#set document(title: "15.3 RLC Series Circuits with AC", 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")) == 15.3#h(0.6em)RLC Series Circuits with AC The ac circuit shown in , called an #strong[#emph[RLC] series circuit], is a series combination of a resistor, capacitor, and inductor connected across an ac source. It produces an emf of #math.equation(block: true, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t .")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t .$] #figure(figph[Figure a shows a circuit with an AC voltage source connected to a resistor, a capacitor and an inductor in series. 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. The voltage curve is labeled V parentheses t parentheses equal to V0 sine omega t. The current curve is labeled I parentheses t parentheses equal to I0 sine parentheses omega t minus phi parentheses.], alt: "Figure a shows a circuit with an AC voltage source connected to a resistor, a capacitor and an inductor in series. 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. The voltage curve is labeled V parentheses t parentheses equal to V0 sine omega t. The current curve is labeled I parentheses t parentheses equal to I0 sine parentheses omega t minus phi parentheses.", caption: [(a) An #emph[RLC] series circuit. (b) A comparison of the generator output voltage and the current. The value of the phase difference #math.equation(block: false, alt: "ϕ")[$ϕ$] depends on the values of #emph[R], #emph[C], and #emph[L].]) Since the elements are in series, the same current flows through each element at all points in time. The relative phase between the current and the emf is not obvious when all three elements are present. Consequently, we represent the current by the general expression #math.equation(block: true, alt: "i open parenthesis t close parenthesis equals I sub 0 sin open parenthesis ω t minus ϕ close parenthesis ,")[$i ( t ) = I_(0) #h(0.2em) "sin" #h(0.2em) ( ω t − ϕ ) ,$] where #math.equation(block: false, alt: "I sub 0")[$I_(0)$] is the current amplitude and #math.equation(block: false, alt: "ϕ")[$ϕ$] is the #strong[phase angle] between the current and the applied voltage. The phase angle is thus the amount by which the voltage and current are out of phase with each other in a circuit. Our task is to find #math.equation(block: false, alt: "I sub 0 and ϕ .")[$I_(0) #h(0.2em) "and" #h(0.2em) ϕ .$] A phasor diagram involving #math.equation(block: false, alt: "i open parenthesis t close parenthesis , v sub R open parenthesis t close parenthesis , v sub C open parenthesis t close parenthesis , and v sub L open parenthesis t close parenthesis")[$i ( t ) , v_(R) ( t ) , v_(C) ( t ) , #h(0.2em) "and" #h(0.2em) v_(L) ( t )$] is helpful for analyzing the circuit. As shown in , the phasor representing #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis")[$v_(R) ( t )$] points in the same direction as the phasor for #math.equation(block: false, alt: "i open parenthesis t close parenthesis ;")[$i ( t ) ;$] its amplitude is #math.equation(block: false, alt: "V sub R equals I sub 0 R .")[$V_(R) = I_(0) R .$] The #math.equation(block: false, alt: "v sub C open parenthesis t close parenthesis")[$v_(C) ( t )$] phasor lags the #emph[i] (#emph[t]) phasor by #math.equation(block: false, alt: "π / 2")[$π "/" 2$] rad and has the amplitude #math.equation(block: false, alt: "V sub C equals I sub 0 X sub C .")[$V_(C) = I_(0) X_(C) .$] The phasor for #math.equation(block: false, alt: "v sub L open parenthesis t close parenthesis")[$v_(L) ( t )$] leads the #emph[i] (#emph[t]) phasor by #math.equation(block: false, alt: "π / 2")[$π "/" 2$] rad and has the amplitude #math.equation(block: false, alt: "V sub L equals I sub 0 X sub L .")[$V_(L) = I_(0) X_(L) .$] #figure(figph[Figure shows the coordinate axes, with four arrows starting from the origin. Arrow V subscripts R points up and right, making an angle omega t minus phi with the x axis. Its y intercept is V subscript R parentheses t parentheses. Arrow I0 is along arrow V subscript R, but shorter than it. Arrow V subscript L points up and left and is perpendicular to V subscript R. It makes a y intercept V subscript L parentheses t parentheses. Arrow V subscript C points down and right. It is perpendicular to V subscript R. It makes a y intercept V subscript C parentheses t parentheses. Three arrows labeled omega are each perpendicular to V subscript R, V subscript L and V subscript C, shown near their tips.], alt: "Figure shows the coordinate axes, with four arrows starting from the origin. Arrow V subscripts R points up and right, making an angle omega t minus phi with the x axis. Its y intercept is V subscript R parentheses t parentheses. Arrow I0 is along arrow V subscript R, but shorter than it. Arrow V subscript L points up and left and is perpendicular to V subscript R. It makes a y intercept V subscript L parentheses t parentheses. Arrow V subscript C points down and right. It is perpendicular to V subscript R. It makes a y intercept V subscript C parentheses t parentheses. Three arrows labeled omega are each perpendicular to V subscript R, V subscript L and V subscript C, shown near their tips.", caption: [The phasor diagram for the #emph[RLC] series circuit of .]) At any instant, the voltage across the #emph[RLC] combination is #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis plus v sub L open parenthesis t close parenthesis plus v sub C open parenthesis t close parenthesis equals v open parenthesis t close parenthesis ,")[$v_(R) ( t ) + v_(L) ( t ) + v_(C) ( t ) = v ( t ) ,$] the emf of the source. Since a component of a sum of vectors is the sum of the components of the individual vectors—for example, #math.equation(block: false, alt: "open parenthesis A plus B close parenthesis sub y equals A sub y plus B sub y")[$attach(( A + B ), b: y) = A_(y) + B_(y)$] —the projection of the vector sum of phasors onto the vertical axis is the sum of the vertical projections of the individual phasors. Hence, if we add vectorially the phasors representing #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis , v sub L open parenthesis t close parenthesis , and v sub C open parenthesis t close parenthesis")[$v_(R) ( t ) , v_(L) ( t ) , #h(0.2em) "and" #h(0.2em) v_(C) ( t )$] and then find the projection of the resultant onto the vertical axis, we obtain #math.equation(block: true, alt: "v sub R open parenthesis t close parenthesis plus v sub L open parenthesis t close parenthesis plus v sub C open parenthesis t close parenthesis equals v open parenthesis t close parenthesis equals V sub 0 sin ω t .")[$v_(R) ( t ) + v_(L) ( t ) + v_(C) ( t ) = v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t .$] The vector sum of the phasors is shown. The resultant phasor has an amplitude #math.equation(block: false, alt: "V sub 0")[$V_(0)$] and is directed at an angle #math.equation(block: false, alt: "ϕ")[$ϕ$] with respect to the #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis ,")[$v_(R) ( t ) ,$] or #math.equation(block: false, alt: "i open parenthesis t close parenthesis")[$i ( t )$], phasor. The projection of this resultant phasor onto the vertical axis is #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t .")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t .$] We can easily determine the unknown quantities #math.equation(block: false, alt: "I sub 0")[$I_(0)$] and #math.equation(block: false, alt: "ϕ")[$ϕ$] from the geometry of the phasor diagram. For the phase angle, #math.equation(block: true, alt: "ϕ equals tan to the power −1 the fraction V sub L minus V sub C over V sub R equals tan to the power −1 the fraction I sub 0 X sub L minus I sub 0 X sub C over I sub 0 R ,")[$ϕ = "tan"^(−1) frac(V_(L) − V_(C), V_(R)) = "tan"^(−1) frac(I_(0) X_(L) − I_(0) X_(C), I_(0) R) ,$] and after cancellation of #math.equation(block: false, alt: "I sub 0 ,")[$I_(0) ,$] this becomes #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "ϕ equals tan to the power −1 the fraction X sub L minus X sub C over R .")[$ϕ = "tan"^(−1) frac(X_(L) − X_(C), R) .$] ] Furthermore, from the Pythagorean theorem, #math.equation(block: true, alt: "V sub 0 equals the square root of V sub R squared plus open parenthesis V sub L minus V sub C close parenthesis squared equals the square root of open parenthesis I sub 0 R close parenthesis squared plus open parenthesis I sub 0 X sub L minus I sub 0 X sub C close parenthesis squared equals I sub 0 the square root of R squared plus open parenthesis X sub L minus X sub C close parenthesis squared .")[$V_(0) = sqrt(V_(R) 2 + attach(( V_(L) − V_(C) ), t: 2)) = sqrt(attach(( I_(0) R ), t: 2) + attach(( I_(0) X_(L) − I_(0) X_(C) ), t: 2)) = I_(0) sqrt(R^(2) + attach(( X_(L) − X_(C) ), t: 2)) .$] #figure(figph[Three arrows start from the origin on the coordinate axis. Arrow V subscript R points up and right, making an angle omega t minus phi with the x axis. Arrow V0 points up and right, making an angle omega t with the x axis. It makes an angle phi with the arrow V subscript R. It makes a y intercept labeled V0 sine omega t. The third arrow is labeled V subscript L minus V subscript C. It points up and left and is perpendicular to arrow V subscript R. Dotted lines indicate that the rectangle formed with its longer side being V subscript R and shorter side being V subscript L minus V subscript C, would have the arrow V0 as a diagonal. An arrow labeled omega is shown near the tip of V subscript R, perpendicular to it.], alt: "Three arrows start from the origin on the coordinate axis. Arrow V subscript R points up and right, making an angle omega t minus phi with the x axis. Arrow V0 points up and right, making an angle omega t with the x axis. It makes an angle phi with the arrow V subscript R. It makes a y intercept labeled V0 sine omega t. The third arrow is labeled V subscript L minus V subscript C. It points up and left and is perpendicular to arrow V subscript R. Dotted lines indicate that the rectangle formed with its longer side being V subscript R and shorter side being V subscript L minus V subscript C, would have the arrow V0 as a diagonal. An arrow labeled omega is shown near the tip of V subscript R, perpendicular to it.", caption: [The resultant of the phasors for #math.equation(block: false, alt: "v sub L open parenthesis t close parenthesis")[$v_(L) ( t )$], #math.equation(block: false, alt: "v sub C open parenthesis t close parenthesis")[$v_(C) ( t )$], and #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis")[$v_(R) ( t )$] is equal to the phasor for #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t .")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t .$] The #emph[i] (#emph[t]) phasor (not shown) is aligned with the #math.equation(block: false, alt: "v sub R open parenthesis t close parenthesis")[$v_(R) ( t )$] phasor.]) The current amplitude is therefore the ac version of Ohm’s law: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "I sub 0 equals the fraction V sub 0 over the square root of R squared plus open parenthesis X sub L minus X sub C close parenthesis squared equals the fraction V sub 0 over Z ,")[$I_(0) = frac(V_(0), sqrt(R^(2) + attach(( X_(L) − X_(C) ), t: 2))) = frac(V_(0), Z) ,$] ] where #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #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")[$Z = sqrt(R^(2) + attach(( X_(L) − X_(C) ), t: 2))$] ] is known as the #strong[impedance] of the circuit. Its unit is the ohm, and it is the ac analog to resistance in a dc circuit, which measures the combined effect of resistance, capacitive reactance, and inductive reactance . #figure(figph[Photograph of power capacitors at a power station.], alt: "Photograph of power capacitors at a power station.", caption: [Power capacitors are used to balance the impedance of the effective inductance in transmission lines.]) The #emph[RLC] circuit is analogous to the wheel of a car driven over a corrugated road . The regularly spaced bumps in the road drive the wheel up and down; in the same way, a voltage source increases and decreases. The shock absorber acts like the resistance of the #emph[RLC] circuit, damping and limiting the amplitude of the oscillation. Energy within the wheel system goes back and forth between kinetic and potential energy stored in the car spring, analogous to the shift between a maximum current, with energy stored in an inductor, and no current, with energy stored in the electric field of a capacitor. The amplitude of the wheel’s motion is at a maximum if the bumps in the road are hit at the resonant frequency, which we describe in more detail in Resonance in an AC Circuit. #figure(figph[Figure shows one wheel of a car. Arrows show the up-down motion of its shock absorber spring.], alt: "Figure shows one wheel of a car. Arrows show the up-down motion of its shock absorber spring.", caption: [On a car, the shock absorber damps motion and dissipates energy. This is much like the resistance in an #emph[RLC] circuit. The mass and spring determine the resonant frequency.]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[AC Circuits] To analyze an ac circuit containing resistors, capacitors, and inductors, it is helpful to think of each device’s reactance and find the equivalent reactance using the rules we used for equivalent resistance in the past. Phasors are a great method to determine whether the emf of the circuit has positive or negative phase (namely, leads or lags other values). A mnemonic device of “ELI the ICE man” is sometimes used to remember that the emf (E) leads the current (I) in an inductor (L) and the current (I) leads the emf (E) in a capacitor (C). Use the following steps to determine the emf of the circuit by phasors: + Draw the phasors for voltage across each device: resistor, capacitor, and inductor, including the phase angle in the circuit. + If there is both a capacitor and an inductor, find the net voltage from these two phasors, since they are antiparallel. + Find the equivalent phasor from the phasor in step 2 and the resistor’s phasor using trigonometry or components of the phasors. The equivalent phasor found is the emf of the circuit. ] #examplebox("Example 1")[An #emph[RLC] Series Circuit][ The output of an ac generator connected to an #emph[RLC] series combination has a frequency of 200 Hz and an amplitude of 0.100 V. If #math.equation(block: false, alt: "R equals 4.00 Ω ,")[$R = 4.00 #h(0.2em) "Ω" ,$] #math.equation(block: false, alt: "L equals 3.00 times 10 to the power −3 H,")[$L = 3.00 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H,"$] and #math.equation(block: false, alt: "C equals 8.00 times 10 to the power −4 F,")[$C = 8.00 #h(0.2em) × #h(0.2em) 10^(−4) #h(0.2em) "F,"$] what are (a) the capacitive reactance, (b) the inductive reactance, (c) the impedance, (d) the current amplitude, and (e) the phase difference between the current and the emf of the generator? Strategy The reactances and impedance in (a)–(c) are found by substitutions into , , and , respectively. The current amplitude is calculated from the peak voltage and the impedance. The phase difference between the current and the emf is calculated by the inverse tangent of the difference between the reactances divided by the resistance. Solution + 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 2 π open parenthesis 200 Hz close parenthesis open parenthesis 8.00 times 10 to the power −4 F close parenthesis equals 0.995 Ω .")[$X_(C) = frac(1, ω C) = frac(1, 2 π ( 200 #h(0.2em) "Hz" ) ( 8.00 #h(0.2em) × #h(0.2em) 10^(−4) #h(0.2em) "F" )) = 0.995 #h(0.2em) "Ω" "."$] + From , the inductive reactance is #math.equation(block: true, alt: "X sub L equals ω L equals 2 π open parenthesis 200 Hz close parenthesis open parenthesis 3.00 times 10 to the power −3 H close parenthesis equals 3.77 Ω .")[$X_(L) = ω L = 2 π ( 200 #h(0.2em) "Hz" ) ( 3.00 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "H" ) = 3.77 #h(0.2em) "Ω" "."$] + Substituting the values of #emph[R], #math.equation(block: false, alt: "X sub C")[$X_(C)$], and #math.equation(block: false, alt: "X sub L")[$X_(L)$] into , we obtain for the impedance #math.equation(block: true, alt: "Z equals the square root of open parenthesis 4.00 Ω close parenthesis squared plus open parenthesis 3.77 Ω minus 0.995 Ω close parenthesis squared equals 4.87 Ω .")[$Z = sqrt(attach(( 4.00 #h(0.2em) "Ω" ), t: 2) + attach(( 3.77 #h(0.2em) "Ω" − 0.995 #h(0.2em) "Ω" ), t: 2)) = 4.87 #h(0.2em) "Ω" "."$] + The current amplitude is #math.equation(block: true, alt: "I sub 0 equals the fraction V sub 0 over Z equals the fraction 0.100 V over 4.87 Ω equals 2.05 times 10 to the power −2 A .")[$I_(0) = frac(V_(0), Z) = frac(0.100 #h(0.2em) "V", 4.87 #h(0.2em) "Ω") = 2.05 #h(0.2em) × #h(0.2em) 10^(−2) "A" "."$] + From , the phase difference between the current and the emf is #math.equation(block: true, alt: "ϕ equals tan to the power −1 the fraction X sub L minus X sub C over R equals tan to the power −1 the fraction 2.77 Ω over 4.00 Ω equals 0.607 rad .")[$ϕ = "tan"^(−1) frac(X_(L) − X_(C), R) = "tan"^(−1) frac(2.77 #h(0.2em) "Ω", 4.00 #h(0.2em) "Ω") = 0.607 #h(0.2em) "rad" "."$] Significance The phase angle is positive because the reactance of the inductor is larger than the reactance of the capacitor. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the voltages across the resistor, the capacitor, and the inductor in the circuit of using #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 sin ω t")[$v ( t ) = V_(0) #h(0.2em) "sin" #h(0.2em) ω t$] as the output of the ac generator. #solutionbox[ #math.equation(block: true, alt: "v sub R equals open parenthesis V sub 0 R / Z close parenthesis sin open parenthesis ω t minus ϕ close parenthesis ; v sub C equals open parenthesis V sub 0 X sub C / Z close parenthesis sin open parenthesis ω t minus ϕ plus π / 2 close parenthesis equals − open parenthesis V sub 0 X sub C / Z close parenthesis cos open parenthesis ω t minus ϕ close parenthesis ;; v sub L equals open parenthesis V sub 0 X sub L / Z close parenthesis sin open parenthesis ω t minus ϕ minus π / 2 close parenthesis equals open parenthesis V sub 0 X sub L / Z close parenthesis cos open parenthesis ω t minus ϕ close parenthesis")[$v_(R) = ( V_(0) R "/" Z ) #h(0.2em) "sin" #h(0.2em) ( ω t − ϕ ) ; v_(C) = ( V_(0) X_(C) "/" Z ) #h(0.2em) "sin" #h(0.2em) ( ω t − ϕ + π "/" 2 ) = "−" ( V_(0) X_(C) "/" Z ) "cos" ( ω t − ϕ ) ; \ v_(L) = ( V_(0) X_(L) "/" Z ) #h(0.2em) "sin" #h(0.2em) ( ω t − ϕ − π "/" 2 ) = ( V_(0) X_(L) "/" Z ) "cos" ( ω t − ϕ )$] ] ] === Summary - An #emph[RLC] series circuit is a resistor, capacitor, and inductor series combination across an ac source. - The same current flows through each element of an #emph[RLC] series circuit at all points in time. - The counterpart of resistance in a dc circuit is impedance, which measures the combined effect of resistors, capacitors, and inductors. The maximum current is defined by the ac version of Ohm’s law. - Impedance has units of ohms and is found using the resistance, the capacitive reactance, and the inductive reactance. === Conceptual Questions In an #emph[RLC] series circuit, can the voltage measured across the capacitor be greater than the voltage of the source? Answer the same question for the voltage across the inductor. #solutionbox[ yes for both ] === Problems What is the impedance of a series combination of a #math.equation(block: false, alt: "50 -Ω")[$50 "-Ω"$] resistor, a #math.equation(block: false, alt: "5.0 - μ F")[$5.0 "-" μ "F"$] capacitor, and a #math.equation(block: false, alt: "10 - μ F")[$10 "-" μ "F"$] capacitor at a frequency of 2.0 kHz? A resistor and capacitor are connected in series across an ac generator. The emf of the generator is given by #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 cos ω t ,")[$v ( t ) = V_(0) #h(0.2em) "cos" #h(0.2em) ω t ,$] where #math.equation(block: false, alt: "V sub 0 equals 120 V,")[$V_(0) = 120 #h(0.2em) "V,"$] #math.equation(block: false, alt: "ω equals 120 π rad/s ,")[$ω = 120 π #h(0.2em) "rad/s" ,$] #math.equation(block: false, alt: "R equals 400 Ω")[$R = 400 #h(0.2em) "Ω"$], and #math.equation(block: false, alt: "C equals 4.0 μ F")[$C = 4.0 μ "F"$]. (a) What is the impedance of the circuit? (b) What is the amplitude of the current through the resistor? (c) Write an expression for the current through the resistor. (d) Write expressions representing the voltages across the resistor and across the capacitor. #solutionbox[ a. #math.equation(block: false, alt: "770 Ω")[$770 #h(0.2em) "Ω"$]; b. 0.16 A; c. #math.equation(block: false, alt: "I equals open parenthesis 0.16 A close parenthesis cos open parenthesis 120 π t minus 0.33 π close parenthesis")[$I = ( 0.16 #h(0.2em) "A" ) "cos" ( 120 π t #h(0.2em) − #h(0.2em) 0.33 π )$]; d. #math.equation(block: false, alt: "v sub R equals 62 cos open parenthesis 120 π t minus 0.33 π close parenthesis")[$v_(R) = 62 #h(0.2em) "cos" ( 120 π t #h(0.2em) − #h(0.2em) 0.33 π )$]; #math.equation(block: false, alt: "v sub C equals 103 cos open parenthesis 120 π t minus 0.83 π close parenthesis")[$v_(C) = 103 #h(0.2em) "cos" ( 120 π t − 0.83 π )$] ] A resistor and inductor are connected in series across an ac generator. The emf of the generator is given by #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals V sub 0 cos ω t ,")[$v ( t ) = V_(0) #h(0.2em) "cos" #h(0.2em) ω t ,$] where #math.equation(block: false, alt: "V sub 0 equals 120 V")[$V_(0) = 120 #h(0.2em) "V"$] and #math.equation(block: false, alt: "ω equals 120 π rad/s;")[$ω = 120 π #h(0.2em) "rad/s;"$] also, #math.equation(block: false, alt: "R equals 400 Ω")[$R = 400 #h(0.2em) "Ω"$] and #math.equation(block: false, alt: "L equals 1.5 H .")[$L = 1.5 #h(0.2em) "H" "."$] (a) What is the impedance of the circuit? (b) What is the amplitude of the current through the resistor? (c) Write an expression for the current through the resistor. (d) Write expressions representing the voltages across the resistor and across the inductor. In an #emph[RLC] series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively, an #math.equation(block: false, alt: "R equals 500 Ω ,")[$R = 500 #h(0.2em) "Ω" ","$] #math.equation(block: false, alt: "L equals 0.20 H,")[$L = 0.20 #h(0.2em) "H,"$] and #math.equation(block: false, alt: "C equals 2.0 μ F .")[$C = 2.0 μ "F" "."$] (a) What is the impedance of the circuit? (b) What is the amplitude of the current from the source? (c) If the emf of the source is given by #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals open parenthesis 100 V close parenthesis sin 1000 π t")[$v ( t ) = ( 100 #h(0.2em) "V" ) #h(0.2em) "sin" #h(0.2em) 1000 π t$], how does the current vary with time? (d) Repeat the calculations with #emph[C] changed to #math.equation(block: false, alt: "0.20 μ F .")[$0.20 μ "F" "."$] #solutionbox[ a. #math.equation(block: false, alt: "690 Ω")[$690 #h(0.2em) "Ω"$]; b. 0.15 A; c. #math.equation(block: false, alt: "I equals open parenthesis 0.15 A close parenthesis sin open parenthesis 1000 π t minus 0.753 close parenthesis")[$I = ( 0.15 "A" ) #h(0.2em) "sin" #h(0.2em) ( 1000 π t − 0.753 )$]; d. #math.equation(block: false, alt: "1100 Ω")[$1100 #h(0.2em) "Ω"$], 0.092 A, #math.equation(block: false, alt: "I equals open parenthesis 0.092 A close parenthesis sin open parenthesis 1000 π t plus 1.09 close parenthesis")[$I = ( 0.092 "A" ) #h(0.2em) "sin" #h(0.2em) ( 1000 π t + 1.09 )$] ] An #emph[RLC] series circuit with #math.equation(block: false, alt: "R equals 600 Ω")[$R = 600 #h(0.2em) "Ω"$], #math.equation(block: false, alt: "L equals 30 mH,")[$L = 30 #h(0.2em) "mH,"$] and #math.equation(block: false, alt: "C equals 0.050 μ F")[$C = 0.050 μ "F"$] is driven by an ac source whose frequency and voltage amplitude are 500 Hz and 50 V, respectively. (a) What is the impedance of the circuit? (b) What is the amplitude of the current in the circuit? (c) What is the phase angle between the emf of the source and the current? For the circuit shown below, what are (a) the total impedance and (b) the phase angle between the current and the emf? (c) Write an expression for #math.equation(block: false, alt: "i open parenthesis t close parenthesis .")[$i ( t ) .$] #figure(figph[Figure shows a circuit with a voltage source 170 V, sine 120 pi t, a resistor of 5 ohm, a capacitor of 400 microfarad and an inductor of 25 milihenry all connected in series.], alt: "Figure shows a circuit with a voltage source 170 V, sine 120 pi t, a resistor of 5 ohm, a capacitor of 400 microfarad and an inductor of 25 milihenry all connected in series.", caption: none) #solutionbox[ a. #math.equation(block: false, alt: "5.7 Ω")[$5.7 #h(0.2em) "Ω"$]; b. #math.equation(block: false, alt: "29 °")[$29 "°"$]; c. #math.equation(block: false, alt: "I ​ equals open parenthesis 30. A close parenthesis cos open parenthesis 120 π t plus 0.51 close parenthesis")[$I "​" = ( 30. #h(0.2em) "A" ) "cos" ( 120 π t + 0.51 )$] ]