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Ch 32: AC Circuits
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 32, Problem 67d

A motor attached to a 120 V/60 Hz power line draws an 8.0 A current. Its average energy dissipation is 800 W. How much series capacitance needs to be added to increase the power factor to 1.0?

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Step 1: Understand the problem. The motor operates with a power factor less than 1.0, meaning there is reactive power due to inductance. To increase the power factor to 1.0, we need to add a series capacitance to cancel out the inductive reactance. This will make the circuit purely resistive.
Step 2: Calculate the apparent power (S) using the formula: \( S = V \cdot I \), where \( V \) is the voltage (120 V) and \( I \) is the current (8.0 A). This gives the total power in the circuit, including both real and reactive components.
Step 3: Determine the reactive power (Q) using the relationship between apparent power (S), real power (P), and reactive power (Q): \( S^2 = P^2 + Q^2 \). Rearrange to solve for \( Q \): \( Q = \sqrt{S^2 - P^2} \), where \( P \) is the average energy dissipation (800 W).
Step 4: Calculate the inductive reactance (\( X_L \)) using the formula \( Q = V^2 / X_L \). Rearrange to solve for \( X_L \): \( X_L = V^2 / Q \). This gives the reactance due to the inductance in the circuit.
Step 5: Determine the required capacitance (C) to cancel out the inductive reactance. The capacitive reactance (\( X_C \)) is given by \( X_C = 1 / (2 \pi f C) \), where \( f \) is the frequency (60 Hz). Set \( X_C = X_L \) and solve for \( C \): \( C = 1 / (2 \pi f X_L) \). This capacitance will increase the power factor to 1.0.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power Factor

The power factor is a measure of how effectively electrical power is being converted into useful work output. It is defined as the cosine of the phase angle between the voltage and current waveforms in an AC circuit. A power factor of 1.0 indicates that all the power supplied is being used effectively, with no reactive power. Improving the power factor can reduce energy losses and improve the efficiency of the electrical system.
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Reactive Power and Capacitance

Reactive power is the power that oscillates between the source and the load in an AC circuit, primarily due to inductive and capacitive components. Capacitors can be added to a circuit to counteract the effects of inductive loads, thereby improving the power factor. The amount of capacitance required to achieve a desired power factor can be calculated using the relationship between reactive power, voltage, and capacitance.
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Capacitors & Capacitance (Intro)

Impedance in AC Circuits

Impedance is the total opposition that a circuit presents to the flow of alternating current, combining both resistance and reactance. It is represented as a complex number, where resistance is the real part and reactance is the imaginary part. Understanding impedance is crucial for calculating the current, voltage, and power factor in AC circuits, especially when capacitors are added to adjust the power factor.
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Related Practice
Textbook Question

FIGURE CP32.68 shows voltage and current graphs for a series RLC circuit. What is the resistance R?

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Textbook Question

Prove that the energy dissipation is a maximum at ω = ω₀.

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Textbook Question

A motor attached to a 120 V/60 Hz power line draws an 8.0 A current. Its average energy dissipation is 800 W. What is the motor's resistance?

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Textbook Question

A motor attached to a 120 V/60 Hz power line draws an 8.0 A current. Its average energy dissipation is 800 W. What is the rms resistor voltage?

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Textbook Question

Commercial electricity is generated and transmitted as three-phase electricity. Instead of a single emf ε = ε0 cos ωt, three separate wires carry currents for the emfs ε1 = ε0 cos ωt, ε2 = ε0 cos(ωt+120°), and ε3 = ε0 cos(ωt−120°). 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. Show that the potential difference between any two of the phases has the rms value 3–√ εrms, where εrms is the familiar single-phase rms voltage. Evaluate this potential difference for εrms = 120 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.

Textbook Question

Show that the peak inductor voltage in a series RLC circuit is maximum at frequency ωL=(1ω0212R2C2)1/2\(\omega\)_L = \(\left\)( \(\frac{1}{\omega_0^2}\) - \(\frac{1}{2}\) R^2 C^2 \(\right\))^{-1/2}.

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