms}^{2}R$. Inductors and capacitors ideally do not consume energy over time. Instead, they store energy temporarily and return it to the system, leading to zero average power over a complete cycle: $P_C=P_L=0$.
Understanding average power helps in efficient energy management, allowing for the design of circuits that minimize wasted energy, ensuring consumer loads receive necessary power for optimal functionality.