/*! 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 20 An ac series circuit has an impe... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(X_{1},\) whichever is appropriate.

Short Answer

Expert verified
Resistance \(R \approx 49.71\ \Omega\), Capacitive reactance \(X_C \approx -185.47\ \Omega\).

Step by step solution

01

Understand Circuit Components

The circuit consists of a resistor and a capacitor or inductor. The impedance of an AC circuit is composed of resistance \(R\) and reactance \(X\) (either inductive or capacitive). Impedance \(Z\) is given by:\[Z = \sqrt{R^2 + X^2}\]Given \(Z = 192\ \Omega\).
02

Determine Reactance Type

The phase angle \(\phi = -75^\circ\) indicates the type of reactance. A negative phase angle implies the circuit has a capacitive reactance \(X_C\), as capacitive reactance causes the current to lead the voltage.
03

Utilize Phase Angle Relation

The phase angle \(\phi\) in an AC circuit is given by:\[\tan(\phi) = \frac{X}{R}\]Substitute \(\phi = -75^\circ\):\[\tan(-75^\circ) = \frac{X_C}{R}\]Using a calculator, \(\tan(-75^\circ) \approx -3.732\).
04

Relate Impedance and Components

We have two equations: 1. \(Z = \sqrt{R^2 + X_C^2} = 192\ \Omega\)2. \(X_C = -3.732 \cdot R\) from \(\tan(\phi)\).
05

Solve for Resistance \(R\)

Substitute \(X_C = -3.732 \cdot R\) into the impedance equation:\[192 = \sqrt{R^2 + (-3.732 \cdot R)^2}\]Squaring both sides results in:\[192^2 = R^2 + (3.732 \cdot R)^2\]\[36864 = R^2 + 13.9158R^2\]\[36864 = 14.9158R^2\]Solving for \(R\):\[R^2 = \frac{36864}{14.9158}\]\[R^2 \approx 2471.78\]\[R \approx 49.71\ \Omega\]
06

Calculate Capacitive Reactance \(X_C\)

Using \(X_C = -3.732 \cdot R\):\[X_C = -3.732 \times 49.71\]\[X_C \approx -185.47\ \Omega\]
07

Conclusion

The resistance \(R\) is approximately \(49.71\ \Omega\) and the capacitive reactance \(X_C\) is approximately \(-185.47\ \Omega\).

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Key Concepts

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

Impedance
Impedance is a key concept in understanding AC series circuits. It measures the total opposition a circuit presents to the flow of alternating current. In a series circuit, impedance combines both resistance and reactance. Imagine it like a traffic jam slowing down the flow of current. The formula for impedance, denoted as \[Z = \sqrt{R^2 + X^2},\] reveals how it's made up of two parts: resistance (R) and reactance (X).
  • Resistance refers to the opposition to current flow in a material. It's straightforward and only changes with the type of material and its temperature.
  • Reactance varies with frequency and comes in two types: capacitive reactance and inductive reactance, depending on the components being a capacitor or an inductor.
In an AC series circuit like in this exercise, the total impedance is the vector sum of these components. Here, it's given as 192 Ω, which helps in breaking down and identifying the roles each segment plays in the circuit.
Phase Angle
The phase angle in an AC circuit indicates the shift between the voltage and the current waveforms. It's an essential feature because it reveals what kind of reactance is present—capacitive or inductive.
A negative phase angle (like the -75° in our exercise) suggests a capacitive reactance. Exactly like saying, "current leads voltage," it helps identify which component predominantly influences the impedance. This phase relation is calculated with\[\tan(\phi) = \frac{X}{R},\]where
  • \(X\) represents the reactance (either capacitive or inductive),
  • \(R\) is the resistance.
Understanding this relationship is crucial in breaking down the circuit's behavior, as it directly relates to how the components work together to resist the current flow.
Resistance
Resistance might sound familiar as it is the most straightforward concept. It is the opposition a material presents to the flow of direct current, measured in ohms (Ω). In our AC circuit, even though alternating current runs through, resistance works the same way, representing the non-reactive segment of impedance.
In our exercise, the resistance \(R\) was calculated using the impedance formula: \[192 = \sqrt{R^2 + (-3.732 \cdot R)^2}.\]By rearranging and simplifying, we find:
  • \(R^2 = \frac{36864}{14.9158}\)
  • \(R \approx 49.71\ \,\Omega\)
This gives us the actual measure of resistance that adds onto the capacitive reactance to form total impedance. It's an important figure because it helps break down and understand the balance between resisting and reactive forces in the circuit.
Capacitive Reactance
Capacitive reactance is a specific type of opposition to the current in circuits containing capacitors. It arises due to the time it takes to charge and discharge a capacitor. Simply put, it influences how much the current "leads" the voltage.
In our AC series circuit problem, the negative phase angle immediately points to capacitive reactance. We calculated the capacitive reactance \(X_C\) using the tangent of the phase angle \(\tan(-75^\circ)\) as:
  • \(X_C = -3.732 \cdot R\)
  • With \(R \approx 49.71\ \,\Omega\):
  • \(X_C \approx -185.47\ \,\Omega\)
This negative value underscores how capacitive reactance operates in opposition to resistance and leads to the current being ahead of voltage. This vital measurement lets engineers understand and manipulate how circuits react to changes in frequency.

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Most popular questions from this chapter

An inductor is to be connected to the terminals of a generator (rms voltage 15.0 V) so that the resulting rms current will be 0.610 A. Determine the required inductive reactance.

The capacitance in a series RCL circuit is \(C_{1}=2.60 \mu \mathrm{F}\) , and the corresponding resonant frequency is \(f_{01}=7.30 \mathrm{kHz}\) . The generator frequency is 5.60 \(\mathrm{kHz}\) . What is the value of the capacitance \(C_{2}\) that should be added to the circuit so that the circuit will have a resonant frequency that matches the generator frequency? Note that you must decide whether \(C_{2}\) is added in series or in parallel with \(C_{1} .\)

A capacitor is connected across the terminals of an ac generator that has a frequency of 440 Hz and supplies a voltage of 24 V. When a second capacitor is connected in parallel with the first one, the current from the generator increases by 0.18 A. Find the capacitance of the second capacitor.

A series RCL circuit contains only a capacitor \((C=6.60 \mu \mathrm{F})\) an inductor \((L=7.20 \mathrm{mH})\) , and a generator \((\text {peak voltage}=32.0 \mathrm{V}\) frequency \(=1.50 \times 10^{3} \mathrm{Hz}\) . When \(t=0\) s, the instantaneous value of the voltage is zero, and it rises to a maximum one- quarter of a period later. (a) Find the instantaneous value of the voltage across the capacitor/inductor combination when \(t=1.20 \times 10^{-4}\) s. (b) What is the instantaneous value of the current when \(t=1.20 \times 10^{-4}\) s? \((\text { Hint: }\) The instantaneous values of the voltage and current are, respectively, the vertical components of the voltage and current phasors.)

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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