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A series \(R L C\) circuit consists of a \(100 \Omega\) resistor, a \(0.15 \mathrm{H}\) inductor, and a \(30 \mu\) F capacitor. It is attached to a \(120 \mathrm{V} / 60 \mathrm{Hz}\) power line. What are (a) the peak current \(I,\) (b) the phase angle \(\phi,\) and \((\mathrm{c})\) the average power loss?

Short Answer

Expert verified
The peak current I, phase angle φ, and average power loss in the circuit can be calculated by using the mentioned formulas. Substituting the given values into the formulas, we can obtain the required results.

Step by step solution

01

Calculate the impedance

Impedance \(Z\) in a RLC circuit is given by the formula \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). But before that, Reactance \(X_L\) of the inductor and \(X_C\) of the capacitor have to be found out. \(X_L = 2\pi fL\) and \(X_C = 1/(2\pi fC)\), where \(f\) is frequency. For the given, R=100 Ohm, L=0.15H, C=30μF=30x10^-6 F and f=60Hz. After substituting these in the formulas, calculate \(X_L\) and \(X_C\) and then substitute them in the equation of \(Z\).
02

Find the peak current

Now, the peak current \(I\) can be obtained from the formula \(I = V/Z\), where \(V\) is voltage. Here, V=120V. Substitute the value of \(Z\) obtained from step 1 and calculate the value of I.
03

Calculate the phase angle

The phase angle \(φ\) in a RLC circuit is given by the formula \(φ=tan^{-1}[(X_L - X_C)/R]\). Substitute the obtained values of \(X_L\), \(X_C\) and \(R\) into this formula and calculate the phase angle.
04

Determine the average power loss

Finally, the power loss in the circuit can be found using the formula \(P = (I^2 * R) / 2\), where \(I\) is the peak current obtained from step 2, and \(R\) is the resistance. Calculate the power loss by substituting the values into the equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Impedance in RLC Circuits
When tackling the topic of impedance in RLC (Resistor-Inductor-Capacitor) circuits, imagining water flowing through pipes can be quite helpful. The impedance, represented by the symbol \(Z\), acts much like resistance to water flow, but for alternating current (AC) in electrical circuits. In an RLC circuit, there's resistance (\(R\)), but we also consider the inductive reactance (\(X_L\)) and capacitive reactance (\(X_C\)).

These reactances depend on the frequency (\(f\)) of the AC supply. The inductive reactance increases with higher frequency, while capacitive reactance decreases. To find the overall impedance, we compute the square root of the sum of resistance squared and the square of the difference between inductive and capacitive reactances, as shown by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).

Remember, impedance is not only about the magnitude of opposition to the AC but also includes the phase difference between the current and voltage, affecting how the circuit behaves over time. Understanding impedance helps in designing circuits with desired properties, like filtering certain frequencies out or in.
AC Circuit Phase Angle
The AC circuit phase angle, symbolized by \(\phi\), is a concept that reflects the time difference between the voltage and current waves in a circuit. Imagine two swimmers racing in a pool— they might start at the same time (in phase), but if one swims faster, they will get out of sync (out of phase). Similarly, in an RLC circuit, the phase angle tells us how much the current lags or leads the voltage.

To determine the phase angle, we use the equation \(\phi = \tan^{-1}[(X_L - X_C)/R]\). A positive angle indicates that the current lags the voltage (common in inductor-heavy circuits), while a negative angle indicates that the current leads the voltage (typical for capacitor-heavy circuits). By knowing the phase angle, experts can predict how the circuit will respond to changes and ensure components work effectively within the system.
Average Power Loss in AC Circuits
Discussing average power loss in AC circuits takes us to the practical side of electrical systems—how much energy is being consumed and potentially wasted. In simple terms, power loss is the energy that doesn't do useful work, turning into heat instead. To access this energy usage, we use the formula \(P = (I^2 * R) / 2\), where \(P\) is the average power loss, \(I\) is the peak current, and \(R\) is the resistance.

In the context of our problem, after calculating the peak current, we use it to determine the power loss. This number has practical implications, like understanding how efficient a circuit is, and it helps in the design of electric systems to avoid excess energy consumption. Knowing the power loss also offers insights into safety, as excessive power loss might lead to overheating and potential damage to the circuit components.

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Most popular questions from this chapter

The tuning circuit in an FM radio receiver is a series \(RLC\) circuit with a 0.200 \(\mu\)H inductor. a. The receiver is tuned to a station at \(104.3 \mathrm{MHz}\). What is the value of the capacitor in the tuning circuit? b. FM radio stations are assigned frequencies every \(0.2 \mathrm{MHz}\) but two nearby stations cannot use adjacent frequencies. What is the maximum resistance the tuning circuit can have if the peak current at a frequency of \(103.9 \mathrm{MHz}\), the closest frequency that can be used by a nearby station, is to be no more than \(0.10 \%\) of the peak current at \(104.3 \mathrm{MHz}\) ? The radio is still tuned to \(104.3 \mathrm{MHz}\), and you can assume the two stations have equal strength.

A low-pass RC filter with a crossover frequency of \(1000 \mathrm{Hz}\) uses a \(100 \Omega\) resistor. What is the value of the capacitor?

Commercial electricity is generated and transmitted as three phase electricity. Instead of a single emf, three separate wires carry currents for the emfs \(\mathcal{E}_{1}=\mathcal{E}_{0} \cos \omega t, \mathcal{E}_{2}=\mathcal{E}_{0} \cos (\omega t+\) \(\left.120^{\circ}\right),\) and \(\mathcal{E}_{3}=\mathcal{E}_{0} \cos \left(\omega t-120^{\circ}\right)\) over three parallel wires, each of which supplies one-third of the power. This is why the long-distance transmission lines you see in the countryside have three wires. Suppose the transmission lines into a city supply a total of 450 MW of electric power, a realistic value. a. What would be the current in each wire if the transmission voltage were \(\mathcal{E}_{0}=120 \mathrm{V} \mathrm{rms} ?\) b. In fact, transformers are used to step the transmission-line voltage up to \(500 \mathrm{kV}\) ms. What is the current in each wire? c. Big transformers are expensive. Why does the electric company use step-up transformers?

Commercial electricity is generated and transmitted as threephase electricity. Instead of a single emf \(\mathcal{E}=\mathcal{E}_{0} \cos \omega t\), three separate wires carry currents for the emfs \(\mathcal{E}_{1}=\mathcal{E}_{0} \cos \omega t\) \(\mathcal{E}_{2}=\mathcal{E}_{0} \cos \left(\omega t+120^{\circ}\right),\) and \(\mathcal{E}_{3}=\mathcal{E}_{0} \cos \left(\omega t-120^{\circ}\right) .\) This is why the long-distance transmission lines you see in the countryside have three parallel wires, as do many distribution lines within a city. a. Draw a phasor diagram showing phasors for all three phases of a three-phase emf. b. Show that the sum of the three phases is zero, producing what is referred to as neutral. In single-phase electricity, provided by the familiar \(120 \mathrm{V} / 60 \mathrm{Hz}\) electric outlets in your home, one side of the outlet is neutral, as established at a nearby electrical substation. The other, called the hot side, is one of the three phases. (The round opening is connected to ground.) c. Show that the potential difference between any two of the phases has the rms value \(\sqrt{3} \mathcal{E}_{\text {rms }},\) where \(\mathcal{E}_{\text {rms }}\) is the familiar single-phase rms voltage. Evaluate this potential difference for \(\mathcal{E}_{\text {rms }}=120 \mathrm{V}\) Some high-power home appliances, especially electric clothes dryers and hot-water heaters, are designed to operate between two of the phases rather than between one phase and neutral. Heavy-duty industrial motors are designed to operate from all three phases, but full three phase power is rare in residential or office use.

A high-pass filter consists of a \(1.59 \mu \mathrm{F}\) capacitor in series with a \(100 \Omega\) resistor. The circuit is driven by an AC source with a peak voltage of \(5.00 \mathrm{V}\) a. What is the crossover frequency \(f_{c} ?\) b. What is \(V_{\mathrm{R}}\) when \(f=\frac{1}{2} f_{\mathrm{c}}, f_{\mathrm{c}},\) and \(2 f_{\mathrm{c}} ?\)

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