#set document(title: "15.4 Power in an AC Circuit", 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")) == 15.4#h(0.6em)Power in an AC Circuit A circuit element dissipates or produces power according to #math.equation(block: false, alt: "P equals I V ,")[$P = I V ,$] where #emph[I] is the current through the element and #emph[V] is the voltage across it. Since the current and the voltage both depend on time in an ac circuit, the instantaneous power #math.equation(block: false, alt: "p open parenthesis t close parenthesis equals i open parenthesis t close parenthesis v open parenthesis t close parenthesis")[$p ( t ) = i ( t ) v ( t )$] is also time dependent. A plot of #emph[p] (#emph[t]) for various circuit elements is shown. For a resistor, #emph[i] (#emph[t]) and #emph[v] (#emph[t]) are in phase and therefore always have the same sign. For a capacitor or inductor, the relative signs of #emph[i] (#emph[t]) and #emph[v] (#emph[t]) vary over a cycle due to their phase differences (see and ). Consequently, #emph[p] (#emph[t]) is positive at some times and negative at others, indicating that capacitive and inductive elements produce power at some instants and absorb it at others. #figure(figph[Figures a through d show sine waves on graphs of P versus t. All have the same amplitude and frequency. Figure a is labeled resistor. P bar is equal to half I0 V0. The sine wave is above the x axis, with the minimum y value being 0. It starts from a trough. Figure b is labeled capacitor. P bar is equal to 0. The equilibrium position of the sine wave is along the x axis. It starts at equilibrium with a positive slope. Figure c is labeled inductor. P bar is equal to 0. The equilibrium position of the sine wave is along the x axis. It starts at equilibrium with a negative slope. Figure d is labeled AC source. P bar is equal to half I0 V0 cos phi. The equilibrium position of the sine wave is above the x axis, with the minimum y-value of the wave being negative.], alt: "Figures a through d show sine waves on graphs of P versus t. All have the same amplitude and frequency. Figure a is labeled resistor. P bar is equal to half I0 V0. The sine wave is above the x axis, with the minimum y value being 0. It starts from a trough. Figure b is labeled capacitor. P bar is equal to 0. The equilibrium position of the sine wave is along the x axis. It starts at equilibrium with a positive slope. Figure c is labeled inductor. P bar is equal to 0. The equilibrium position of the sine wave is along the x axis. It starts at equilibrium with a negative slope. Figure d is labeled AC source. P bar is equal to half I0 V0 cos phi. The equilibrium position of the sine wave is above the x axis, with the minimum y-value of the wave being negative.", caption: [Graph of instantaneous power for various circuit elements. (a) For the resistor, #math.equation(block: false, alt: "P sub ave equals I sub 0 V sub 0 / 2 ,")[$P_("ave") = I_(0) V_(0) "/" 2 ,$] whereas for (b) the capacitor and (c) the inductor, #math.equation(block: false, alt: "P sub ave equals 0 .")[$P_("ave") = 0 .$] (d) For the source, #math.equation(block: false, alt: "P sub ave equals I sub 0 V sub 0 open parenthesis cos ϕ close parenthesis / 2 ,")[$P_("ave") = I_(0) V_(0) ( "cos" #h(0.2em) ϕ ) "/" 2 ,$] which may be positive, negative, or zero, depending on #math.equation(block: false, alt: "ϕ .")[$ϕ .$]]) Because instantaneous power varies in both magnitude and sign over a cycle, it seldom has any practical importance. What we’re almost always concerned with is the power averaged over time, which we refer to as the #strong[average power]. It is defined by the time average of the instantaneous power over one cycle: #math.equation(block: true, alt: "P sub ave equals the fraction 1 over T ∫ 0 T p open parenthesis t close parenthesis d t ,")[$P_("ave") = frac(1, T) display(∫_(0)^(T) p ( t ) d t ,)$] where #math.equation(block: false, alt: "T equals 2 π / ω")[$T = 2 π "/" ω$] is the period of the oscillations. With the substitutions #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t$] and #math.equation(block: false, alt: "i open parenthesis t close parenthesis equals I sub 0 sin open parenthesis ω t minus ϕ close parenthesis ,")[$i ( t ) = I_(0) #h(0.2em) "sin" #h(0.2em) ( ω t − ϕ ) ,$] this integral becomes #math.equation(block: true, alt: "P sub ave equals the fraction I sub 0 V sub 0 over T ∫ 0 T sin open parenthesis ω t minus ϕ close parenthesis sin ω t d t .")[$P_("ave") = frac(I_(0) V_(0), T) display(∫_(0)^(T) #h(0.2em) "sin" #h(0.2em) ( ω t − ϕ )) #h(0.2em) "sin" #h(0.2em) ω t #h(0.2em) d t .$] Using the trigonometric relation #math.equation(block: false, alt: "sin open parenthesis A minus B close parenthesis equals sin A cos B minus sin B cos A ,")[$#h(0.2em) "sin" #h(0.2em) ( A − B ) = #h(0.2em) "sin" #h(0.2em) A "cos" #h(0.2em) B − #h(0.2em) "sin" #h(0.2em) B #h(0.2em) "cos" #h(0.2em) A ,$] we obtain #math.equation(block: true, alt: "P sub ave equals the fraction I sub 0 V sub 0 cos ϕ over T ∫ 0 T sin squared ω t d t minus the fraction I sub 0 V sub 0 sin ϕ over T ∫ 0 T sin ω t cos ω t d t .")[$P_("ave") = frac(I_(0) V_(0) #h(0.2em) "cos" #h(0.2em) ϕ, T) display(∫_(0)^(T) #h(0.2em) "sin"^(2) #h(0.2em) ω t d t −) frac(I_(0) V_(0) #h(0.2em) "sin" #h(0.2em) ϕ, T) display(∫_(0)^(T) #h(0.2em) "sin" #h(0.2em) ω t "cos" #h(0.2em) ω t d t) .$] Evaluation of these two integrals yields #math.equation(block: true, alt: "the fraction 1 over T ∫ 0 T sin squared ω t d t equals the fraction 1 over 2")[$frac(1, T) display(∫_(0)^(T) attach(#h(0.2em) "sin" #h(0.2em), t: 2) ω t d t) = frac(1, 2)$] and #math.equation(block: true, alt: "the fraction 1 over T ∫ 0 T sin ω t cos ω t d t equals 0 .")[$frac(1, T) display(∫_(0)^(T) #h(0.2em) "sin" #h(0.2em) ω t "cos" #h(0.2em) ω t d t) = 0 .$] Hence, the average power associated with a circuit element is given by #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "P sub ave equals the fraction 1 over 2 I sub 0 V sub 0 cos ϕ .")[$P_("ave") = frac(1, 2) I_(0) V_(0) #h(0.2em) "cos" #h(0.2em) ϕ .$] ] In engineering applications, #math.equation(block: false, alt: "cos ϕ")[$"cos" #h(0.2em) ϕ$] is known as the #strong[power factor], which is the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase. For a resistor, #math.equation(block: false, alt: "ϕ equals 0 ,")[$ϕ = 0 ,$] so the average power dissipated is #math.equation(block: true, alt: "P sub ave equals the fraction 1 over 2 I sub 0 V sub 0 .")[$P_("ave") = frac(1, 2) I_(0) V_(0) .$] A comparison of #emph[p] (#emph[t]) and #math.equation(block: false, alt: "P sub ave")[$P_("ave")$] is shown in (d). To make #math.equation(block: false, alt: "P sub ave equals open parenthesis 1 / 2 close parenthesis I sub 0 V sub 0")[$P_("ave") = ( 1 "/" 2 ) I_(0) V_(0)$] look like its dc counterpart, we use the rms values #math.equation(block: false, alt: "I sub rms and V sub rms")[$I_("rms") #h(0.2em) "and" #h(0.2em) V_("rms")$] of the current and the voltage. By definition, these are #math.equation(block: true, alt: "I sub rms equals the square root of i ave 2 and V sub rms equals the square root of v ave 2 ,")[$I_("rms") = sqrt(i_("ave")^(2)) #h(0.2em) "and" #h(0.2em) V_("rms") = sqrt(v_("ave")^(2)) ,$] where #math.equation(block: true, alt: "i ave 2 equals the fraction 1 over T ∫ 0 T i squared open parenthesis t close parenthesis d t and v ave 2 equals the fraction 1 over T ∫ 0 T v squared open parenthesis t close parenthesis d t .")[$i_("ave")^(2) = frac(1, T) display(∫_(0)^(T) i^(2) ( t ) d t #h(0.2em) "and " #h(0.2em)) v_("ave")^(2) = frac(1, T) display(∫_(0)^(T) v^(2) ( t ) d t) .$] With #math.equation(block: false, alt: "i open parenthesis t close parenthesis equals I sub 0 sin open parenthesis ω t minus ϕ close parenthesis and v open parenthesis t close parenthesis equals V sub 0 sin ω t ,")[$i ( t ) = I_(0) #h(0.2em) "sin" #h(0.2em) ( ω t − ϕ ) #h(0.2em) "and" #h(0.2em) v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t ,$] we obtain #math.equation(block: true, alt: "I sub rms equals the fraction 1 over the square root of 2 I sub 0 and V sub rms equals the fraction 1 over the square root of 2 V sub 0 .")[$I_("rms") = frac(1, sqrt(2)) I_(0) #h(0.2em) "and" #h(0.2em) V_("rms") = frac(1, sqrt(2)) V_(0) .$] We may then write for the average power dissipated by a resistor, #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "P sub ave equals the fraction 1 over 2 I sub 0 V sub 0 equals I sub rms V sub rms equals I rms 2 R .")[$P_("ave") = frac(1, 2) I_(0) V_(0) = I_("rms") V_("rms") = I_("rms")^(2) R .$] ] This equation further emphasizes why the rms value is chosen in discussion rather than peak values. Both equations for average power are correct for , but the rms values in the formula give a cleaner representation, so the extra factor of 1/2 is not necessary. Alternating voltages and currents are usually described in terms of their rms values. For example, the 110 V from a household outlet is an rms value. The amplitude of this source is #math.equation(block: false, alt: "110 the square root of 2 V equals 156 V .")[$110 sqrt(2) #h(0.2em) "V" = "156 V" "."$] Because most ac meters are calibrated in terms of rms values, a typical ac voltmeter placed across a household outlet will read 110 V. For a capacitor and an inductor, #math.equation(block: false, alt: "ϕ equals π / 2 and minus π / 2 rad,")[$ϕ = π "/" 2 #h(0.2em) "and" #h(0.2em) − π "/" 2 #h(0.2em) "rad,"$] respectively. Since #math.equation(block: false, alt: "cos π / 2 equals cos open parenthesis − π / 2 close parenthesis equals 0 ,")[$"cos" π "/" 2 = "cos" ( "−" π "/" 2 ) = 0 ,$] we find from that the average power dissipated by either of these elements is #math.equation(block: false, alt: "P sub ave equals 0 .")[$P_("ave") = 0 .$] Capacitors and inductors absorb energy from the circuit during one half-cycle and then discharge it back to the circuit during the other half-cycle. This behavior is illustrated in the plots of , (b) and (c), which show #emph[p(]t) oscillating sinusoidally about zero. The phase angle for an ac generator may have any value. If #math.equation(block: false, alt: "cos ϕ greater than 0 ,")[$"cos" #h(0.2em) ϕ > 0 ,$] the generator produces power; if #math.equation(block: false, alt: "cos ϕ less than 0 ,")[$"cos" #h(0.2em) ϕ < 0 ,$] it absorbs power. In terms of rms values, the average power of an ac generator is written as #math.equation(block: true, alt: "P sub ave equals I sub rms V sub rms cos ϕ .")[$P_("ave") = I_("rms") V_("rms") #h(0.2em) "cos" #h(0.2em) ϕ .$] For the generator in an #emph[RLC] circuit, #math.equation(block: true, alt: "tan ϕ equals the fraction X sub L minus X sub C over R")[$"tan" #h(0.2em) ϕ = frac(X_(L) − X_(C), R)$] and #math.equation(block: true, alt: "cos ϕ equals the fraction R over the square root of R squared plus open parenthesis X sub L minus X sub C close parenthesis squared equals the fraction R over Z .")[$"cos" #h(0.2em) ϕ = frac(R, sqrt(R^(2) + attach(( X_(L) − X_(C) ), t: 2))) = frac(R, Z) .$] Hence the average power of the generator is #math.equation(block: true, alt: "P sub ave equals I sub rms V sub rms cos ϕ equals the fraction V sub rms over Z V sub rms the fraction R over Z equals the fraction V rms 2 R over Z squared .")[$P_("ave") = I_("rms") V_("rms") #h(0.2em) "cos" #h(0.2em) ϕ = frac(V_("rms"), Z) V_("rms") frac(R, Z) = frac(V_("rms")^(2) R, Z^(2)) .$] This can also be written as #math.equation(block: true, alt: "P sub ave equals I rms 2 R ,")[$P_("ave") = I_("rms")^(2) R ,$] which designates that the power produced by the generator is dissipated in the resistor. As we can see, Ohm’s law for the rms ac is found by dividing the rms voltage by the impedance. #examplebox("Example 1")[Power Output of a Generator][ An ac generator whose emf is given by #math.equation(block: true, alt: "v open parenthesis t close parenthesis equals open parenthesis 4.00 V close parenthesis sin [ open parenthesis 1.00 times 10 to the power 4 rad / s close parenthesis t ]")[$v ( t ) = ( 4.00 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) [ ( 1.00 #h(0.2em) × #h(0.2em) 10^(4) "rad" "/" "s" ) t ]$] is connected to an #emph[RLC] circuit for which #math.equation(block: false, alt: "L equals 2.00 times 10 to the power −3 H")[$L = 2.00 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H"$], #math.equation(block: false, alt: "C equals 4.00 times 10 to the power −6 F")[$C = 4.00 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "F"$], and #math.equation(block: false, alt: "R equals 5.00 Ω")[$R = 5.00 #h(0.2em) "Ω"$]. (a) What is the rms voltage across the generator? (b) What is the impedance of the circuit? (c) What is the average power output of the generator? Strategy The rms voltage is the amplitude of the voltage times #math.equation(block: false, alt: "1 / the square root of 2")[$1 "/" sqrt(2)$]. The impedance of the circuit involves the resistance and the reactances of the capacitor and the inductor. The average power is calculated by , or more specifically, the last part of the equation, because we have the impedance of the circuit #emph[Z], the rms voltage #math.equation(block: false, alt: "V sub rms")[$V_("rms")$], and the resistance #emph[R]. Solution + Since #math.equation(block: false, alt: "V sub 0 equals 4.00 V,")[$V_(0) = 4.00 #h(0.2em) "V,"$] the rms voltage across the generator is #math.equation(block: true, alt: "V sub rms equals the fraction 1 over the square root of 2 open parenthesis 4.00 V close parenthesis equals 2.83 V .")[$V_("rms") = frac(1, sqrt(2)) ( 4.00 #h(0.2em) "V" ) = 2.83 #h(0.2em) "V" "."$] + The impedance of the circuit is #math.equation(block: true, alt: "Z, equals the square root of R squared plus open parenthesis X sub L minus X sub C close parenthesis squared; equals { open parenthesis 5.00 Ω close parenthesis squared plus [ open parenthesis 1.00 times 10 to the power 4 rad/s close parenthesis open parenthesis 2.00 times 10 to the power −3 H close parenthesis minus the fraction 1 over open parenthesis 1.00 times 10 to the power 4 rad/s close parenthesis open parenthesis 4.00 times 10 to the power −6 F close parenthesis ] squared } to the power 1 / 2; equals 7.07 Ω .")[$Z & = sqrt(R^(2) + attach(( X_(L) − X_(C) ), t: 2)) \ & = attach(\{ attach(( 5.00 #h(0.2em) "Ω" ), t: 2) + attach([ ( 1.00 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "rad/s" ) ( 2.00 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H" ) − frac(1, ( 1.00 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "rad/s" ) ( 4.00 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "F" )) ], t: 2) \}, t: 1 "/" 2) \ & = 7.07 #h(0.2em) "Ω" "."$] + From , the average power transferred to the circuit is #math.equation(block: true, alt: "P sub ave equals the fraction V rms 2 R over Z squared equals the fraction open parenthesis 2.83 V close parenthesis squared open parenthesis 5.00 Ω close parenthesis over open parenthesis 7.07 Ω close parenthesis squared equals 0.801 W .")[$P_("ave") = frac(V_("rms")^(2) R, Z^(2)) = frac(attach(( 2.83 #h(0.2em) "V" ), t: 2) ( 5.00 #h(0.2em) "Ω" ), attach(( 7.07 #h(0.2em) "Ω" ), t: 2)) = 0.801 #h(0.2em) "W" "."$] Significance If the resistance is much larger than the reactance of the capacitor or inductor, the average power is a dc circuit equation of #math.equation(block: false, alt: "P equals V squared / R ,")[$P = V^(2) "/" R ,$] where #emph[V] replaces the rms voltage. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ An ac voltmeter attached across the terminals of a 45-Hz ac generator reads 7.07 V. Write an expression for the emf of the generator. #solutionbox[ #math.equation(block: true, alt: "v open parenthesis t close parenthesis equals open parenthesis 10.0 V close parenthesis sin 90 π t")[$v ( t ) = ( 10.0 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) 90 π t$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Show that the rms voltages across a resistor, a capacitor, and an inductor in an ac circuit where the rms current is #math.equation(block: false, alt: "I sub rms")[$I_("rms")$] are given by #math.equation(block: false, alt: "I sub rms R , I sub rms X sub C , and I sub rms X sub L ,")[$I_("rms") R , I_("rms") X_(C) , #h(0.2em) "and" #h(0.2em) I_("rms") X_(L) ,$] respectively. Determine these values for the components of the #emph[RLC] circuit of . #solutionbox[ 2.00 V; 10.01 V; 8.01 V ] ] === Summary - The average ac power is found by multiplying the rms values of current and voltage. - Ohm’s law for the rms ac is found by dividing the rms voltage by the impedance. - In an ac circuit, there is a phase angle between the source voltage and the current, which can be found by dividing the resistance by the impedance. - The average power delivered to an #emph[RLC] circuit is affected by the phase angle. - The power factor ranges from –1 to 1. === Conceptual Questions For what value of the phase angle #math.equation(block: false, alt: "ϕ")[$ϕ$] between the voltage output of an ac source and the current is the average power output of the source a maximum? Discuss the differences between average power and instantaneous power. #solutionbox[ The instantaneous power is the power at a given instant. The average power is the power averaged over a cycle or number of cycles. ] The average ac current delivered to a circuit is zero. Despite this, power is dissipated in the circuit. Explain. Can the instantaneous power output of an ac source ever be negative? Can the average power output be negative? #solutionbox[ The instantaneous power can be negative, but the power output can’t be negative. ] The power rating of a resistor used in ac circuits refers to the maximum average power dissipated in the resistor. How does this compare with the maximum instantaneous power dissipated in the resistor? === Problems The emf of an ac source is given by #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t ,")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t ,$] where #math.equation(block: false, alt: "V sub 0 equals 100 V")[$V_(0) = 100 #h(0.2em) "V"$] and #math.equation(block: false, alt: "ω equals 200 π rad/s .")[$ω = 200 π #h(0.2em) "rad/s" "."$] Calculate the average power output of the source if it is connected across (a) a #math.equation(block: false, alt: "20 - μ F")[$20 "-" μ "F"$] capacitor, (b) a 20-mH inductor, and (c) a #math.equation(block: false, alt: "50 - Ω")[$50 "-" "Ω"$] resistor. Calculate the rms currents for an ac source is given by #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t ,")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t ,$] where #math.equation(block: false, alt: "V sub 0 equals 100 V")[$V_(0) = 100 #h(0.2em) "V"$] and #math.equation(block: false, alt: "ω equals 200 π rad/s")[$ω = 200 π #h(0.2em) "rad/s"$] when connected across (a) a #math.equation(block: false, alt: "20 - μ F")[$20 "-" μ "F"$] capacitor, (b) a 20-mH inductor, and (c) a #math.equation(block: false, alt: "50 - Ω")[$50 "-" "Ω"$] resistor. #solutionbox[ a. 0.89 A; b. 5.6A; c. 1.4 A ] A 40-mH inductor is connected to a 60-Hz AC source whose voltage amplitude is 50 V. If an AC voltmeter is placed across the inductor, what does it read? For an #emph[RLC] series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively; #math.equation(block: false, alt: "R equals 500 Ω")[$R = 500 #h(0.2em) "Ω"$]; and #math.equation(block: false, alt: "L equals 0.20 H")[$L = 0.20 #h(0.2em) "H"$]. Find the average power dissipated in the resistor for the following values for the capacitance: (a) #math.equation(block: false, alt: "C equals 2.0 μ F")[$C = 2.0 μ "F"$] and (b) #math.equation(block: false, alt: "C equals 0.20 μ F .")[$C = 0.20 #h(0.2em) μ "F" "."$] #solutionbox[ a. 5.3 W; b. 2.1 W ] An ac source of voltage amplitude 10 V delivers electric energy at a rate of 0.80 W when its current output is 2.5 A. What is the phase angle #math.equation(block: false, alt: "ϕ")[$ϕ$] between the emf and the current? An #emph[RLC] series circuit has an impedance of #math.equation(block: false, alt: "60 Ω")[$60 #h(0.2em) "Ω"$] and a power factor of 0.50, with the voltage lagging the current. (a) Should a capacitor or an inductor be placed in series with the elements to raise the power factor of the circuit? (b) What is the value of the reactance across the inductor that will raise the power factor to unity? #solutionbox[ a. inductor; b. #math.equation(block: false, alt: "X sub L equals 52 Ω")[$X_(L) = 52 #h(0.2em) "Ω"$] ]