/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 A circuit consists of a resistor... [FREE SOLUTION] | 91Ó°ÊÓ

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A circuit consists of a resistor in series with an inductor and an ac generator that supplies a voltage of 115 V. The inductive reactance is \(52.0 \Omega\), and the current in the circuit is 1.75 A. Find the average power delivered to the circuit.

Short Answer

Expert verified
Under pure inductive conditions, average power is zero. Practically, need resistance for non-zero power.

Step by step solution

01

Identify the Given Values

The problem provides several values: the voltage of the AC generator is 115 V, the inductive reactance is \(X_L = 52.0 \, \Omega\), and the current in the circuit is \(I = 1.75 \, A\).
02

Recall the Power Formula

The average power delivered to the circuit is given by the formula \(P = VI\cos(\phi)\), where \(\phi\) is the phase angle between the voltage and the current.
03

Calculate the Phase Angle Cosine

The formula for the cosine of the phase angle \(\phi\) in a circuit with resistance and inductance is \(\cos(\phi) = \frac{R}{Z}\), where \(Z = \sqrt{R^2 + X_L^2}\) is the impedance. Since the power associated with pure inductance is zero on average, \(\cos(\phi)\) is essentially zero for an ideal inductor-resistor series circuit.
04

Approximate Power Calculation

Given that there is no resistance \(R\) specifically mentioned, we use the approximation \(\cos(\phi) = 0\) (assuming a highly inductive circuit with minimal resistance). Thus the power equation simplifies to a negligible average power if it's ideally inductive. However, practically, it's often combined resistance that contributes the effective power; calculation would then require knowing that resistance value. Using \(V = IZ\) allows finding total impedance but without resistance, we assume \(P \approx 0\) under pure inductive assumption.
05

Consider Practical Implication

In practice, if resistance were provided or measurable, you could determine an average power using \(P = I^2R\). Since the problem assumes ideal conditions without explicit resistance, an average practical outcome can be closest to zero.

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Key Concepts

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

Resistor
In AC circuits, resistors play a crucial part as they oppose the flow of electric current, just like they do in DC circuits. However, one important aspect to note is that they do not cause any phase shift between voltage and current. This means that the current through a resistor is directly proportional to the voltage across it, according to Ohm's Law, given by the equation \( V = IR \). In the context of AC circuits, resistors contribute to the overall impedance, which affects how much current will flow through the circuit.
  • Resistors in an AC circuit dissipate energy in the form of heat.
  • They provide a real power component which can be calculated as \( P = I^2R \), where \( I \) is the current and \( R \) is the resistance.
  • This power is known as "active" or "real" power, and it differs from reactive power, which is associated with components like inductors and capacitors.
Understanding the role of the resistor is essential, especially in circuits with multiple elements, where its resistance plays a key part in calculating total impedance.
Inductor
Inductors are another essential component in AC circuits, characterized by their ability to store energy in a magnetic field when electrical current flows through them. Differing from resistors, inductors can cause a phase shift between voltage and current. Specifically, the current lags the voltage by 90 degrees in a purely inductive circuit. This happens due to the energy required to build up the magnetic field around the inductor.
  • The impedance of an inductor is known as inductive reactance, represented by \( X_L \), calculated using the formula \( X_L = 2\pi f L \), where \( f \) is the frequency of the AC supply and \( L \) is the inductance.
  • Inductive reactance increases with higher frequency, explaining why inductors are considered frequency dependent components.
  • Inductors do not dissipate energy; instead, they store and then release it, which is why the real power associated with an ideal inductor is zero.
Recognizing how inductors behave in AC circuits helps in understanding phenomena like signal filtering and electromagnetic interference.
Average Power Calculation
Calculating the average power in AC circuits requires understanding the relationship between voltage and current. Generally, this involves the phase angle \( \phi \) between these two signals, giving rise to the power formula:\[ P = VI\cos(\phi) \]In circuits with only resistors, the phase angle \( \phi \) is zero, leading to the maximum power output because \( \cos(\phi) = 1 \). However, in circuits with inductors, \( \phi \) is non-zero, hence affecting the power calculations.
  • The real or average power is a reflection of the actual work done while supplying energy within the circuit, not to be confused with the reactive power mainly associated with storage and return of energy.
  • In practical circuits, having some known resistive effects or components allows for calculating an accurate average power using parameters like current and observed resistance.
  • For a mainly inductive circuit without explicit resistance, one might assume power is near zero due to low cosines involved in the calculations.
Being able to determine average power is critical in designing efficient circuits and helps in managing power consumption effectively.
Inductive Reactance
Inductive reactance is a measure of the opposition that an inductor presents to the change of current within an AC circuit. It plays a significant role in determining the overall impedance, which impacts the current flow in the circuit.Given by the formula \[ X_L = 2\pi f L \]where \( f \) represents the frequency, and \( L \) is the inductance, it's clear why the frequency is crucial; as frequency increases, so does the inductive reactance.
  • Inductive reactance is vital in tuning circuits, signal processing, and attenuation of unwanted frequencies.
  • Its behavior helps in filtering applications, where specific frequencies need to be blocked or allowed, thus differentiating it from resistance which provides uniform opposition irrespective of frequency.
  • In practice, it's essential for engineers to measure inductive reactance to ensure proper functioning of high-frequency circuits and manage the unintended effects of electromagnetic fields.
Understanding inductive reactance assists us not only in dealing with AC circuits but also in applications involving alternating magnetic fields.

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Most popular questions from this chapter

Part \(a\) of the drawing shows a resistor and a charged capacitor wired in series. When the switch is closed, the capacitor discharges as charge moves from one plate to the other. Part \(b\) shows the amount \(q\) of charge remaining on each plate of the capacitor as a function of time. In part \(c\) of the drawing, the switch has been removed and an ac generator has been inserted into the circuit. The circuit elements in the drawing have the following values: \(R=18 \Omega, V_{\mathrm{rms}}=24 \mathrm{V}\) for the generator, and \(f=380 \mathrm{Hz}\) for the generator. The time constant for the circuit in part \(a\) is \(\tau=3.0 \times 10^{-4} \mathrm{s} .\) What is the rms current in the circuit in part \(c ?\)

Two parallel plate capacitors are identical, except that one of them is empty and the other contains a material with a dielectric constant of 4.2 in the space between the plates. The empty capacitor is connected between the terminals of an ac generator that has a fixed frequency and rms voltage. The generator delivers a current of 0.22 A. What current does the generator deliver after the other capacitor is connected in parallel with the first one?

A series RCL circuit contains a \(5.10-\mu \mathrm{F}\) capacitor and a generator whose voltage is \(11.0 \mathrm{V}\). At a resonant frequency of \(1.30 \mathrm{kHz}\) the power delivered to the circuit is \(25.0 \mathrm{W}\). Find the values of (a) the inductance and (b) the resistance. (c) Calculate the power factor when the generator frequency is \(2.31 \mathrm{kHz}\)

A capacitor is connected across an ac generator whose frequency is \(750 \mathrm{Hz}\) and whose peak output voltage is \(140 \mathrm{V} .\) The rms current in the circuit is \(3.0 \mathrm{A}\). (a) What is the capacitance of the capacitor? (b) What is the magnitude of the maximum charge on one plate of the capacitor?

In a series circuit, a generator \((1350 \mathrm{Hz}, 15.0 \mathrm{V})\) is connected to a \(16.0-\Omega\) resistor, a \(4.10-\mu \mathrm{F}\) capacitor, and a \(5.30-\mathrm{mH}\) inductor. Find the voltage across each circuit element.

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