#set document(title: "15.2 Simple AC Circuits", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 15.2#h(0.6em)Simple AC Circuits 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 #math.equation(block: true, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t ,")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t ,$] as shown. This sine function assumes we start recording the voltage when it is #math.equation(block: false, alt: "v equals 0 V")[$v = 0 #h(0.2em) "V"$] at a time of #math.equation(block: false, alt: "t equals 0 s .")[$t = 0 #h(0.2em) "s" "."$] 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(figph[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.], alt: "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.", caption: [(a) The output #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t$] of an ac generator. (b) Symbol used to represent an ac voltage source in a circuit diagram.]) === Resistor First, consider a #strong[resistor] connected across an ac voltage source. From Kirchhoff’s loop rule, the instantaneous voltage across the resistor of (a) is #math.equation(block: true, alt: "v sub R open parenthesis t close parenthesis equals V sub 0 sin ω t")[$v_(R) ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t$] and the instantaneous current through the resistor is #math.equation(block: true, alt: "i sub R open parenthesis t close parenthesis equals the fraction v sub R open parenthesis t close parenthesis over R equals the fraction V sub 0 over R sin ω t equals I sub 0 sin ω t .")[$i_(R) ( t ) = frac(v_(R) ( t ), R) = frac(V_(0), R) #h(0.2em) "sin" #h(0.2em) ω t = I_(0) #h(0.2em) "sin" #h(0.2em) ω t .$] #figure(figph[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.], alt: "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.", caption: [(a) A resistor connected across an ac voltage source. (b) The current #math.equation(block: false, alt: "i sub R open parenthesis t close parenthesis")[$i_(R) ( t )$] through the resistor and the voltage #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis")[$v_(R) ( t )$] across the resistor. The two quantities are in phase.]) Here, #math.equation(block: false, alt: "I sub 0 equals V sub 0 / R")[$I_(0) = V_(0) "/" R$] is the amplitude of the time-varying current. Plots of #math.equation(block: false, alt: "i sub R open parenthesis t close parenthesis")[$i_(R) ( t )$] and #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis")[$v_(R) ( t )$] are shown in (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 #strong[#emph[phasor diagrams]]. The phasor diagram for #math.equation(block: false, alt: "i sub R open parenthesis t close parenthesis")[$i_(R) ( t )$] is shown in (a), with the current on the vertical axis. The arrow (or phasor) is rotating counterclockwise at a constant angular frequency #math.equation(block: false, alt: "ω ,")[$ω ,$] so we are viewing it at one instant in time. If the length of the arrow corresponds to the current amplitude #math.equation(block: false, alt: "I sub 0 ,")[$I_(0) ,$] the projection of the rotating arrow onto the vertical axis is #math.equation(block: false, alt: "i sub R open parenthesis t close parenthesis equals I sub 0 sin ω t ,")[$i_(R) ( t ) = I_(0) #h(0.2em) "sin" #h(0.2em) ω t ,$] which is the instantaneous current. #figure(figph[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.], alt: "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.", caption: [(a) The phasor diagram representing the current through the resistor of . (b) The phasor diagram representing both #math.equation(block: false, alt: "i sub R open parenthesis t close parenthesis")[$i_(R) ( t )$] and #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis")[$v_(R) ( t )$].]) 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 #math.equation(block: false, alt: "i sub R open parenthesis t close parenthesis")[$i_(R) ( t )$] and the voltage #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis")[$v_(R) ( t )$] are shown in the diagram of (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 #strong[capacitor] connected across an ac voltage source. From Kirchhoff’s loop rule, the instantaneous voltage across the capacitor of (a) is #math.equation(block: true, alt: "v sub C open parenthesis t close parenthesis equals V sub 0 sin ω t .")[$v_(C) ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t .$] Recall that the charge in a capacitor is given by #math.equation(block: false, alt: "Q equals C V .")[$Q = C V .$] This is true at any time measured in the ac cycle of voltage. Consequently, the instantaneous charge on the capacitor is #math.equation(block: true, alt: "q open parenthesis t close parenthesis equals C v sub C open parenthesis t close parenthesis equals C V sub 0 sin ω t .")[$q ( t ) = C v_(C) ( t ) = C V_(0) #h(0.2em) "sin" #h(0.2em) ω t .$] Since the current in the circuit is the rate at which charge enters (or leaves) the capacitor, #math.equation(block: true, alt: "i sub C open parenthesis t close parenthesis equals the fraction d q open parenthesis t close parenthesis over d t equals ω C V sub 0 cos ω t equals I sub 0 cos ω t ,")[$i_(C) ( t ) = frac(d q ( t ), d t) = ω C V_(0) #h(0.2em) "cos" #h(0.2em) ω t = I_(0) #h(0.2em) "cos" #h(0.2em) ω t ,$] where #math.equation(block: false, alt: "I sub 0 equals ω C V sub 0")[$I_(0) = ω C V_(0)$] is the current amplitude. Using the trigonometric relationship #math.equation(block: false, alt: "cos ω t equals sin open parenthesis ω t plus π / 2 close parenthesis ,")[$"cos" #h(0.2em) ω t = #h(0.2em) "sin" #h(0.2em) ( ω t + π "/" 2 ) ,$] we may express the instantaneous current as #math.equation(block: true, alt: "i sub C open parenthesis t close parenthesis equals I sub 0 sin open parenthesis ω t plus the fraction π over 2 close parenthesis .")[$i_(C) ( t ) = I_(0) #h(0.2em) "sin" #h(0.2em) ( ω t + frac(π, 2) ) .$] Dividing #math.equation(block: false, alt: "V sub 0")[$V_(0)$] by #math.equation(block: false, alt: "I sub 0")[$I_(0)$], we obtain an equation that looks similar to Ohm’s law: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "the fraction V sub 0 over I sub 0 equals the fraction 1 over ω C equals X sub C .")[$frac(V_(0), I_(0)) = frac(1, ω C) = X_(C) .$] ] The quantity #math.equation(block: false, alt: "X sub C")[$X_(C)$] 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 #math.equation(block: false, alt: "X sub C")[$X_(C)$] is known as the #strong[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(figph[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.], alt: "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.", caption: [(a) A capacitor connected across an ac generator. (b) The current #math.equation(block: false, alt: "i sub C open parenthesis t close parenthesis")[$i_(C) ( t )$] through the capacitor and the voltage #math.equation(block: false, alt: "v sub C open parenthesis t close parenthesis")[$v_(C) ( t )$] across the capacitor. Notice that #math.equation(block: false, alt: "i sub C open parenthesis t close parenthesis")[$i_(C) ( t )$] leads #math.equation(block: false, alt: "v sub C open parenthesis t close parenthesis")[$v_(C) ( t )$] by #math.equation(block: false, alt: "π / 2")[$π "/" 2$] rad.]) A comparison of the expressions for #math.equation(block: false, alt: "v sub C open parenthesis t close parenthesis")[$v_(C) ( t )$] and #math.equation(block: false, alt: "i sub C open parenthesis t close parenthesis")[$i_(C) ( t )$] shows that there is a phase difference of #math.equation(block: false, alt: "π / 2 rad")[$π "/" 2 #h(0.2em) "rad"$] between them. When these two quantities are plotted together, the current peaks a quarter cycle (or #math.equation(block: false, alt: "π / 2 rad")[$π "/" 2 #h(0.2em) "rad"$]) ahead of the voltage, as illustrated in (b). The current through a capacitor leads the voltage across a capacitor by #math.equation(block: false, alt: "π / 2 rad ,")[$π "/" 2 #h(0.2em) "rad" ,$] or a quarter of a cycle. The corresponding phasor diagram is shown. Here, the relationship between #math.equation(block: false, alt: "i sub C open parenthesis t close parenthesis")[$i_(C) ( t )$] and #math.equation(block: false, alt: "v sub C open parenthesis t close parenthesis")[$v_(C) ( t )$] is represented by having their phasors rotate at the same angular frequency, with the current phasor leading by #math.equation(block: false, alt: "π / 2 rad .")[$π "/" 2 #h(0.2em) "rad" .$] #figure(figph[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.], alt: "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.", caption: [The phasor diagram for the capacitor of . The current phasor leads the voltage phasor by #math.equation(block: false, alt: "π / 2")[$π "/" 2$] 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, #math.equation(block: false, alt: "I sub 0")[$I_(0)$] and #math.equation(block: false, alt: "V sub 0 .")[$V_(0) .$] 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 #math.equation(block: false, alt: "I sub rms")[$I_("rms")$]). Although a capacitor is basically an open circuit, an #strong[rms current], or the root mean square of the current, appears in a circuit with an ac voltage applied to a capacitor. Consider that #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "I sub rms equals the fraction I sub 0 over the square root of 2 ,")[$I_("rms") = frac(I_(0), sqrt(2)) ,$] ] where #math.equation(block: false, alt: "I sub 0")[$I_(0)$] is the peak current in an ac system. The #strong[rms voltage], or the root mean square of the voltage, is #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "V sub rms equals the fraction V sub 0 over the square root of 2 ,")[$V_("rms") = frac(V_(0), sqrt(2)) ,$] ] where #math.equation(block: false, alt: "V sub 0")[$V_(0)$] 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, #math.equation(block: false, alt: "X sub C")[$X_(C)$] 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 #strong[inductor] connected to an ac voltage source. From Kirchhoff’s loop rule, the voltage across the inductor #emph[L] of (a) is #math.equation(block: true, alt: "v sub L open parenthesis t close parenthesis equals V sub 0 sin ω t .")[$v_(L) ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t .$] The emf across an inductor is equal to #math.equation(block: false, alt: "ε equals − L open parenthesis d i sub L / d t close parenthesis ;")[$ε = "−" L ( d i_(L) "/" d t ) ;$] however, the potential difference across the inductor is #math.equation(block: false, alt: "v sub L open parenthesis t close parenthesis equals L d i sub L open parenthesis t close parenthesis / d t")[$v_(L) ( t ) = L d i_(L) ( t ) "/" d t$], 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 #math.equation(block: true, alt: "the fraction d i sub L open parenthesis t close parenthesis over d t equals the fraction V sub 0 over L sin ω t .")[$frac(d i_(L) ( t ), d t) = frac(V_(0), L) #h(0.2em) "sin" #h(0.2em) ω t .$] #figure(figph[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.], alt: "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.", caption: [(a) An inductor connected across an ac generator. (b) The current #math.equation(block: false, alt: "i sub L open parenthesis t close parenthesis")[$i_(L) ( t )$] through the inductor and the voltage #math.equation(block: false, alt: "v sub L open parenthesis t close parenthesis")[$v_(L) ( t )$] across the inductor. Here #math.equation(block: false, alt: "i sub L open parenthesis t close parenthesis")[$i_(L) ( t )$] lags #math.equation(block: false, alt: "v sub L open parenthesis t close parenthesis")[$v_(L) ( t )$] by #math.equation(block: false, alt: "π / 2")[$π "/" 2$] rad.]) The current #math.equation(block: false, alt: "i sub L open parenthesis t close parenthesis")[$i_(L) ( t )$] 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 #math.equation(block: true, alt: "i sub L open parenthesis t close parenthesis equals minus the fraction V sub 0 over ω L cos ω t equals the fraction V sub 0 over ω L sin open parenthesis ω t minus the fraction π over 2 close parenthesis equals I sub 0 sin open parenthesis ω t minus the fraction π over 2 close parenthesis ,")[$i_(L) ( t ) = − frac(V_(0), ω L) "cos" #h(0.2em) ω t = frac(V_(0), ω L) #h(0.2em) "sin" #h(0.2em) ( ω t − frac(π, 2) ) = I_(0) #h(0.2em) "sin" #h(0.2em) ( ω t − frac(π, 2) ) ,$] where #math.equation(block: false, alt: "I sub 0 equals V sub 0 / ω L .")[$I_(0) = V_(0) "/" ω L .$] The relationship between #math.equation(block: false, alt: "V sub 0")[$V_(0)$] and #math.equation(block: false, alt: "I sub 0")[$I_(0)$] may also be written in a form analogous to Ohm’s law: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "the fraction V sub 0 over I sub 0 equals ω L equals X sub L .")[$frac(V_(0), I_(0)) = ω L = X_(L) .$] ] The quantity #math.equation(block: false, alt: "X sub L")[$X_(L)$] is known as the #strong[inductive reactance] of the inductor, or the opposition of an inductor to a change in current; its unit is also the ohm. Note that #math.equation(block: false, alt: "X sub L")[$X_(L)$] varies directly as the frequency of the ac source—high frequency causes high inductive reactance. A phase difference of #math.equation(block: false, alt: "π / 2")[$π "/" 2$] rad occurs between the current through and the voltage across the inductor. From and , the current through an inductor lags the potential difference across an inductor by #math.equation(block: false, alt: "π / 2 rad")[$π "/" 2 #h(0.2em) "rad"$], or a quarter of a cycle. The phasor diagram for this case is shown. #figure(figph[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.], alt: "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.", caption: [The phasor diagram for the inductor of . The current phasor lags the voltage phasor by #math.equation(block: false, alt: "π / 2")[$π "/" 2$] rad as they both rotate with the same angular frequency.]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ An animation from the University of New South Wales #link("https://openstax.org/l/21accircuits")[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. ] #examplebox("Example 1")[Simple AC Circuits][ An ac generator produces an emf of amplitude 10 V at a frequency #math.equation(block: false, alt: "f equals 60 Hz .")[$f = 60 #h(0.2em) "Hz" "."$] Determine the voltages across and the currents through the circuit elements when the generator is connected to (a) a #math.equation(block: false, alt: "100 minus Ω")[$100 − "Ω"$] resistor, (b) a #math.equation(block: false, alt: "10 minus μ F")[$10 − μ "F"$] 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 #emph[X] of each device and solving for the peak current using #math.equation(block: false, alt: "I sub 0 equals V sub 0 / X .")[$I_(0) = V_(0) "/" X .$] Solution The voltage across the terminals of the source is #math.equation(block: true, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t equals open parenthesis 10 V close parenthesis sin 120 π t ,")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t = ( 10 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) 120 π t ,$] where #math.equation(block: false, alt: "ω equals 2 π f equals 120 π rad/s")[$ω = 2 π f = 120 π #h(0.2em) "rad/s"$] is the angular frequency. Since #emph[v] (#emph[t]) is also the voltage across each of the elements, we have #math.equation(block: true, alt: "v open parenthesis t close parenthesis equals v sub R open parenthesis t close parenthesis equals v sub C open parenthesis t close parenthesis equals v sub L open parenthesis t close parenthesis equals open parenthesis 10 V close parenthesis sin 120 π t .")[$v ( t ) = v_(R) ( t ) = v_(C) ( t ) = v_(L) ( t ) = ( 10 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) 120 π t .$] a. When #math.equation(block: false, alt: "R equals 100 Ω ,")[$R = 100 #h(0.2em) "Ω" ,$] the amplitude of the current through the resistor is #math.equation(block: true, alt: "I sub 0 equals V sub 0 / R equals 10 V / 100 Ω equals 0.10 A,")[$I_(0) = V_(0) "/" R = 10 #h(0.2em) "V" "/" 100 #h(0.2em) "Ω" = 0.10 #h(0.2em) "A,"$] so #math.equation(block: true, alt: "i sub R open parenthesis t close parenthesis equals open parenthesis 0.10 A close parenthesis sin 120 π t .")[$i_(R) ( t ) = ( 0.10 #h(0.2em) "A" ) #h(0.2em) "sin" #h(0.2em) 120 π t .$] b. From , the capacitive reactance is #math.equation(block: true, alt: "X sub C equals the fraction 1 over ω C equals the fraction 1 over open parenthesis 120 π rad/s close parenthesis open parenthesis 10 times 10 to the power −6 F close parenthesis equals 265 Ω ,")[$X_(C) = frac(1, ω C) = frac(1, ( 120 π #h(0.2em) "rad/s" ) ( 10 #h(0.2em) × #h(0.2em) 10^(−6) "F" )) = 265 #h(0.2em) "Ω" ,$] so the maximum value of the current is #math.equation(block: true, alt: "I sub 0 equals the fraction V sub 0 over X sub C equals the fraction 10 V over 265 Ω equals 3.8 times 10 to the power −2 A")[$I_(0) = frac(V_(0), X_(C)) = frac(10 #h(0.2em) "V", 265 #h(0.2em) "Ω") = 3.8 #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "A"$] and the instantaneous current is given by #math.equation(block: true, alt: "i sub C open parenthesis t close parenthesis equals open parenthesis 3.8 times 10 to the power −2 A close parenthesis sin open parenthesis 120 π t plus the fraction π over 2 close parenthesis .")[$i_(C) ( t ) = ( 3.8 #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "A" ) #h(0.2em) "sin" #h(0.2em) ( 120 π t + frac(π, 2) ) .$] c. From , the inductive reactance is #math.equation(block: true, alt: "X sub L equals ω L equals open parenthesis 120 π rad/s close parenthesis open parenthesis 15 times 10 to the power −3 H close parenthesis equals 5.7 Ω .")[$X_(L) = ω L = ( 120 π #h(0.2em) "rad/s" ) ( 15 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H" ) = 5.7 #h(0.2em) "Ω" "."$] The maximum current is therefore #math.equation(block: true, alt: "I sub 0 equals the fraction 10 V over 5.7 Ω equals 1.8 A")[$I_(0) = frac(10 #h(0.2em) "V", 5.7 #h(0.2em) "Ω") = 1.8 #h(0.2em) "A"$] and the instantaneous current is #math.equation(block: true, alt: "i sub L open parenthesis t close parenthesis equals open parenthesis 1.8 A close parenthesis sin open parenthesis 120 π t minus the fraction π over 2 close parenthesis .")[$i_(L) ( t ) = ( 1.8 #h(0.2em) "A" ) #h(0.2em) "sin" #h(0.2em) ( 120 π t − frac(π, 2) ) .$] 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. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Repeat for an ac source of amplitude 20 V and frequency 100 Hz. #solutionbox[ a. #math.equation(block: false, alt: "open parenthesis 20 V close parenthesis sin 200 π t , open parenthesis 0.20 A close parenthesis sin 200 π t")[$( 20 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) 200 π t "," #h(0.2em) ( 0.20 #h(0.2em) "A" ) #h(0.2em) "sin" #h(0.2em) 200 π t$]; b. #math.equation(block: false, alt: "open parenthesis 20 V close parenthesis sin 200 π t , open parenthesis 0.13 A close parenthesis sin open parenthesis 200 π t plus π / 2 close parenthesis")[$( 20 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) 200 π t "," #h(0.2em) ( 0.13 #h(0.2em) "A" ) #h(0.2em) "sin" #h(0.2em) ( 200 π t + π "/" 2 )$]; c. #math.equation(block: false, alt: "open parenthesis 20 V close parenthesis sin 200 π t , open parenthesis 2.1 A close parenthesis sin open parenthesis 200 π t minus π / 2 close parenthesis")[$( 20 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) 200 π t "," #h(0.2em) ( 2.1 #h(0.2em) "A" ) #h(0.2em) "sin" #h(0.2em) ( 200 π t − π "/" 2 )$] ] ] === Summary - For resistors, the current through and the voltage across are in phase. - For capacitors, we find that when a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle. Since a capacitor can stop current when fully charged, it limits current and offers another form of ac resistance, called capacitive reactance, which has units of ohms. - For inductors in ac circuits, we find that when a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle. - The opposition of an inductor to a change in current is expressed as a type of ac reactance. This inductive reactance, which has units of ohms, varies with the frequency of the ac source. === Conceptual Questions Explain why at high frequencies a capacitor acts as an ac short, whereas an inductor acts as an open circuit. === Problems Calculate the reactance of a #math.equation(block: false, alt: "5.0 - μ F")[$5.0 "-" μ "F"$] capacitor at (a) 60 Hz, (b) 600 Hz, and (c) 6000 Hz. #solutionbox[ a. #math.equation(block: false, alt: "530 Ω")[$530 #h(0.2em) "Ω"$]; b. #math.equation(block: false, alt: "53 Ω")[$53 #h(0.2em) "Ω"$]; c. #math.equation(block: false, alt: "5.3 Ω")[$5.3 #h(0.2em) "Ω"$] ] What is the capacitance of a capacitor whose reactance is #math.equation(block: false, alt: "10 Ω")[$10 #h(0.2em) "Ω"$] at 60 Hz? Calculate the reactance of a 5.0-mH inductor at (a) 60 Hz, (b) 600 Hz, and (c) 6000 Hz. #solutionbox[ a. #math.equation(block: false, alt: "1.9 Ω")[$1.9 #h(0.2em) "Ω"$]; b. #math.equation(block: false, alt: "19 Ω")[$19 #h(0.2em) "Ω"$]; c. #math.equation(block: false, alt: "190 Ω")[$190 #h(0.2em) "Ω"$] ] What is the self-inductance of a coil whose reactance is #math.equation(block: false, alt: "10 Ω")[$10 #h(0.2em) "Ω"$] at 60 Hz? At what frequency is the reactance of a #math.equation(block: false, alt: "20 - μ F")[$20 "-" μ "F"$] capacitor equal to that of a 10-mH inductor? #solutionbox[ 360 Hz ] At 1000 Hz, the reactance of a 5.0-mH inductor is equal to the reactance of a particular capacitor. What is the capacitance of the capacitor? A #math.equation(block: false, alt: "50 - Ω")[$50 "-" "Ω"$] resistor is connected across the emf #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals open parenthesis 160 V close parenthesis sin open parenthesis 120 π t close parenthesis")[$v ( t ) = ( 160 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) ( 120 π t )$]. Write an expression for the current through the resistor. #solutionbox[ #math.equation(block: true, alt: "i open parenthesis t close parenthesis equals open parenthesis 3.2 A close parenthesis sin open parenthesis 120 π t close parenthesis")[$i ( t ) = ( 3.2 #h(0.2em) "A" ) #h(0.2em) "sin" #h(0.2em) ( 120 π t )$] ] A #math.equation(block: false, alt: "25 - μ F")[$25 "-" μ "F"$] capacitor is connected to an emf given by #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals open parenthesis 160 V close parenthesis sin open parenthesis 120 π t close parenthesis")[$v ( t ) = ( 160 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) ( 120 π t )$]. (a) What is the reactance of the capacitor? (b) Write an expression for the current output of the source. A 100-mH inductor is connected across the emf of the preceding problem. (a) What is the reactance of the inductor? (b) Write an expression for the current through the inductor. #solutionbox[ a. #math.equation(block: false, alt: "38 Ω")[$38 #h(0.2em) "Ω"$]; b. #math.equation(block: false, alt: "i open parenthesis t close parenthesis equals open parenthesis 4.24 A close parenthesis sin open parenthesis 120 π t minus π / 2 close parenthesis")[$i ( t ) = ( 4.24 "A" ) #h(0.2em) "sin" #h(0.2em) ( 120 π t − π "/" 2 )$] ]