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Chapter 23: Problem 20

An ac series circuit has an impedance of \(192 \Omega,\) and the phase angle between the current and the voltage of the generator is \(\phi=-75^{\circ} .\) The circuit contains a resistor and either a capacitor or an inductor. Find the resistance \(R\) and the capacitive reactance \(X_{\mathrm{C}}\) or the inductive reactance \(X_{\mathrm{L}}\) whichever is appropriate.

Short Answer

Expert verified
Resistance \(R\) is approximately \(49.45 \, \Omega\), and capacitive reactance \(X_C\) is approximately \(184.51 \, \Omega\).

Step by step solution

01

Understand the Problem

We are given an AC series circuit with impedance magnitude \(Z = 192 \, \Omega\) and a phase angle \(\phi = -75^{\circ}\). The negative phase angle indicates that the circuit is capacitive. We need to find the resistance \(R\) and the capacitive reactance \(X_C\).
02

Use the Impedance Formula

In an AC circuit with a resistor \(R\) and a capacitor, the impedance \(Z\) is given by \(Z = \sqrt{R^2 + X_C^2}\). Here, we know \(Z = 192 \, \Omega\). We need additional equations to solve for \(R\) and \(X_C\).
03

Apply the Phase Angle Formula

The phase angle \(\phi\) between current and voltage is given by \(\tan \phi = \frac{-X_C}{R}\). Given \(\phi = -75^{\circ}\), we have:\[\tan(-75^{\circ}) = \frac{-X_C}{R}\]This simplifies to \(\tan(-75^{\circ}) = -3.732\).
04

Solve for Ratios of \(R\) and \(X_C\)

From the phase angle equation, we have:\[ -X_C = -3.732R \]Therefore, \(X_C = 3.732R\).
05

Substitute into the Impedance Equation

Substitute \(X_C = 3.732R\) into the impedance equation:\[Z = \sqrt{R^2 + (3.732R)^2}\]\[192 = \sqrt{R^2 + 13.914R^2}\]\[192 = \sqrt{14.914R^2}\]
06

Solve for \(R\)

Square both sides:\[192^2 = 14.914R^2\]\[36864 = 14.914R^2\]Solve for \(R\):\[R = \sqrt{\frac{36864}{14.914}}\approx 49.45 \, \Omega\]
07

Find \(X_c\)

Now substitute \(R = 49.45\, \Omega\) into \(X_C = 3.732R\):\[X_C = 3.732 \times 49.45 \approx 184.51 \, \Omega\]
08

Final Step: Verify Results

Verify that the calculated values satisfy both the impedance and phase angle conditions:1. Impedance: \(\sqrt{49.45^2 + 184.51^2} \approx 192\)2. Phase angle: \(\tan^{-1}\left(-\frac{184.51}{49.45}\right) \approx -75^{\circ}\).Both conditions are satisfied, confirming the solution.

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Key Concepts

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

Impedance
Let's delve into the concept of impedance in AC circuits. Impedance is the resistance to the flow of alternating current (AC) in a circuit. It is similar to resistance in direct current circuits but takes into account the additional effects of inductance and capacitance. Impedance is a complex quantity, consisting of a real part (resistance) and an imaginary part (reactance).
  • Impedance is represented by the symbol \( Z \) and measured in ohms (\(\Omega\)).
  • In our example, the impedance \( Z \) is given as \( 192 \, \Omega \).
  • Mathematically, impedance is expressed as \( Z = \sqrt{R^2 + X_C^2} \) for circuits involving capacitive reactance.
Understanding impedance is crucial for analyzing how AC circuits respond to different frequencies and how they affect current and voltage waveforms.
Phase Angle
The phase angle in an AC circuit signifies the difference in phase between the voltage and the current. This angle, denoted as \( \phi \), indicates how much the current waveform is leading or lagging the voltage waveform.
  • A negative phase angle, as in our problem \( \phi = -75^{\circ} \), suggests that the circuit is capacitive.
  • Capacitive circuits cause the current to lead the voltage.
  • The phase angle equation is \( \tan \phi = \frac{-X_C}{R} \), which we use to solve for the relationship between the capacitive reactance \( X_C \) and resistance \( R \).
The phase angle helps in understanding power flow, as it impacts power factor鈥攁 key element in real-world applications of AC circuits.
Capacitive Reactance
Capacitive reactance is a form of reactance that introduces a phase difference between current and voltage in capacitive elements of an AC circuit. It is represented by the symbol \( X_C \) and is measured in ohms (\(\Omega\)).
  • In capacitive circuits, \( X_C \) causes the current to lead the voltage by 90 degrees.
  • Capacitive reactance is inversely proportional to the frequency and capacitance: \( X_C = \frac{1}{2 \pi f C} \).
  • In our problem, \( X_C = 3.732R \), which shows how the reactance depends on resistance.
Understanding capacitive reactance helps in analyzing how capacitors affect the circuit's behavior, particularly in terms of filtering and timing applications.
Resistor in AC Circuit
A resistor in an AC circuit behaves much like it does in a DC circuit, offering resistance to the flow of current. However, in AC circuits, resistors do not affect the phase angle between voltage and current.
  • The role of the resistor is to dissipate energy in the form of heat.
  • In our example, the resistance \( R \) is approximately \( 49.45 \, \Omega \).
  • Resistors help determine the total impedance \( Z \) in combination with reactance (either capacitive or inductive).
Resistors are crucial for controlling current flow and dividing voltage within AC circuits, making them integral components in both simple and complex circuit designs.

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Most popular questions from this chapter

Two identical capacitors are connected in parallel to an ac generator that has a frequency of \(610 \mathrm{Hz}\) and produces a voltage of \(24 \mathrm{V}\). The current in the circuit is 0.16 A. What is the capacitance of each capacitor?

The resonant frequency of an \(\mathrm{RCL}\) circuit is \(1.3 \mathrm{kHz},\) and the value of the inductance is \(7.0 \mathrm{mH}\). What is the resonant frequency (in \(\mathrm{kHz}\) ) when the value of the inductance is \(1.5 \mathrm{mH} ?\)

Part \(a\) of the figure shows a heterodyne metal detector being used. As part \(b\) of the figure illustrates, this device utilizes two capacitor/ inductor oscillator circuits, A and B. Each produces its own resonant frequency, \(f_{0 \mathrm{A}}=1 /\left[2 \pi\left(L_{\mathrm{A}} C\right)^{1 / 2}\right]\) and \(f_{0 \mathrm{B}}=1 /\left[2 \pi\left(L_{\mathrm{B}} C\right)^{1 / 2}\right]\). Any difference between these frequencies is detected through earphones as a beat frequency \(\mid f_{0 \mathrm{B}}-\) \(f_{0 A} \mid .\) In the absence of any nearby metal object, the inductances \(L_{\mathrm{A}}\) and \(L_{\mathrm{B}}\) are identical. When inductor \(\mathrm{B}\) (the search coil) comes near a piece of metal, the inductance \(L_{\mathrm{B}}\) increases, the corresponding oscillator frequency \(f_{\mathrm{oB}}\) decreases, and a beat frequency is heard. Suppose that initially each inductor is adjusted so that \(L_{\mathrm{B}}=L_{\mathrm{A}},\) and each oscillator has a resonant frequency of \(855.5 \mathrm{kHz}\). Assuming that the inductance of search coil \(\mathrm{B}\) increases by \(1.000 \%\) due to a nearby piece of metal, determine the beat frequency heard through the earphones.

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.

An ac generator has a frequency of \(2.2 \mathrm{kHz}\) and a voltage of \(240 \mathrm{V} .\) An inductance \(L_{1}=6.0 \mathrm{mH}\) is connected across its terminals. Then a second inductance \(L_{2}=9.0 \mathrm{mH}\) is connected in parallel with \(L_{1} .\) Find the current that the generator delivers to \(L_{1}\) and to the parallel combination.
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