#set document(title: "15.6 Transformers", 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")) == 15.6#h(0.6em)Transformers Although ac electric power is produced at relatively low voltages, it is sent through #strong[transmission lines] at very high voltages (as high as 500 kV). The same power can be transmitted at different voltages because power is the product #math.equation(block: false, alt: "I sub rms V sub rms .")[$I_("rms") V_("rms") .$] (For simplicity, we ignore the phase factor #math.equation(block: false, alt: "cos ϕ . close parenthesis")[$"cos" #h(0.2em) ϕ . \)$] A particular power requirement can therefore be met with a low voltage and a high current or with a high voltage and a low current. The advantage of the high-voltage/low-current choice is that it results in lower #math.equation(block: false, alt: "I rms 2 R")[$I_("rms")^(2) R$] ohmic losses in the transmission lines, which can be significant in lines that are many kilometers long . #figure(figph[Figure shows a power plant on the left. This is connected to a step up transformer through a 12 kV line. The transformer is connected to a high voltage transmission line of 400 kV. This is connected to a step down transformer at a substation. From here, a 13 kV line goes to a step down transformer on an electric pole. From here a 240 V line goes to a house.], alt: "Figure shows a power plant on the left. This is connected to a step up transformer through a 12 kV line. The transformer is connected to a high voltage transmission line of 400 kV. This is connected to a step down transformer at a substation. From here, a 13 kV line goes to a step down transformer on an electric pole. From here a 240 V line goes to a house.", caption: [The rms voltage from a power plant eventually needs to be stepped down from 12 kV to 240 V so that it can be safely introduced into a home. A high-voltage transmission line allows a low current to be transmitted via a substation over long distances.]) Typically, the alternating emfs produced at power plants are “stepped up” to very high voltages before being transmitted through power lines; then, they must be “stepped down” to relatively safe values (110 or 220 V rms) before they are introduced into homes. The device that transforms voltages from one value to another using induction is the #strong[transformer] . #figure(figph[Photograph of transformers on an electric pole. There are three transformers, each encased in a cylindrical container.], alt: "Photograph of transformers on an electric pole. There are three transformers, each encased in a cylindrical container.", caption: [Transformers are used to step down the high voltages in transmission lines to the 110 to 220 V used in homes.]) As illustrates, a transformer basically consists of two separated coils, or windings, wrapped around a soft iron core. The primary winding has #math.equation(block: false, alt: "N sub P")[$N_("P")$] loops, or turns, and is connected to an alternating voltage #math.equation(block: false, alt: "v sub P open parenthesis t close parenthesis .")[$v_("P") ( t ) .$] The secondary winding has #math.equation(block: false, alt: "N sub S")[$N_("S")$] turns and is connected to a load resistor #math.equation(block: false, alt: "R sub S .")[$R_("S") .$] We assume the ideal case for which all magnetic field lines are confined to the core so that the same magnetic flux permeates each turn of both the primary and the secondary windings. We also neglect energy losses to magnetic hysteresis, to ohmic heating in the windings, and to ohmic heating of the induced eddy currents in the core. A good transformer can have losses as low as 1% of the transmitted power, so this is not a bad assumption. #figure(figph[Figure shows a soft iron core in the center. This is in the form of a rectangular ring. There are windings on its left arm, connected to a voltage source. These are labeled N subscript p turns. The current through them is i subscript p parentheses t parentheses. The voltage across two ends of the windings is v subscript p parentheses t parentheses. The windings on the right arm of the core are connected to a resistor R subscript s. The windings are labeled N subscript s turns. These are more in number than the windings on the left arm. The current in the right circuit is i subscript s parentheses t parentheses. The voltage across the windings is v subscript s parentheses t parentheses. The current in the left circuit flows into the windings from the top. The current in the right circuit flows out of the winding from the top.], alt: "Figure shows a soft iron core in the center. This is in the form of a rectangular ring. There are windings on its left arm, connected to a voltage source. These are labeled N subscript p turns. The current through them is i subscript p parentheses t parentheses. The voltage across two ends of the windings is v subscript p parentheses t parentheses. The windings on the right arm of the core are connected to a resistor R subscript s. The windings are labeled N subscript s turns. These are more in number than the windings on the left arm. The current in the right circuit is i subscript s parentheses t parentheses. The voltage across the windings is v subscript s parentheses t parentheses. The current in the left circuit flows into the windings from the top. The current in the right circuit flows out of the winding from the top.", caption: [A step-up transformer (more turns in the secondary winding than in the primary winding). The two windings are wrapped around a soft iron core.]) To analyze the transformer circuit, we first consider the primary winding. The input voltage #math.equation(block: false, alt: "v sub P open parenthesis t close parenthesis")[$v_("P") ( t )$] is equal to the potential difference induced across the primary winding. From Faraday’s law, the induced potential difference is #math.equation(block: false, alt: "− N sub P open parenthesis d Φ / d t close parenthesis ,")[$"−" N_("P") ( d "Φ" "/" d t ) ,$] where #math.equation(block: false, alt: "Φ")[$"Φ"$] is the flux through one turn of the primary winding. Thus, #math.equation(block: true, alt: "v sub P open parenthesis t close parenthesis equals − N sub P the fraction d Φ over d t .")[$v_("P") ( t ) = "−" N_("P") frac(d "Φ", d t) .$] Similarly, the output voltage #math.equation(block: false, alt: "v sub S open parenthesis t close parenthesis")[$v_("S") ( t )$] delivered to the load resistor must equal the potential difference induced across the secondary winding. Since the transformer is ideal, the flux through every turn of the secondary winding is also #math.equation(block: false, alt: "Φ ,")[$Φ ,$] and #math.equation(block: true, alt: "v sub S open parenthesis t close parenthesis equals − N sub S the fraction d Φ over d t .")[$v_("S") ( t ) = "−" N_("S") frac(d "Φ", d t) .$] Combining the last two equations, we have #math.equation(block: true, alt: "v sub S open parenthesis t close parenthesis equals the fraction N sub S over N sub P v sub P open parenthesis t close parenthesis .")[$v_("S") ( t ) = frac(N_("S"), N_("P")) v_("P") ( t ) .$] Hence, with appropriate values for #math.equation(block: false, alt: "N sub S and N sub P ,")[$N_("S") #h(0.2em) "and" #h(0.2em) N_("P") ,$] the input voltage #math.equation(block: false, alt: "v sub P open parenthesis t close parenthesis")[$v_("P") ( t )$] may be “stepped up” #math.equation(block: false, alt: "open parenthesis N sub S greater than N sub P close parenthesis")[$( N_("S") > N_("P") )$] or “stepped down” (#math.equation(block: false, alt: "N sub S less than N sub P")[$N_("S") < N_("P")$]) to #math.equation(block: false, alt: "v sub S open parenthesis t close parenthesis ,")[$v_("S") ( t ) ,$] the output voltage. This is often abbreviated as the #strong[transformer equation], #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "the fraction V sub S over V sub P equals the fraction N sub S over N sub P ,")[$frac(V_("S"), V_("P")) = frac(N_("S"), N_("P")) ,$] ] which shows that the ratio of the secondary to primary voltages in a transformer equals the ratio of the number of turns in their windings. For a #strong[step-up transformer], which increases voltage and decreases current, this ratio is greater than one; for a #strong[step-down transformer], which decreases voltage and increases current, this ratio is less than one. From the law of energy conservation, the power introduced at any instant by #math.equation(block: false, alt: "v sub P open parenthesis t close parenthesis")[$v_("P") ( t )$] to the primary winding must be equal to the power dissipated in the resistor of the secondary circuit; thus, #math.equation(block: true, alt: "i sub P open parenthesis t close parenthesis v sub P open parenthesis t close parenthesis equals i sub S open parenthesis t close parenthesis v sub S open parenthesis t close parenthesis .")[$i_("P") ( t ) v_("P") ( t ) = i_("S") ( t ) v_("S") ( t ) .$] When combined with , this gives #math.equation(block: true, alt: "i sub S open parenthesis t close parenthesis equals the fraction N sub P over N sub S i sub P open parenthesis t close parenthesis .")[$i_("S") ( t ) = frac(N_("P"), N_("S")) i_("P") ( t ) .$] If the voltage is stepped up, the current is stepped down, and vice versa. Finally, we can use #math.equation(block: false, alt: "i sub S open parenthesis t close parenthesis equals v sub S open parenthesis t close parenthesis / R sub S")[$i_("S") ( t ) = v_("S") ( t ) "/" R_("S")$], along with and , to obtain #math.equation(block: true, alt: "v sub P open parenthesis t close parenthesis equals i sub P [ open parenthesis the fraction N sub P over N sub S close parenthesis squared R sub S ] ,")[$v_("P") ( t ) = i_("P") [ attach(( frac(N_("P"), N_("S")) ), t: 2) R_("S") ] ,$] which tells us that the input voltage #math.equation(block: false, alt: "v sub P open parenthesis t close parenthesis")[$v_("P") ( t )$] “sees” not a resistance #math.equation(block: false, alt: "R sub S")[$R_("S")$] but rather a resistance #math.equation(block: true, alt: "R sub P equals open parenthesis the fraction N sub P over N sub S close parenthesis squared R sub S .")[$R_("P") = attach(( frac(N_("P"), N_("S")) ), t: 2) R_("S") .$] Our analysis has been based on instantaneous values of voltage and current. However, the resulting equations are not limited to instantaneous values; they hold also for maximum and rms values. #examplebox("Example 1")[A Step-Down Transformer][ A transformer on a utility pole steps the rms voltage down from 12 kV to 240 V. The transmission line resistance is 200 Ω. (a) What is the ratio of the number of secondary turns to the number of primary turns? (b) If the input current to the transformer is 2.0 A, what is the output current? (c) Determine the power loss in the transmission line. (d) Find the power loss if there were no transformer and the output power were kept the same. Strategy The number of turns related to the voltages is found from . The output current is calculated using . Solution + Using with rms values #math.equation(block: false, alt: "V sub P")[$V_("P")$] and #math.equation(block: false, alt: "V sub S ,")[$V_("S") ,$] we have #math.equation(block: true, alt: "the fraction N sub S over N sub P equals the fraction 240 V over 12 times 10 cubed V equals the fraction 1 over 50 ,")[$frac(N_("S"), N_("P")) = frac(240 #h(0.2em) "V", 12 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "V") = frac(1, 50) ,$] so the primary winding has 50 times the number of turns in the secondary winding. + From , the output rms current #math.equation(block: false, alt: "I sub S")[$I_("S")$] is found using the transformer equation with current #linebreak() #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "I sub S equals the fraction N sub P over N sub S I sub P")[$I_("S") = frac(N_("P"), N_("S")) I_("P")$] ] #linebreak() such that #math.equation(block: true, alt: "I sub S equals the fraction N sub P over N sub S I sub P equals open parenthesis 50 close parenthesis open parenthesis 2.0 A close parenthesis equals 100 A .")[$I_("S") = frac(N_("P"), N_("S")) I_("P") = ( 50 ) ( 2.0 #h(0.2em) "A" ) = 100 #h(0.2em) "A" "."$] + The power loss in the transmission line is calculated to be #math.equation(block: true, alt: "P sub loss equals I P 2 R equals open parenthesis 2.0 A close parenthesis squared open parenthesis 200 Ω close parenthesis equals 800 W .")[$P_("loss") = I_("P")^(2) R = attach(( 2.0 #h(0.2em) "A" ), t: 2) ( 200 #h(0.2em) "Ω" ) = 800 #h(0.2em) "W" "."$] + If there were no transformer, the power would have to be sent at 240 V to work for these houses, and the power loss would be #math.equation(block: true, alt: "P sub loss equals I S 2 R equals open parenthesis 100 A close parenthesis squared open parenthesis 200 Ω close parenthesis equals 2 times 10 to the power 6 W .")[$P_("loss") = I_("S")^(2) R = attach(( 100 #h(0.2em) "A" ), t: 2) ( 200 #h(0.2em) "Ω" ) = 2 #h(0.2em) × #h(0.2em) 10^(6) #h(0.2em) "W" "."$] Therefore, when power needs to be transmitted, we want to avoid power loss. Thus, lines are sent with high voltages and low currents and adjusted with a transformer before power is sent into homes. Significance This application of a step-down transformer allows a home that uses 240-V outlets to have 100 A available to draw upon. This can power many devices in the home. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A transformer steps the line voltage down from 110 to 9.0 V so that a current of 0.50 A can be delivered to a doorbell. (a) What is the ratio of the number of turns in the primary and secondary windings? (b) What is the current in the primary winding? (c) What is the resistance seen by the 110-V source? #solutionbox[ a. 12:1; b. 0.042 A; c. #math.equation(block: false, alt: "2.6 times 10 cubed Ω")[$2.6 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "Ω"$] ] ] === Summary - Power plants transmit high voltages at low currents to achieve lower ohmic losses in their many kilometers of transmission lines. - Transformers use induction to transform voltages from one value to another. - For a transformer, the voltages across the primary and secondary coils, or windings, are related by the transformer equation. - The currents in the primary and secondary windings are related by the number of primary and secondary loops, or turns, in the windings of the transformer. - A step-up transformer increases voltage and decreases current, whereas a step-down transformer decreases voltage and increases current. === Key Equations #figure(table( columns: 2, align: left, inset: 6pt, table.header([AC voltage], [#math.equation(block: false, alt: "v equals V sub 0 sin ω t")[$v = V_(0) #h(0.2em) "sin" #h(0.2em) ω t$]]), [AC current], [#math.equation(block: false, alt: "i equals I sub 0 sin ω t")[$i = I_(0) #h(0.2em) "sin" #h(0.2em) ω t$]], [capacitive reactance], [#math.equation(block: false, alt: "the fraction V sub 0 over I sub 0 equals the fraction 1 over ω C equals X sub C")[$frac(V_(0), I_(0)) = frac(1, ω C) = X_(C)$]], [rms voltage], [#math.equation(block: false, alt: "V sub rms equals the fraction V sub 0 over the square root of 2")[$V_("rms") = frac(V_(0), sqrt(2))$]], [rms current], [#math.equation(block: false, alt: "I sub rms equals the fraction I sub 0 over the square root of 2")[$I_("rms") = frac(I_(0), sqrt(2))$]], [inductive reactance], [#math.equation(block: false, alt: "the fraction V sub 0 over I sub 0 equals ω L equals X sub L")[$frac(V_(0), I_(0)) = ω L = X_(L)$]], [Phase angle of an RLC series circuit], [#math.equation(block: false, alt: "ϕ equals tan to the power −1 the fraction X sub L minus X sub C over R")[$ϕ = "tan"^(−1) frac(X_(L) − X_(C), R)$]], [AC version of Ohm’s law], [#math.equation(block: false, alt: "I sub 0 equals the fraction V sub 0 over Z")[$I_(0) = frac(V_(0), Z)$]], [Impedance of an RLC series circuit], [#math.equation(block: false, alt: "Z equals the square root of R squared plus open parenthesis X sub L minus X sub C close parenthesis squared")[$Z = sqrt(R^(2) + attach(( X_(L) − X_(C) ), t: 2))$]], [Average power associated with a circuit element], [#math.equation(block: false, alt: "P sub ave equals the fraction 1 over 2 I sub 0 V sub 0 cos ϕ")[$P_("ave") = frac(1, 2) I_(0) V_(0) #h(0.2em) "cos" #h(0.2em) ϕ$]], [Average power dissipated by a resistor], [#math.equation(block: false, 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")[$P_("ave") = frac(1, 2) I_(0) V_(0) = I_("rms") V_("rms") = I_("rms")^(2) R$]], [Resonant angular frequency of a circuit], [#math.equation(block: false, alt: "ω sub 0 equals the square root of the fraction 1 over L C")[$ω_(0) = sqrt(frac(1, L C))$]], [Quality factor of a circuit], [#math.equation(block: false, alt: "Q equals the fraction ω sub 0 over Δ ω")[$Q = frac(ω_(0), "Δ" ω)$]], [Quality factor of a circuit in terms of the circuit parameters], [#math.equation(block: false, alt: "Q equals the fraction ω sub 0 L over R")[$Q = frac(ω_(0) L, R)$]], [Transformer equation with voltage], [#math.equation(block: false, alt: "the fraction V sub S over V sub P equals the fraction N sub S over N sub P")[$frac(V_("S"), V_("P")) = frac(N_("S"), N_("P"))$]], [Transformer equation with current], [#math.equation(block: false, alt: "I sub S equals the fraction N sub P over N sub S I sub P")[$I_("S") = frac(N_("P"), N_("S")) I_("P")$]], )) === Conceptual Questions Why do transmission lines operate at very high voltages while household circuits operate at fairly small voltages? #solutionbox[ There is less thermal loss if the transmission lines operate at low currents and high voltages. ] How can you distinguish the primary winding from the secondary winding in a step-up transformer? Battery packs in some electronic devices are charged using an adapter connected to a wall socket. Speculate as to the purpose of the adapter. #solutionbox[ The adapter has a step-down transformer to have a lower voltage and possibly higher current at which the device can operate. ] Will a transformer work if the input is a dc voltage? Why are the primary and secondary coils of a transformer wrapped around the same closed loop of iron? #solutionbox[ so each loop can experience the same changing magnetic flux ] === Problems A step-up transformer is designed so that the output of its secondary winding is 2000 V (rms) when the primary winding is connected to a 110-V (rms) line voltage. (a) If there are 100 turns in the primary winding, how many turns are there in the secondary winding? (b) If a resistor connected across the secondary winding draws an rms current of 0.75 A, what is the current in the primary winding? A step-up transformer connected to a 110-V line is used to supply a hydrogen-gas discharge tube with 5.0 kV (rms). The tube dissipates 75 W of power. (a) What is the ratio of the number of turns in the secondary winding to the number of turns in the primary winding? (b) What are the rms currents in the primary and secondary windings? (c) What is the effective resistance seen by the 110-V source? #solutionbox[ a. 45:1; b. 0.68 A, 0.015 A; c. #math.equation(block: false, alt: "160 Ω")[$160 #h(0.2em) "Ω"$] ] An ac source of emf delivers 5.0 mW of power at an rms current of 2.0 mA when it is connected to the primary coil of a transformer. The rms voltage across the secondary coil is 20 V. (a) What are the voltage across the primary coil and the current through the secondary coil? (b) What is the ratio of secondary to primary turns for the transformer? A transformer is used to step down 110 V from a wall socket to 9.0 V for a radio. (a) If the primary winding has 500 turns, how many turns does the secondary winding have? (b) If the radio operates at a current of 500 mA, what is the current through the primary winding? #solutionbox[ a. 41 turns; b. 40.9 mA ] A transformer is used to supply a 12-V model train with power from a 110-V wall plug. The train operates at 50 W of power. (a) What is the rms current in the secondary coil of the transformer? (b) What is the rms current in the primary coil? (c) What is the ratio of the number of primary to secondary turns? (d) What is the resistance of the train? (e) What is the resistance seen by the 110-V source? === Additional 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 .")[$ω = 200 π #h(0.2em) "rad/s" "."$] Find an expression that represents the output current 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. #solutionbox[ a. #math.equation(block: false, alt: "i open parenthesis t close parenthesis equals open parenthesis 1.26 A close parenthesis sin open parenthesis 200 π t plus π / 2 close parenthesis")[$i ( t ) = ( 1.26 "A" ) #h(0.2em) "sin" #h(0.2em) ( 200 π t + π "/" 2 )$]; b. #math.equation(block: false, alt: "i open parenthesis t close parenthesis equals open parenthesis 7.96 A close parenthesis sin open parenthesis 200 π t minus π / 2 close parenthesis")[$i ( t ) = ( 7.96 "A" ) #h(0.2em) "sin" #h(0.2em) ( 200 π t − π "/" 2 )$]; c. #math.equation(block: false, alt: "i open parenthesis t close parenthesis equals open parenthesis 2 A close parenthesis sin open parenthesis 200 π t close parenthesis")[$i ( t ) = ( 2 "A" ) #h(0.2em) "sin" #h(0.2em) ( 200 π t )$] ] A 700-pF capacitor is connected across an ac source with a voltage amplitude of 160 V and a frequency of 20 kHz. (a) Determine the capacitive reactance of the capacitor and the amplitude of the output current of the source. (b) If the frequency is changed to 60 Hz while keeping the voltage amplitude at 160 V, what are the capacitive reactance and the current amplitude? A 20-mH inductor is connected across an AC source with a variable frequency and a constant-voltage amplitude of 9.0 V. (a) Determine the reactance of the circuit and the maximum current through the inductor when the frequency is set at 20 kHz. (b) Do the same calculations for a frequency of 60 Hz. #solutionbox[ a. #math.equation(block: false, alt: "2.5 times 10 cubed Ω , 3.6 times 10 to the power −3 A")[$2.5 #h(0.2em) × #h(0.2em) 10^(3) "Ω" , 3.6 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "A"$]; b. #math.equation(block: false, alt: "7.5 Ω , 1.2 A")[$7.5 #h(0.2em) "Ω" , 1.2 "A"$] ] A #math.equation(block: false, alt: "30 - μ F")[$30 "-" μ "F"$] capacitor is connected across a 60-Hz ac source whose voltage amplitude is 50 V. (a) What is the maximum charge on the capacitor? (b) What is the maximum current into the capacitor? (c) What is the phase relationship between the capacitor charge and the current in the circuit? A 7.0-mH inductor is connected across a 60-Hz ac source whose voltage amplitude is 50 V. (a) What is the maximum current through the inductor? (b) What is the phase relationship between the current through and the potential difference across the inductor? #solutionbox[ a. 19 A; b. inductor leads by #math.equation(block: false, alt: "90 °")[$90 "°"$] ] What is the impedance of an #emph[RLC] series circuit at the resonant frequency? What is the resistance #emph[R] in the circuit shown below if the amplitude of the ac through the inductor is 4.24 A? #figure(figph[Figure shows a circuit with an AC source of 50 V, sine 120 pi t. This is connected to an inductor of 8 mH, a capacitor of 400 mu F and a resistor R. Another capacitor is connected in parallel with the first one. The value of this is 200 mu F.], alt: "Figure shows a circuit with an AC source of 50 V, sine 120 pi t. This is connected to an inductor of 8 mH, a capacitor of 400 mu F and a resistor R. Another capacitor is connected in parallel with the first one. The value of this is 200 mu F.", caption: none) #solutionbox[ #math.equation(block: true, alt: "11.7 Ω")[$11.7 #h(0.2em) "Ω"$] ] An ac source of voltage amplitude 100 V and frequency 1.0 kHz drives 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 4.0 mH")[$L = 4.0 #h(0.2em) "mH"$], and #math.equation(block: false, alt: "C equals 50 μ F")[$C = 50 μ "F"$]. (a) Determine the rms current through the circuit. (b) What are the rms voltages across the three elements? (c) What is the phase angle between the emf and the current? (d) What is the power output of the source? (e) What is the power dissipated in the resistor? In an #emph[RLC] series circuit, #math.equation(block: false, alt: "R equals 200 Ω")[$R = 200 #h(0.2em) "Ω"$], #math.equation(block: false, alt: "L equals 1.0 H")[$L = 1.0 #h(0.2em) "H"$], #math.equation(block: false, alt: "C equals 50 μ F,")[$C = 50 μ "F,"$] #math.equation(block: false, alt: "V sub 0 equals 120 V")[$V_(0) = 120 #h(0.2em) "V"$], and #math.equation(block: false, alt: "f equals 50 Hz")[$f = 50 #h(0.2em) "Hz"$]. What is the power output of the source? #solutionbox[ 14 W ] A power plant generator produces 100 A at 15 kV (rms). A transformer is used to step up the transmission line voltage to 150 kV (rms). (a) What is rms current in the transmission line? (b) If the resistance per unit length of the line is #math.equation(block: false, alt: "8.6 times 10 to the power −8 Ω /m,")[$8.6 #h(0.2em) × #h(0.2em) 10^(−8) #h(0.2em) "Ω" "/m,"$] what is the power loss per meter in the line? (c) What would the power loss per meter be if the line voltage were 15 kV (rms)? Consider a power plant located 25 km outside a town delivering 50 MW of power to the town. The transmission lines are made of aluminum cables with a #math.equation(block: false, alt: "7 cm squared")[$7 #h(0.2em) "cm"^(2)$] cross-sectional area. Find the loss of power in the transmission lines if it is transmitted at (a) 200 kV (rms) and (b) 120 V (rms). #solutionbox[ a. #math.equation(block: false, alt: "5.9 times 10 to the power 4 W")[$5.9 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "W"$]; b. #math.equation(block: false, alt: "1.64 times 10 to the power 11 W")[$1.64 #h(0.2em) × #h(0.2em) 10^(11) #h(0.2em) "W"$] ] Neon signs require 12-kV for their operation. A transformer is to be used to change the voltage from 220-V (rms) ac to 12-kV (rms) ac. What must the ratio be of turns in the secondary winding to the turns in the primary winding? (b) What is the maximum rms current the neon lamps can draw if the fuse in the primary winding goes off at 0.5 A? (c) How much power is used by the neon sign when it is drawing the maximum current allowed by the fuse in the primary winding? === Challenge Problems The 335-kV ac electricity from a power transmission line is fed into the primary winding of a transformer. The ratio of the number of turns in the secondary winding to the number in the primary winding is #math.equation(block: false, alt: "N sub s / N sub p equals 1000")[$N_("s") "/" N_("p") = 1000$]. (a) What voltage is induced in the secondary winding? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? #solutionbox[ a. 335 MV; b. the result is way too high, well beyond the breakdown voltage of air over reasonable distances; c. the input voltage is too high ] A #math.equation(block: false, alt: "1.5 -k Ω")[$1.5 "-k" "Ω"$] resistor and 30-mH inductor are connected in series, as shown below, across a 120-V (rms) ac power source oscillating at 60-Hz frequency. (a) Find the current in the circuit. (b) Find the voltage drops across the resistor and inductor. (c) Find the impedance of the circuit. (d) Find the power dissipated in the resistor. (e) Find the power dissipated in the inductor. (f) Find the power produced by the source. #figure(figph[Series circuit with voltage source V parentheses t parentheses, a 30 mH inductor and a 1.5 kilo ohm resistor], alt: "Series circuit with voltage source V parentheses t parentheses, a 30 mH inductor and a 1.5 kilo ohm resistor", caption: none) A #math.equation(block: false, alt: "20 - Ω")[$20 "-" "Ω"$] resistor, #math.equation(block: false, alt: "50 - μ F")[$50 "-" μ "F"$] capacitor, and 30-mH inductor are connected in series with an ac source of amplitude 10 V and frequency 125 Hz. (a) What is the impedance of the circuit? (b) What is the amplitude of the current in the circuit? (c) What is the phase constant of the current? Is it leading or lagging the source voltage? (d) Write voltage drops across the resistor, capacitor, and inductor and the source voltage as a function of time. (e) What is the power factor of the circuit? (f) How much energy is used by the resistor in 2.5 s? #solutionbox[ a. #math.equation(block: false, alt: "20 Ω")[$20 #h(0.2em) "Ω"$]; b. 0.5 A; c. #math.equation(block: false, alt: "5.4 °")[$5.4 "°"$], lagging; #linebreak() d. #math.equation(block: false, alt: "V sub R equals open parenthesis 9.96 V close parenthesis cos open parenthesis 250 π t plus 5.4 ° close parenthesis , V sub C equals open parenthesis 12.7 V close parenthesis cos open parenthesis 250 π t plus 5.4 ° minus 90 ° close parenthesis ,; V sub L equals open parenthesis 11.8 V close parenthesis cos open parenthesis 250 π t plus 5.4 ° plus 90 ° close parenthesis , V sub source equals open parenthesis 10.0 ​ V close parenthesis cos open parenthesis 250 π t close parenthesis ;")[$V_(R) = ( 9.96 #h(0.2em) "V" ) "cos" ( 250 π t + 5.4 "°" ) "," #h(0.2em) V_(C) = ( 12.7 #h(0.2em) "V" ) "cos" ( 250 π t + 5.4 "°" − 90 "°" ) "," \ V_(L) = ( 11.8 #h(0.2em) "V" ) "cos" ( 250 π t + 5.4 "°" + 90 "°" ) "," #h(0.2em) V_("source") = ( 10.0 "​" #h(0.2em) "V" ) "cos" ( 250 π t ) ;$] e. 0.995; f. 6.25 J ] A #math.equation(block: false, alt: "200 - Ω")[$200 "-" "Ω"$] resistor, #math.equation(block: false, alt: "150 - μ F")[$150 "-" μ "F"$] capacitor, and 2.5-H inductor are connected in series with an ac source of amplitude 10 V and variable angular frequency #math.equation(block: false, alt: "ω")[$ω$]. (a) What is the value of the resonance frequency #math.equation(block: false, alt: "ω sub R")[$ω_(R)$]? (b) What is the amplitude of the current if #math.equation(block: false, alt: "ω equals ω sub R")[$ω = ω_(R)$]? (c) What is the phase constant of the current when #math.equation(block: false, alt: "ω equals ω sub R")[$ω = ω_(R)$]? Is it leading or lagging the source voltage, or is it in phase? (d) Write an equation for the voltage drop across the resistor as a function of time when #math.equation(block: false, alt: "ω equals ω sub R")[$ω = ω_(R)$]. (e) What is the power factor of the circuit when #math.equation(block: false, alt: "ω equals ω sub R")[$ω = ω_(R)$]? (f) How much energy is used up by the resistor in 2.5 s when #math.equation(block: false, alt: "ω equals ω sub R")[$ω = ω_(R)$]? Find the reactances of the following capacitors and inductors in ac circuits with the given frequencies in each case: (a) 2-mH inductor with a frequency 60-Hz of the ac circuit; (b) 2-mH inductor with a frequency 600-Hz of the ac circuit; (c) 20-mH inductor with a frequency 6-Hz of the ac circuit; (d) 20-mH inductor with a frequency 60-Hz of the ac circuit; (e) 2-mF capacitor with a frequency 60-Hz of the ac circuit; and (f) 2-mF capacitor with a frequency 600-Hz of the AC circuit. #solutionbox[ a. #math.equation(block: false, alt: "0.75 Ω")[$0.75 #h(0.2em) "Ω"$]; b. #math.equation(block: false, alt: "7.5 Ω")[$7.5 #h(0.2em) "Ω"$]; c. #math.equation(block: false, alt: "0.75 Ω")[$0.75 #h(0.2em) "Ω"$]; d. #math.equation(block: false, alt: "7.5 Ω")[$7.5 #h(0.2em) "Ω"$]; e. #math.equation(block: false, alt: "1.3 Ω")[$1.3 #h(0.2em) "Ω"$]; f. #math.equation(block: false, alt: "0.13 Ω")[$0.13 #h(0.2em) "Ω"$] ] An output impedance of an audio amplifier has an impedance of #math.equation(block: false, alt: "500 Ω")[$500 #h(0.2em) "Ω"$] and has a mismatch with a low-impedance #math.equation(block: false, alt: "8 - Ω")[$8 "-" "Ω"$] loudspeaker. You are asked to insert an appropriate transformer to match the impedances. What turns ratio will you use, and why? Use the simplified circuit shown below. #figure(figph[Figure shows a transformer with more windings in the primary coil. The primary coil is connected to a voltage source through an impedance Z equal to 500 ohm. The voltage across the windings is labeled amplifier output V subscript P. The two ends of the secondary coil of the transformer are connected across a resistance of 8 ohm.], alt: "Figure shows a transformer with more windings in the primary coil. The primary coil is connected to a voltage source through an impedance Z equal to 500 ohm. The voltage across the windings is labeled amplifier output V subscript P. The two ends of the secondary coil of the transformer are connected across a resistance of 8 ohm.", caption: none) Show that the SI unit for capacitive reactance is the ohm. Show that the SI unit for inductive reactance is also the ohm. #solutionbox[ The units as written for inductive reactance are #math.equation(block: false, alt: "the fraction rad over s H")[$frac("rad", "s") "H"$]. Radians can be ignored in unit analysis. The Henry can be defined as #math.equation(block: false, alt: "H equals the fraction V times s over A equals Ω times s")[$"H" = frac("V" · "s", "A") = "Ω" · "s"$]. Combining these together results in a unit of #math.equation(block: false, alt: "Ω")[$"Ω"$] for reactance. ] A coil with a self-inductance of 16 mH and a resistance of #math.equation(block: false, alt: "6.0 Ω")[$6.0 #h(0.2em) "Ω"$] is connected to an ac source whose frequency can be varied. At what frequency will the voltage across the coil lead the current through the coil by #math.equation(block: false, alt: "45 ° ?")[$45 "°" ?$] An #emph[RLC] series circuit consists of a #math.equation(block: false, alt: "50 - Ω")[$50 "-" "Ω"$] resistor, a #math.equation(block: false, alt: "200 - μ F")[$200 "-" μ "F"$] capacitor, and a 120-mH inductor whose coil has a resistance of #math.equation(block: false, alt: "20 Ω")[$20 #h(0.2em) "Ω"$]. The source for the circuit has an rms emf of 240 V at a frequency of 60 Hz. Calculate the rms voltages across the (a) resistor, (b) capacitor, and (c) inductor. #solutionbox[ a. 156 V; b. 42 V; c. 154 V ] An #emph[RLC] series circuit consists of a #math.equation(block: false, alt: "10 - Ω")[$10 "-" "Ω"$] resistor, an #math.equation(block: false, alt: "8.0 - μ F")[$8.0 "-" μ "F"$] capacitor, and a 50-mH inductor. A 110-V (rms) source of variable frequency is connected across the combination. What is the power output of the source when its frequency is set to one-half the resonant frequency of the circuit? Shown below are two circuits that act as crude high-pass filters. The input voltage to the circuits is #math.equation(block: false, alt: "v sub in")[$v_("in")$], and the output voltage is #math.equation(block: false, alt: "v sub out .")[$v_("out") .$] (a) Show that for the capacitor circuit, #math.equation(block: true, alt: "the fraction v sub out over v sub in equals the fraction 1 over the square root of 1 plus 1 / ω squared R squared C squared ,")[$frac(v_("out"), v_("in")) = frac(1, sqrt(1 + 1 "/" ω^(2) R^(2) C^(2))) ,$] and for the inductor circuit, #math.equation(block: true, alt: "the fraction v sub out over v sub in equals the fraction ω L over the square root of R squared plus ω squared L squared .")[$frac(v_("out"), v_("in")) = frac(ω L, sqrt(R^(2) + ω^(2) L^(2))) .$] (b) Show that for high frequencies, #math.equation(block: false, alt: "v sub out approximately equals v sub in ,")[$v_("out") ≈ v_("in") ,$] but for low frequencies, #math.equation(block: false, alt: "v sub out approximately equals 0 .")[$v_("out") ≈ 0 .$] #figure(figph[Figure shows two circuits. The first shows a capacitor and resistor in series with a voltage source labeled V in. V out is measured across the resistor. The second circuit shows an inductor and resistor in series with a voltage source labeled V in. V out is measured across the inductor.], alt: "Figure shows two circuits. The first shows a capacitor and resistor in series with a voltage source labeled V in. V out is measured across the resistor. The second circuit shows an inductor and resistor in series with a voltage source labeled V in. V out is measured across the inductor.", caption: none) #solutionbox[ a. #math.equation(block: false, alt: "the fraction v sub out over v sub in equals the fraction 1 over the square root of 1 plus 1 / ω squared R squared C squared")[$frac(v_("out"), v_("in")) = frac(1, sqrt(1 + 1 "/" ω^(2) R^(2) C^(2)))$] and #math.equation(block: false, alt: "the fraction v sub out over v sub in equals the fraction ω L over the square root of R squared plus ω squared L squared")[$frac(v_("out"), v_("in")) = frac(ω L, sqrt(R^(2) + ω^(2) L^(2)))$]; b. #math.equation(block: false, alt: "v sub out approximately equals v sub in")[$v_("out") ≈ v_("in")$] and #math.equation(block: false, alt: "v sub out approximately equals 0")[$v_("out") ≈ 0$] ] The two circuits shown below act as crude low-pass filters. The input voltage to the circuits is #math.equation(block: false, alt: "v sub in")[$v_("in")$], and the output voltage is #math.equation(block: false, alt: "v sub out .")[$v_("out") .$] (a) Show that for the capacitor circuit, #math.equation(block: true, alt: "the fraction v sub out over v sub in equals the fraction 1 over the square root of 1 plus ω squared R squared C squared ,")[$frac(v_("out"), v_("in")) = frac(1, sqrt(1 + ω^(2) R^(2) C^(2))) ,$] and for the inductor circuit, #math.equation(block: true, alt: "the fraction v sub out over v sub in equals the fraction R over the square root of R squared plus ω squared L squared .")[$frac(v_("out"), v_("in")) = frac(R, sqrt(R^(2) + ω^(2) L^(2))) .$] (b) Show that for low frequencies, #math.equation(block: false, alt: "v sub out approximately equals v sub in ,")[$v_("out") ≈ v_("in") ,$] but for high frequencies, #math.equation(block: false, alt: "v sub out approximately equals 0 .")[$v_("out") ≈ 0 .$] #figure(figph[Figure shows two circuits. The first shows a capacitor and resistor in series with a voltage source labeled V in. V out is measured across the capacitor. The second circuit shows an inductor and resistor in series with a voltage source labeled V in. V out is measured across the resistor.], alt: "Figure shows two circuits. The first shows a capacitor and resistor in series with a voltage source labeled V in. V out is measured across the capacitor. The second circuit shows an inductor and resistor in series with a voltage source labeled V in. V out is measured across the resistor.", caption: none)