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Chapter 31: Problem 45

A series circuit has an impedance of 60.0\(\Omega\) and a power factor of 0.720 at 50.0 \(\mathrm{Hz}\) . The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

Short Answer

Expert verified
Add a capacitor of approximately 75.72 碌F to raise the power factor to unity.

Step by step solution

01

Identify the circuit element needed

Since the source voltage lags the current, the circuit is more inductive. To raise the power factor to unity (1.0), a capacitive element should be added to offset the inductance.
02

Calculate the current phase angle

The power factor is defined as the cosine of the phase angle (\( \phi \)) between voltage and current. Given a power factor of 0.720, we find the phase angle using: \[ \phi = \cos^{-1}(0.720) \approx 43.94^\circ \]
03

Determine the reactance of the circuit

For an impedance \( Z = 60.0 \Omega \) with a phase angle \( \phi = 43.94^\circ \), calculate the total reactance \( X \):\[ X = Z \sin(\phi) = 60.0 \cdot \sin(43.94^\circ) \approx 41.95 \Omega \]
04

Calculate the capacitive reactance needed

To achieve a unity power factor, the capacitive reactance (\( X_C \)) should neutralize the total reactance: \[ X_C = X = 41.95 \Omega \]
05

Find the capacitance required

Using the formula for capacitive reactance \( X_C = \frac{1}{2 \pi f C} \), solve for the capacitance \( C \):\[ C = \frac{1}{2 \pi f X_C} = \frac{1}{2 \pi \cdot 50.0 \cdot 41.95} = \frac{1}{13198} \approx 75.72 \mu F \]

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Key Concepts

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

Inductive Reactance
In electrical circuits, inductive reactance is a property that arises when an inductor is present. Inductance is caused by the tendency of an electrical conductor to oppose changes in the electric current that flows through it. This opposition is due to the creation of a magnetic field around the conductor. Inductive reactance (X_L) depends on the frequency of the alternating current and the inductance of the coil.
  • The inductive reactance is calculated using the formula: \( X_L = 2\pi f L \), where \( f \) is the frequency in hertz and \( L \) is the inductance in henrys.
  • At higher frequencies, \( X_L \) increases, meaning greater opposition to current changes.
  • In a circuit where the voltage lags the current, it indicates the circuit is dominated by inductive reactance.
To correct such a circuit's phase angle and raise its power factor, adding a capacitive component can counteract this inductance.
Capacitive Reactance
Capacitive reactance works in opposition to inductive reactance. It arises in circuits containing capacitors, which store and release electrical energy. When an alternating current flows through a capacitor, the current leads the voltage, opposite to what happens with inductors.
  • The formula to calculate capacitive reactance \( X_C \) is: \( X_C = \frac{1}{2\pi f C} \), where \( f \) is the frequency in hertz and \( C \) is the capacitance in farads.
  • As the frequency increases, \( X_C \) decreases, resulting in lower opposition to the current.
  • Capacitive reactance can be used to 'cancel out' the effects of inductive reactance in a circuit, improving the power factor and optimizing the circuit's efficiency.
By carefully adding the right amount of capacitance, like in the described exercise, it's possible to neutralize excess inductive reactance and balance the circuit.
Circuit Impedance
Impedance is a measure of how much a circuit resists the flow of alternating current. It combines both resistance (opposition to direct current) and reactance (opposition to alternating current), making the total impedance a complex quantity usually expressed in ohms (\( \Omega \)).
  • In calculations, impedance is the vector sum of both resistance and the total reactance: \( Z = R + jX \), where \( j \) is the imaginary unit.
  • The given impedance of a circuit, 60.0 \( \Omega \), encompasses both the resistive and reactive components.
  • Changes in circuit impedance affect the overall behavior of the circuit, including its power factor, which needs correction when dealing with power transmission.
Understanding impedance allows us to properly adjust the circuit for optimal performance, ensuring energy is consumed efficiently.
Phase Angle
The phase angle is crucial in defining the relationship between different waveforms in AC circuits, particularly voltage and current. It is the angular difference in degrees or radians between the peaks of the voltage wave and the current wave.
  • A positive phase angle means the current lags the voltage, signifying inductive dominance in the circuit.
  • A negative phase angle shows the current leads the voltage, hinting at capacitive properties.
  • It's calculated using the cosine of the power factor: \( \phi = \cos^{-1}(\text{power factor}) \).
In our scenario, correcting the phase angle involves adding capacitance to move the phase angle towards zero, reflecting a unity power factor where current and voltage waveforms are in synch. Converting an inductive circuit to one with a balanced or zero phase angle optimizes its performance.

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Most popular questions from this chapter

A series ac circuit contains a \(250-\Omega\) resistor, a 15 -m \(\mathrm{H}\) 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?

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}(\mathrm{rms})\) to a portable electronic device. The load resistance in the secondary is 5.00\(\Omega .\) (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120-\mathrm{V}\) line would draw the same power as the transformer? Show that this is equal to 5.00\(\Omega\) times the square of the ratio of primary to secondary turns.

In an \(L-R-C\) series circuit, \(R=400 \Omega, L=0.350 \mathrm{H},\) and \(C=0.0120 \mu \mathrm{F}\) (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 550 \(\mathrm{V}\) . If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

An \(L-R-C\) series circuit has \(C=4.80 \mu \mathrm{F}, L=0.520 \mathrm{H}\) and source voltage amplitude \(V=56.0 \mathrm{V}\) . The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude \(80.0 \mathrm{V},\) what is the value of \(R\) for the resistor in the circuit?

In an \(L-R-C\) series circuit, \(L=0.280 \mathrm{H}\) and \(C=\) 4.00\(\mu \mathrm{F}\) . The voltage amplitude of the source is 120 \(\mathrm{V}\) . (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 A. What is the resistance \(R\) of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?
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