#set document(title: "15.5 Resonance 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")) == 15.5#h(0.6em)Resonance in an AC Circuit In the #emph[RLC] series circuit of , the current amplitude is, from , #math.equation(block: true, alt: "I sub 0 equals the fraction V sub 0 over the square root of R squared plus open parenthesis ω L minus 1 / ω C close parenthesis squared .")[$I_(0) = frac(V_(0), sqrt(R^(2) + attach(( "ω" L − 1 "/" "ω" C ), t: 2))) .$] If we can vary the frequency of the ac generator while keeping the amplitude of its output voltage constant, then the current changes accordingly. A plot of #math.equation(block: false, alt: "I sub 0")[$I_(0)$] versus #math.equation(block: false, alt: "ω")[$"ω"$] is shown. The current amplitude this section plots, I₀ = V₀/√(R² + (ωL − 1/ωC)²), for the circuit the example Resonance in an RLC Series Circuit uses: V₀ = 0.100 V, R = 4.00 Ω, L = 3.00 × 10⁻³ H and C = 8.00 × 10⁻⁴ F. The horizontal axis is the driving angular frequency in hundreds of rad/s and the vertical axis is the current amplitude in milliamperes. At the default the peak sits on the dashed line at ω₀ = 1/√(LC) = 645 rad/s — the example's f₀ = 1.03 × 10² Hz — and reaches I₀ = V₀/R = 25.0 mA, because at resonance the two reactances cancel and the circuit is purely resistive. The slider is the tuning capacitor, and it does exactly what the section says tuning does: the peak slides along the frequency axis as 1/√(LC), from 408 rad/s at 2000 μF out to 2582 rad/s at 50 μF, while its height never changes. That is the point worth taking away — C picks the frequency the circuit selects, R alone fixes how much current it passes there, and the dashed line stays behind to show how far you have tuned away from the example.{"functions":\[{"color":"\#1f4e79","domain":\[0.4,48\],"expression":"100/sqrt(16 + (0.3\*x - 10000/(x\*Cu))^2)","expression2":"","id":"current-amplitude","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#b8442a","domain":null,"expression":"6.455","expression2":"t","id":"example-resonance","inequality":"lt","lineDash":\[5,4\],"lineWidth":1.5,"mode":"parametric","tMax":27,"tMin":0,"variable":"t","visible":true}\],"grid":{"axisColor":"\#333333","gridColor":"\#cccccc","majorSpacing":5,"minorGridColor":"\#eeeeee","showAxes":true,"showLabels":true},"parameters":\[{"id":"up2-15-5-Cu","label":"Tuning capacitance C (the worked example uses 800 \\u03bcF)","max":2000,"min":50,"name":"Cu","step":5,"unit":"\\u03bcF","value":800}\],"title":"RLC series resonance curve, V0 = 0.100 V, R = 4.00 \\u03a9, L = 3.00 mH","version":1,"viewport":{"centerX":24,"centerY":13.5,"scale":15.83}}#figure(figph[Figure shows a graph of I0 versus omega. The curve ascends gradually, has one blunt peak at the centre and then gradually descends to its original value. The y-value at the peak is V0 by R and the x-value is omega 0.], alt: "Figure shows a graph of I0 versus omega. The curve ascends gradually, has one blunt peak at the centre and then gradually descends to its original value. The y-value at the peak is V0 by R and the x-value is omega 0.", caption: [At an #emph[RLC] circuit’s resonant frequency, #math.equation(block: false, alt: "ω sub 0 equals the square root of 1 / L C ,")[$"ω"_(0) = sqrt(1 "/" L C) ,$] the current amplitude is at its maximum value.]) In Oscillations, we encountered a similar graph where the amplitude of a damped harmonic oscillator was plotted against the angular frequency of a sinusoidal driving force (see Forced Oscillations). This similarity is more than just a coincidence, as shown earlier by the application of Kirchhoff’s loop rule to the circuit of . This yields #math.equation(block: true, alt: "L the fraction d i over d t plus i R plus the fraction q over C equals V sub 0 sin ω t ,")[$L frac(d i, d t) + i R + frac(q, C) = V_(0) #h(0.2em) "sin" #h(0.2em) "ω" t ,$] or #math.equation(block: true, alt: "L the fraction d squared q over d t squared plus R the fraction d q over d t plus the fraction 1 over C q equals V sub 0 sin ω t ,")[$L frac(d^(2) q, d t^(2)) + R frac(d q, d t) + frac(1, C) q = V_(0) #h(0.2em) "sin" #h(0.2em) "ω" t ,$] where we substituted #emph[dq] (t)/#emph[dt] for #emph[i] (t). A comparison of and, from Oscillations, Damped Oscillations for damped harmonic motion clearly demonstrates that the driven #emph[RLC] series circuit is the electrical analog of the driven damped harmonic oscillator. The #strong[resonant frequency] #math.equation(block: false, alt: "f sub 0")[$f_(0)$] of the #emph[RLC] circuit is the frequency at which the amplitude of the current is a maximum and the circuit would oscillate if not driven by a voltage source. By inspection, this corresponds to the angular frequency #math.equation(block: false, alt: "ω sub 0 equals 2 π f sub 0")[$"ω"_(0) = 2 π f_(0)$] at which the impedance #emph[Z] in is a minimum, or when #math.equation(block: true, alt: "ω sub 0 L equals the fraction 1 over ω sub 0 C")[$"ω"_(0) L = frac(1, "ω"_(0) C)$] and #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "ω sub 0 equals the square root of the fraction 1 over L C .")[$"ω"_(0) = sqrt(frac(1, L C)) .$] ] This is the resonant angular frequency of the circuit. Substituting #math.equation(block: false, alt: "ω sub 0")[$"ω"_(0)$] into , , and , we find that at resonance, #math.equation(block: true, alt: "ϕ equals tan to the power −1 open parenthesis 0 close parenthesis equals 0 , I sub 0 equals V sub 0 / R , and Z equals R .")[$ϕ = "tan"^(−1) ( 0 ) = 0 "," " " I_(0) = V_(0) "/" R , " " #h(0.2em) "and" #h(0.2em) " " Z = R .$] Therefore, at resonance, an #emph[RLC] circuit is purely resistive, with the applied emf and current in phase. What happens to the power at resonance? tells us how the average power transferred from an ac generator to the #emph[RLC] combination varies with frequency. In addition, #math.equation(block: false, alt: "P sub ave")[$P_("ave")$] reaches a maximum when #emph[Z], which depends on the frequency, is a minimum, that is, when #math.equation(block: false, alt: "X sub L equals X sub C and Z equals R .")[$X_(L) = X_(C) #h(0.2em) "and" #h(0.2em) Z = R .$] Thus, at resonance, the average power output of the source in an #emph[RLC] series circuit is a maximum. From , this maximum is #math.equation(block: false, alt: "V rms 2 / R .")[$V_("rms")^(2) "/" R .$] is a typical plot of #math.equation(block: false, alt: "P sub ave")[$P_("ave")$] versus #math.equation(block: false, alt: "ω")[$"ω"$] in the region of maximum power output. The #strong[bandwidth] #math.equation(block: false, alt: "Δ ω")[$"Δ" "ω"$] of the resonance peak is defined as the range of angular frequencies #math.equation(block: false, alt: "ω")[$"ω"$] over which the average power #math.equation(block: false, alt: "P sub ave")[$P_("ave")$] is greater than one-half the maximum value of #math.equation(block: false, alt: "P sub ave .")[$P_("ave") .$] The sharpness of the peak is described by a dimensionless quantity known as the #strong[quality factor] #emph[Q] of the circuit. By definition, #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "Q equals the fraction ω sub 0 over Δ ω ,")[$Q = frac("ω"_(0), "Δ" "ω") ,$] ] where #math.equation(block: false, alt: "ω sub 0")[$"ω"_(0)$] is the resonant angular frequency. A high #emph[Q] indicates a sharp resonance peak. We can give #emph[Q] in terms of the circuit parameters as #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "Q equals the fraction ω sub 0 L over R .")[$Q = frac("ω"_(0) L, R) .$] ] #figure(figph[Figure shows a graph of P bar versus omega. The curve ascends gradually, has one blunt peak at the centre and then gradually descends to its original value. The y-value at the peak is V squared subscript rms by R and the x-value is omega 0. The y value near the middle of the curve is V squared subscript rms by 2R. The width of the curve near the middle is labeled delta omega.], alt: "Figure shows a graph of P bar versus omega. The curve ascends gradually, has one blunt peak at the centre and then gradually descends to its original value. The y-value at the peak is V squared subscript rms by R and the x-value is omega 0. The y value near the middle of the curve is V squared subscript rms by 2R. The width of the curve near the middle is labeled delta omega.", caption: [Like the current, the average power transferred from an ac generator to an #emph[RLC] circuit peaks at the resonant frequency.]) Resonant circuits are commonly used to pass or reject selected frequency ranges. This is done by adjusting the value of one of the elements and hence “tuning” the circuit to a particular resonant frequency. For example, in radios, the receiver is tuned to the desired station by adjusting the resonant frequency of its circuitry to match the frequency of the station. If the tuning circuit has a high #emph[Q], it will have a small bandwidth, so signals from other stations at frequencies even slightly different from the resonant frequency encounter a high impedance and are not passed by the circuit. Cell phones work in a similar fashion, communicating with signals of around 1 GHz that are tuned by an inductor-capacitor circuit. One of the most common applications of capacitors is their use in ac-timing circuits, based on attaining a resonant frequency. A metal detector also uses a shift in resonance frequency in detecting metals . #figure(figph[Photograph of an underwater diver using a metal detector.], alt: "Photograph of an underwater diver using a metal detector.", caption: [When a metal detector comes near a piece of metal, the self-inductance of one of its coils changes. This causes a shift in the resonant frequency of a circuit containing the coil. That shift is detected by the circuitry and transmitted to the diver by means of the headphones.]) #examplebox("Example 1")[Resonance in an #emph[RLC] Series Circuit][ (a) What is the resonant frequency of a circuit using the voltage and LRC values all wired in series from ? (b) If the ac generator is set to this frequency without changing the amplitude of the output voltage, what is the amplitude of the current? Strategy The resonant frequency for a #emph[RLC] circuit is calculated from , which comes from a balance between the reactances of the capacitor and the inductor. Since the circuit is at resonance, the impedance is equal to the resistor. Then, the peak current is calculated by the voltage divided by the resistance. Solution + The resonant frequency is found from : #math.equation(block: true, alt: "f sub 0, equals the fraction 1 over 2 π the square root of the fraction 1 over L C equals the fraction 1 over 2 π the square root of the fraction 1 over open parenthesis 3.00 times 10 to the power −3 H close parenthesis open parenthesis 8.00 times 10 to the power −4 F close parenthesis; equals 1.03 times 10 squared Hz .")[$f_(0) & = frac(1, 2 π) sqrt(frac(1, L C)) = frac(1, 2 π) #h(0.2em) sqrt(frac(1, ( 3.00 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H" ) ( 8.00 #h(0.2em) × #h(0.2em) 10^(−4) #h(0.2em) "F" ))) \ & = 1.03 #h(0.2em) × #h(0.2em) 10^(2) #h(0.2em) "Hz" "."$] + At resonance, the impedance of the circuit is purely resistive, and the current amplitude is #math.equation(block: true, alt: "I sub 0 equals the fraction 0.100 V over 4.00 Ω equals 2.50 times 10 to the power −2 A .")[$I_(0) = frac(0.100 #h(0.2em) "V", 4.00 #h(0.2em) "Ω") = 2.50 #h(0.2em) × #h(0.2em) 10^(−2) "A" "."$] Significance If the circuit were not set to the resonant frequency, we would need the impedance of the entire circuit to calculate the current. ] #examplebox("Example 2")[Power Transfer in an #emph[RLC] Series Circuit at Resonance][ (a) What is the resonant angular frequency of an #emph[RLC] circuit with #math.equation(block: false, alt: "R equals 0.200 Ω ,")[$R = 0.200 #h(0.2em) "Ω" ","$] #math.equation(block: false, alt: "L equals 4.00 times 10 to the power −3 H,")[$L = 4.00 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H,"$] and #math.equation(block: false, alt: "C equals 2.00 times 10 to the power −6 F?")[$C = 2.00 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "F?"$] (b) If an ac source of constant amplitude 4.00 V is set to this frequency, what is the average power transferred to the circuit? (c) Determine #emph[Q] and the bandwidth of this circuit. Strategy The resonant angular frequency is calculated from . The average power is calculated from the rms voltage and the resistance in the circuit. The quality factor is calculated from and by knowing the resonant frequency. The bandwidth is calculated from and by knowing the quality factor. Solution + The resonant angular frequency is #math.equation(block: true, alt: "ω sub 0, equals the square root of the fraction 1 over L C equals the square root of the fraction 1 over open parenthesis 4.00 times 10 to the power −3 H close parenthesis open parenthesis 2.00 times 10 to the power −6 F close parenthesis; equals 1.12 times 10 to the power 4 rad/s .")[$"ω"_(0) & = sqrt(frac(1, L C)) = sqrt(frac(1, ( 4.00 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H" ) ( 2.00 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "F" ))) \ & = 1.12 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "rad/s" "."$] + At this frequency, the average power transferred to the circuit is a maximum. It is #math.equation(block: true, alt: "P sub ave equals the fraction V rms 2 over R equals the fraction [ open parenthesis 1 / the square root of 2 close parenthesis open parenthesis 4.00 V close parenthesis ] squared over 0.200 Ω equals 40.0 W .")[$P_("ave") = frac(V_("rms")^(2), R) = frac(attach([ ( 1 "/" sqrt(2) ) ( 4.00 #h(0.2em) "V" ) ], t: 2), 0.200 #h(0.2em) "Ω") = 40.0 #h(0.2em) "W" "."$] + The quality factor of the circuit is #math.equation(block: true, alt: "Q equals the fraction ω sub 0 L over R equals the fraction open parenthesis 1.12 times 10 to the power 4 rad/s close parenthesis open parenthesis 4.00 times 10 to the power −3 H close parenthesis over 0.200 Ω equals 224 .")[$Q = frac("ω"_(0) L, R) = frac(( 1.12 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "rad/s" ) ( 4.00 #h(0.2em) × #h(0.2em) 10^("−3") #h(0.2em) "H" ), 0.200 #h(0.2em) "Ω") = 224 .$] We then find for the bandwidth #math.equation(block: true, alt: "Δ ω equals the fraction ω sub 0 over Q equals the fraction 1.12 times 10 to the power 4 rad/s over 224 equals 50.0 rad/s .")[$"Δ" "ω" = frac("ω"_(0), Q) = frac(1.12 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "rad/s", 224) = 50.0 #h(0.2em) "rad/s" "."$] Significance If a narrower bandwidth is desired, a lower resistance or higher inductance would help. However, a lower resistance increases the power transferred to the circuit, which may not be desirable, depending on the maximum power that could possibly be transferred. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the circuit of , #math.equation(block: false, alt: "L equals 2.0 times 10 to the power −3 H,")[$L = 2.0 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H,"$] #math.equation(block: false, alt: "C equals 5.0 times 10 to the power −4 F,")[$C = 5.0 #h(0.2em) × #h(0.2em) 10^(−4) #h(0.2em) "F,"$] and #math.equation(block: false, alt: "R equals 40 Ω .")[$R = 40 #h(0.2em) "Ω" "."$] (a) What is the resonant frequency? (b) What is the impedance of the circuit at resonance? (c) If the voltage amplitude is 10 V, what is #emph[i] (#emph[t]) at resonance? (d) The frequency of the AC generator is now changed to 200 Hz. Calculate the phase difference between the current and the emf of the generator. #solutionbox[ a. 160 Hz; b. #math.equation(block: false, alt: "40 Ω")[$40 #h(0.2em) "Ω"$]; c. #math.equation(block: false, alt: "open parenthesis 0.25 A close parenthesis sin 10 cubed t")[$( 0.25 #h(0.2em) "A" ) #h(0.2em) "sin" #h(0.2em) 10^(3) t$]; d. 0.023 rad ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ What happens to the resonant frequency of an #emph[RLC] series circuit when the following quantities are increased by a factor of 4: (a) the capacitance, (b) the self-inductance, and (c) the resistance? #solutionbox[ a. halved; b. halved; c. same ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The resonant angular frequency of an #emph[RLC] series circuit is #math.equation(block: false, alt: "4.0 times 10 squared rad/s .")[$4.0 #h(0.2em) × #h(0.2em) 10^(2) #h(0.2em) "rad/s" "."$] An ac source operating at this frequency transfers an average power of #math.equation(block: false, alt: "2.0 times 10 to the power −2 W")[$2.0 #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "W"$] to the circuit. The resistance of the circuit is #math.equation(block: false, alt: "0.50 Ω .")[$0.50 #h(0.2em) "Ω" "."$] Write an expression for the emf of the source. #solutionbox[ #math.equation(block: true, alt: "v open parenthesis t close parenthesis equals open parenthesis 0.14 V close parenthesis sin open parenthesis 4.0 times 10 squared t close parenthesis")[$v ( t ) = ( 0.14 #h(0.2em) "V" ) #h(0.2em) #h(0.2em) "sin" #h(0.2em) ( 4.0 #h(0.2em) × #h(0.2em) 10^(2) t )$] ] ] === Summary - At the resonant frequency, inductive reactance equals capacitive reactance. - The average power versus angular frequency plot for a #emph[RLC] circuit has a peak located at the resonant frequency; the sharpness or width of the peak is known as the bandwidth. - The bandwidth is related to a dimensionless quantity called the quality factor. A high quality factor value is a sharp or narrow peak. === Problems (a) Calculate the resonant angular frequency of an #emph[RLC] series circuit for which #math.equation(block: false, alt: "R equals 20 Ω")[$R = 20 #h(0.2em) "Ω"$], #math.equation(block: false, alt: "L equals 7 5 mH")[$L = 7 "5 mH"$], and #math.equation(block: false, alt: "C equals 4.0 μ F .")[$C = 4.0 μ "F" "."$] (b) If #emph[R] is changed to #math.equation(block: false, alt: "300 Ω ,")[$300 #h(0.2em) "Ω" ,$] what happens to the resonant angular frequency? The resonant frequency of an #emph[RLC] series circuit is #math.equation(block: false, alt: "2.0 times 10 cubed Hz .")[$2.0 #h(0.2em) × #h(0.2em) 10^(3) "Hz" "."$] If the self-inductance in the circuit is 5.0 mH, what is the capacitance in the circuit? #solutionbox[ #math.equation(block: true, alt: "1.3 times 10 to the power −6 F")[$1.3 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "F"$] ] (a) What is the resonant frequency of an #emph[RLC] series circuit with #math.equation(block: false, alt: "R equals 20 Ω")[$R = 20 #h(0.2em) "Ω"$], #math.equation(block: false, alt: "L equals 2.0 mH")[$L = 2.0 #h(0.2em) "mH"$], and #math.equation(block: false, alt: "C equals 4.0 μ F")[$C = 4.0 μ "F"$]? (b) What is the impedance of the circuit at resonance? For an #emph[RLC] series circuit, #math.equation(block: false, alt: "R equals 100 Ω")[$R = 100 #h(0.2em) "Ω"$], #math.equation(block: false, alt: "L equals 150 mH")[$L = 150 #h(0.2em) "mH"$], and #math.equation(block: false, alt: "C equals 0.25 μ F .")[$C = 0.25 μ "F" "."$] (a) If an ac source of variable frequency is connected to the circuit, at what frequency is maximum power dissipated in the resistor? (b) What is the quality factor of the circuit? #solutionbox[ a. 820 Hz; b. 7.8 ] An ac source of voltage amplitude 100 V and variable frequency #emph[f] drives an #emph[RLC] series circuit with #math.equation(block: false, alt: "R equals 10 Ω")[$R = 10 #h(0.2em) "Ω"$], #math.equation(block: false, alt: "L equals 2.0 mH")[$L = 2.0 #h(0.2em) "mH"$], and #math.equation(block: false, alt: "C equals 25 μ F .")[$C = 25 μ "F" "."$] (a) Plot the current through the resistor as a function of the frequency #emph[f]. (b) Use the plot to determine the resonant frequency of the circuit. (a) What is the resonant frequency of a resistor, capacitor, and inductor connected in series if #math.equation(block: false, alt: "R equals 100 Ω,")[$R = 100 #h(0.2em) "Ω,"$] #math.equation(block: false, alt: "L equals 2.0 H")[$L = 2.0 #h(0.2em) "H"$], and #math.equation(block: false, alt: "C equals 5.0 μ F")[$C = 5.0 μ "F"$]? (b) If this combination is connected to a 100-V source operating at the resonant frequency, what is the power output of the source? (c) What is the #emph[Q] of the circuit? (d) What is the bandwidth of the circuit? #solutionbox[ a. 50 Hz; b. 50 W; c. 6.32; d. 50 rad/s ] Suppose a coil has a self-inductance of 20.0 H and a resistance of #math.equation(block: false, alt: "200 Ω")[$200 #h(0.2em) "Ω"$]. What (a) capacitance and (b) resistance must be connected in series with the coil to produce a circuit that has a resonant frequency of 100 Hz and a #emph[Q] of 10? An ac generator is connected to a device whose internal circuits are not known. We only know current and voltage outside the device, as shown below. Based on the information given, what can you infer about the electrical nature of the device and its power usage? #figure(figph[Figure shows an AC source connected to a box labeled Z. The source is 170V, cos 120 pi t. The current through the circuit is 0.5 Amp, cos parentheses 120 pi t plus pi by 4 parentheses.], alt: "Figure shows an AC source connected to a box labeled Z. The source is 170V, cos 120 pi t. The current through the circuit is 0.5 Amp, cos parentheses 120 pi t plus pi by 4 parentheses.", caption: none) #solutionbox[ The reactance of the capacitor is larger than the reactance of the inductor because the current leads the voltage. The power usage is 30 W. ]