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University Physics Volume 2

15.2 Simple AC Circuits

University Physics Volume 215.2 Simple AC Circuits
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  1. Preface
  2. Unit 1. Thermodynamics
    1. 1 Temperature and Heat
      1. Introduction
      2. 1.1 Temperature and Thermal Equilibrium
      3. 1.2 Thermometers and Temperature Scales
      4. 1.3 Thermal Expansion
      5. 1.4 Heat Transfer, Specific Heat, and Calorimetry
      6. 1.5 Phase Changes
      7. 1.6 Mechanisms of Heat Transfer
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 The Kinetic Theory of Gases
      1. Introduction
      2. 2.1 Molecular Model of an Ideal Gas
      3. 2.2 Pressure, Temperature, and RMS Speed
      4. 2.3 Heat Capacity and Equipartition of Energy
      5. 2.4 Distribution of Molecular Speeds
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 The First Law of Thermodynamics
      1. Introduction
      2. 3.1 Thermodynamic Systems
      3. 3.2 Work, Heat, and Internal Energy
      4. 3.3 First Law of Thermodynamics
      5. 3.4 Thermodynamic Processes
      6. 3.5 Heat Capacities of an Ideal Gas
      7. 3.6 Adiabatic Processes for an Ideal Gas
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 The Second Law of Thermodynamics
      1. Introduction
      2. 4.1 Reversible and Irreversible Processes
      3. 4.2 Heat Engines
      4. 4.3 Refrigerators and Heat Pumps
      5. 4.4 Statements of the Second Law of Thermodynamics
      6. 4.5 The Carnot Cycle
      7. 4.6 Entropy
      8. 4.7 Entropy on a Microscopic Scale
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Electricity and Magnetism
    1. 5 Electric Charges and Fields
      1. Introduction
      2. 5.1 Electric Charge
      3. 5.2 Conductors, Insulators, and Charging by Induction
      4. 5.3 Coulomb's Law
      5. 5.4 Electric Field
      6. 5.5 Calculating Electric Fields of Charge Distributions
      7. 5.6 Electric Field Lines
      8. 5.7 Electric Dipoles
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    2. 6 Gauss's Law
      1. Introduction
      2. 6.1 Electric Flux
      3. 6.2 Explaining Gauss’s Law
      4. 6.3 Applying Gauss’s Law
      5. 6.4 Conductors in Electrostatic Equilibrium
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 7 Electric Potential
      1. Introduction
      2. 7.1 Electric Potential Energy
      3. 7.2 Electric Potential and Potential Difference
      4. 7.3 Calculations of Electric Potential
      5. 7.4 Determining Field from Potential
      6. 7.5 Equipotential Surfaces and Conductors
      7. 7.6 Applications of Electrostatics
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 8 Capacitance
      1. Introduction
      2. 8.1 Capacitors and Capacitance
      3. 8.2 Capacitors in Series and in Parallel
      4. 8.3 Energy Stored in a Capacitor
      5. 8.4 Capacitor with a Dielectric
      6. 8.5 Molecular Model of a Dielectric
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 9 Current and Resistance
      1. Introduction
      2. 9.1 Electrical Current
      3. 9.2 Model of Conduction in Metals
      4. 9.3 Resistivity and Resistance
      5. 9.4 Ohm's Law
      6. 9.5 Electrical Energy and Power
      7. 9.6 Superconductors
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 10 Direct-Current Circuits
      1. Introduction
      2. 10.1 Electromotive Force
      3. 10.2 Resistors in Series and Parallel
      4. 10.3 Kirchhoff's Rules
      5. 10.4 Electrical Measuring Instruments
      6. 10.5 RC Circuits
      7. 10.6 Household Wiring and Electrical Safety
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 11 Magnetic Forces and Fields
      1. Introduction
      2. 11.1 Magnetism and Its Historical Discoveries
      3. 11.2 Magnetic Fields and Lines
      4. 11.3 Motion of a Charged Particle in a Magnetic Field
      5. 11.4 Magnetic Force on a Current-Carrying Conductor
      6. 11.5 Force and Torque on a Current Loop
      7. 11.6 The Hall Effect
      8. 11.7 Applications of Magnetic Forces and Fields
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 12 Sources of Magnetic Fields
      1. Introduction
      2. 12.1 The Biot-Savart Law
      3. 12.2 Magnetic Field Due to a Thin Straight Wire
      4. 12.3 Magnetic Force between Two Parallel Currents
      5. 12.4 Magnetic Field of a Current Loop
      6. 12.5 Ampère’s Law
      7. 12.6 Solenoids and Toroids
      8. 12.7 Magnetism in Matter
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    9. 13 Electromagnetic Induction
      1. Introduction
      2. 13.1 Faraday’s Law
      3. 13.2 Lenz's Law
      4. 13.3 Motional Emf
      5. 13.4 Induced Electric Fields
      6. 13.5 Eddy Currents
      7. 13.6 Electric Generators and Back Emf
      8. 13.7 Applications of Electromagnetic Induction
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 14 Inductance
      1. Introduction
      2. 14.1 Mutual Inductance
      3. 14.2 Self-Inductance and Inductors
      4. 14.3 Energy in a Magnetic Field
      5. 14.4 RL Circuits
      6. 14.5 Oscillations in an LC Circuit
      7. 14.6 RLC Series Circuits
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 15 Alternating-Current Circuits
      1. Introduction
      2. 15.1 AC Sources
      3. 15.2 Simple AC Circuits
      4. 15.3 RLC Series Circuits with AC
      5. 15.4 Power in an AC Circuit
      6. 15.5 Resonance in an AC Circuit
      7. 15.6 Transformers
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 16 Electromagnetic Waves
      1. Introduction
      2. 16.1 Maxwell’s Equations and Electromagnetic Waves
      3. 16.2 Plane Electromagnetic Waves
      4. 16.3 Energy Carried by Electromagnetic Waves
      5. 16.4 Momentum and Radiation Pressure
      6. 16.5 The Electromagnetic Spectrum
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
  12. Index

Learning Objectives

By the end of the section, you will be able to:
  • Interpret phasor diagrams and apply them to ac circuits with resistors, capacitors, and inductors
  • Define the reactance for a resistor, capacitor, and inductor to help understand how current in the circuit behaves compared to each of these devices

In this section, we study simple models of ac voltage sources connected to three circuit components: (1) a resistor, (2) a capacitor, and (3) an inductor. The power furnished by an ac voltage source has an emf given by

v(t)=V0sinωt,v(t)=V0sinωt,

as shown in Figure 15.4. This sine function assumes we start recording the voltage when it is v=0Vv=0V at a time of t=0s.t=0s. A phase constant may be involved that shifts the function when we start measuring voltages, similar to the phase constant in the waves we studied in Waves. However, because we are free to choose when we start examining the voltage, we can ignore this phase constant for now. We can measure this voltage across the circuit components using one of two methods: (1) a quantitative approach based on our knowledge of circuits, or (2) a graphical approach that is explained in the coming sections.

Figure shows a sine wave with maximum and minimum values of the voltage being V0 and minus V0 respectively. Each positive slope of the wave, at the x-axis, marks one complete wavelength. These points are labeled in sequence: 2 pi by omega, 4 pi by omega and 6 pi by omega.
Figure 15.4 (a) The output v(t)=V0sinωtv(t)=V0sinωt of an ac generator. (b) Symbol used to represent an ac voltage source in a circuit diagram.

Resistor

First, consider a resistor connected across an ac voltage source. From Kirchhoff’s loop rule, the instantaneous voltage across the resistor of Figure 15.5(a) is

vR(t)=V0sinωtvR(t)=V0sinωt

and the instantaneous current through the resistor is

iR(t)=vR(t)R=V0Rsinωt=I0sinωt.iR(t)=vR(t)R=V0Rsinωt=I0sinωt.
Figure a shows a circuit with an AC voltage source connected to a resistor. The source is labeled V0 sine omega t. Figure b shows sine waves of AC voltage and current on the same graph. Voltage has a greater amplitude than current and its maximum value is marked V0 on the y axis. The maximum value of current is marked I0. The voltage curve is labeled V subscript R parentheses t parentheses equal to V0 sine omega t. The current curve is labeled I subscript R parentheses t parentheses equal to I0 sine omega t.
Figure 15.5 (a) A resistor connected across an ac voltage source. (b) The current iR(t)iR(t) through the resistor and the voltage vR(t)vR(t) across the resistor. The two quantities are in phase.

Here, I0=V0/RI0=V0/R is the amplitude of the time-varying current. Plots of iR(t)iR(t) and vR(t)vR(t) are shown in Figure 15.5(b). Both curves reach their maxima and minima at the same times, that is, the current through and the voltage across the resistor are in phase.

Graphical representations of the phase relationships between current and voltage are often useful in the analysis of ac circuits. Such representations are called phasor diagrams. The phasor diagram for iR(t)iR(t) is shown in Figure 15.6(a), with the current on the vertical axis. The arrow (or phasor) is rotating counterclockwise at a constant angular frequency ω,ω, so we are viewing it at one instant in time. If the length of the arrow corresponds to the current amplitude I0,I0, the projection of the rotating arrow onto the vertical axis is iR(t)=I0sinωt,iR(t)=I0sinωt, which is the instantaneous current.

Figure shows the coordinate axes. An arrow labeled V0 starts from the origin and points up and right making an angle omega t with the x axis. An arrow labeled omega is shown near its tip, perpendicular to it, pointing up and left. The tip of the arrow V0 makes a y-intercept labeled V subscript C parentheses t parentheses. An arrow labeled I0 starts at the origin and points up and left. It is perpendicular to V0. It makes a y intercept labeled i subscript C parentheses t parentheses. A arrow labeled omega is shown near its tip, perpendicular to it, pointing down and left.
Figure 15.6 (a) The phasor diagram representing the current through the resistor of Figure 15.5. (b) The phasor diagram representing both iR(t)iR(t) and vR(t)vR(t).

The vertical axis on a phasor diagram could be either the voltage or the current, depending on the phasor that is being examined. In addition, several quantities can be depicted on the same phasor diagram. For example, both the current iR(t)iR(t) and the voltage vR(t)vR(t) are shown in the diagram of Figure 15.6(b). Since they have the same frequency and are in phase, their phasors point in the same direction and rotate together. The relative lengths of the two phasors are arbitrary because they represent different quantities; however, the ratio of the lengths of the two phasors can be represented by the resistance, since one is a voltage phasor and the other is a current phasor.

Capacitor

Now let’s consider a capacitor connected across an ac voltage source. From Kirchhoff’s loop rule, the instantaneous voltage across the capacitor of Figure 15.7(a) is

vC(t)=V0sinωt.vC(t)=V0sinωt.

Recall that the charge in a capacitor is given by Q=CV.Q=CV. This is true at any time measured in the ac cycle of voltage. Consequently, the instantaneous charge on the capacitor is

q(t)=CvC(t)=CV0sinωt.q(t)=CvC(t)=CV0sinωt.

Since the current in the circuit is the rate at which charge enters (or leaves) the capacitor,

iC(t)=dq(t)dt=ωCV0cosωt=I0cosωt,iC(t)=dq(t)dt=ωCV0cosωt=I0cosωt,

where I0=ωCV0I0=ωCV0 is the current amplitude. Using the trigonometric relationship cosωt=sin(ωt+π/2),cosωt=sin(ωt+π/2), we may express the instantaneous current as

iC(t)=I0sin(ωt+π2).iC(t)=I0sin(ωt+π2).

Dividing V0V0 by I0I0, we obtain an equation that looks similar to Ohm’s law:

V0I0=1ωC=XC.V0I0=1ωC=XC.
(15.3)

The quantity XCXC is analogous to resistance in a dc circuit in the sense that both quantities are a ratio of a voltage to a current. As a result, they have the same unit, the ohm. Keep in mind, however, that a capacitor stores and discharges electric energy, whereas a resistor dissipates it. The quantity XCXC is known as the capacitive reactance of the capacitor, or the opposition of a capacitor to a change in current. It depends inversely on the frequency of the ac source—high frequency leads to low capacitive reactance.

Figure a shows a circuit with an AC voltage source connected to a capacitor. The source is labeled V0 sine omega t. Figure b shows sine waves of AC voltage and current on the same graph. Voltage has a greater amplitude than current and its maximum value is marked V0 on the y axis. The maximum value of current is marked I0. The two curves have the same wavelength but are out of phase by one quarter wavelength. The voltage curve is labeled V subscript C parentheses t parentheses equal to V0 sine omega t. The current curve is labeled I subscript C parentheses t parentheses equal to I0 sine parentheses omega t plus pi by 2 parentheses.
Figure 15.7 (a) A capacitor connected across an ac generator. (b) The current iC(t)iC(t) through the capacitor and the voltage vC(t)vC(t) across the capacitor. Notice that iC(t)iC(t) leads vC(t)vC(t) by π/2π/2 rad.

A comparison of the expressions for vC(t)vC(t) and iC(t)iC(t) shows that there is a phase difference of π/2radπ/2rad between them. When these two quantities are plotted together, the current peaks a quarter cycle (or π/2radπ/2rad) ahead of the voltage, as illustrated in Figure 15.7(b). The current through a capacitor leads the voltage across a capacitor by π/2rad,π/2rad, or a quarter of a cycle.

The corresponding phasor diagram is shown in Figure 15.8. Here, the relationship between iC(t)iC(t) and vC(t)vC(t) is represented by having their phasors rotate at the same angular frequency, with the current phasor leading by π/2rad.π/2rad.

Figure shows the coordinate axes. An arrow labeled V0 starts from the origin and points up and right making an angle omega t with the x axis. An arrow labeled omega is shown near its tip, perpendicular to it, pointing up and left. The tip of the arrow V0 makes a y-intercept labeled V subscript C parentheses t parentheses. An arrow labeled I0 starts at the origin and points up and left. It is perpendicular to V0. It makes a y intercept labeled i subscript C parentheses t parentheses. A arrow labeled omega is shown near its tip, perpendicular to it, pointing down and left.
Figure 15.8 The phasor diagram for the capacitor of Figure 15.7. The current phasor leads the voltage phasor by π/2π/2 rad as they both rotate with the same angular frequency.

To this point, we have exclusively been using peak values of the current or voltage in our discussion, namely, I0I0 and V0.V0. However, if we average out the values of current or voltage, these values are zero. Therefore, we often use a second convention called the root mean square value, or rms value, in discussions of current and voltage. The rms operates in reverse of the terminology. First, you square the function, next, you take the mean, and then, you find the square root. As a result, the rms values of current and voltage are not zero. Appliances and devices are commonly quoted with rms values for their operations, rather than peak values. We indicate rms values with a subscript attached to a capital letter (such as IrmsIrms).

Although a capacitor is basically an open circuit, an rms current, or the root mean square of the current, appears in a circuit with an ac voltage applied to a capacitor. Consider that

Irms=I02,Irms=I02,
(15.4)

where I0I0 is the peak current in an ac system. The rms voltage, or the root mean square of the voltage, is

Vrms=V02,Vrms=V02,
(15.5)

where V0V0 is the peak voltage in an ac system. The rms current appears because the voltage is continually reversing, charging, and discharging the capacitor. If the frequency goes to zero, which would be a dc voltage, XCXC tends to infinity, and the current is zero once the capacitor is charged. At very high frequencies, the capacitor’s reactance tends to zero—it has a negligible reactance and does not impede the current (it acts like a simple wire).

Inductor

Lastly, let’s consider an inductor connected to an ac voltage source. From Kirchhoff’s loop rule, the voltage across the inductor L of Figure 15.9(a) is

vL(t)=V0sinωt.vL(t)=V0sinωt.
(15.6)

The emf across an inductor is equal to ε=L(diL/dt);ε=L(diL/dt); however, the potential difference across the inductor is vL(t)=LdiL(t)/dtvL(t)=LdiL(t)/dt, because if we consider that the voltage around the loop must equal zero, the voltage gained from the ac source must dissipate through the inductor. Therefore, connecting this with the ac voltage source, we have

diL(t)dt=V0Lsinωt.diL(t)dt=V0Lsinωt.
Figure a shows a circuit with an AC voltage source connected to an inductor. The source is labeled V0 sine omega t. Figure b shows sine waves of AC voltage and current on the same graph. Voltage has a smaller amplitude than current and its maximum value is marked V0 on the y axis. The maximum value of current is marked I0. The two curves have the same wavelength but are out of phase by one quarter wavelength. The voltage curve is labeled V subscript L parentheses t parentheses equal to V0 sine omega t. The current curve is labeled I subscript L parentheses t parentheses equal to I0 sine parentheses omega t minus pi by 2 parentheses.
Figure 15.9 (a) An inductor connected across an ac generator. (b) The current iL(t)iL(t) through the inductor and the voltage vL(t)vL(t) across the inductor. Here iL(t)iL(t) lags vL(t)vL(t) by π/2π/2 rad.

The current iL(t)iL(t) is found by integrating this equation. Since the circuit does not contain a source of constant emf, there is no steady current in the circuit. Hence, we can set the constant of integration, which represents the steady current in the circuit, equal to zero, and we have

iL(t)=V0ωLcosωt=V0ωLsin(ωtπ2)=I0sin(ωtπ2),iL(t)=V0ωLcosωt=V0ωLsin(ωtπ2)=I0sin(ωtπ2),
(15.7)

where I0=V0/ωL.I0=V0/ωL. The relationship between V0V0 and I0I0 may also be written in a form analogous to Ohm’s law:

V0I0=ωL=XL.V0I0=ωL=XL.
(15.8)

The quantity XLXL is known as the inductive reactance of the inductor, or the opposition of an inductor to a change in current; its unit is also the ohm. Note that XLXL varies directly as the frequency of the ac source—high frequency causes high inductive reactance.

A phase difference of π/2π/2 rad occurs between the current through and the voltage across the inductor. From Equation 15.6 and Equation 15.7, the current through an inductor lags the potential difference across an inductor by π/2radπ/2rad, or a quarter of a cycle. The phasor diagram for this case is shown in Figure 15.10.

Figure shows the coordinate axes. An arrow labeled V0 starts from the origin and points up and right making an angle omega t with the x axis. An arrow labeled omega is shown near its tip, perpendicular to it, pointing up and left. The tip of the arrow V0 makes a y-intercept labeled V subscript L parentheses t parentheses. An arrow labeled I0 starts at the origin and points down and right. It is perpendicular to V0. Its intercept on the negative y-axis is labeled i subscript L parentheses t parentheses. A arrow labeled omega is shown near its tip, perpendicular to it, pointing up and right.
Figure 15.10 The phasor diagram for the inductor of Figure 15.9. The current phasor lags the voltage phasor by π/2π/2 rad as they both rotate with the same angular frequency.

Interactive

An animation from the University of New South Wales AC Circuits illustrates some of the concepts we discuss in this chapter. They also include wave and phasor diagrams that evolve over time so that you can get a better picture of how each changes over time.

Example 15.1

Simple AC Circuits An ac generator produces an emf of amplitude 10 V at a frequency f=60Hz.f=60Hz. Determine the voltages across and the currents through the circuit elements when the generator is connected to (a) a 100-Ω100-Ω resistor, (b) a 10-μF10-μF capacitor, and (c) a 15-mH inductor.

Strategy The entire AC voltage across each device is the same as the source voltage. We can find the currents by finding the reactance X of each device and solving for the peak current using I0=V0/X.I0=V0/X.

Solution The voltage across the terminals of the source is

v(t)=V0sinωt=(10V)sin120πt,v(t)=V0sinωt=(10V)sin120πt,

where ω=2πf=120πrad/sω=2πf=120πrad/s is the angular frequency. Since v(t) is also the voltage across each of the elements, we have

v(t)=vR(t)=vC(t)=vL(t)=(10V)sin120πt.v(t)=vR(t)=vC(t)=vL(t)=(10V)sin120πt.

a. When R=100Ω,R=100Ω, the amplitude of the current through the resistor is

I0=V0/R=10V/100Ω=0.10A,I0=V0/R=10V/100Ω=0.10A,

so

iR(t)=(0.10A)sin120πt.iR(t)=(0.10A)sin120πt.

b. From Equation 15.3, the capacitive reactance is

XC=1ωC=1(120πrad/s)(10×10−6F)=265Ω,XC=1ωC=1(120πrad/s)(10×10−6F)=265Ω,

so the maximum value of the current is

I0=V0XC=10V265Ω=3.8×10−2AI0=V0XC=10V265Ω=3.8×10−2A

and the instantaneous current is given by

iC(t)=(3.8×10−2A)sin(120πt+π2).iC(t)=(3.8×10−2A)sin(120πt+π2).

c. From Equation 15.8, the inductive reactance is

XL=ωL=(120πrad/s)(15×10−3H)=5.7Ω.XL=ωL=(120πrad/s)(15×10−3H)=5.7Ω.

The maximum current is therefore

I0=10V5.7Ω=1.8AI0=10V5.7Ω=1.8A

and the instantaneous current is

iL(t)=(1.8A)sin(120πtπ2).iL(t)=(1.8A)sin(120πtπ2).

Significance Although the voltage across each device is the same, the peak current has different values, depending on the reactance. The reactance for each device depends on the values of resistance, capacitance, or inductance.

Check Your Understanding 15.2

Repeat Example 15.1 for an ac source of amplitude 20 V and frequency 100 Hz.

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