{"id":1332,"date":"2021-11-24T06:41:13","date_gmt":"2021-11-24T06:41:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1332"},"modified":"2022-11-01T23:04:42","modified_gmt":"2022-11-01T23:04:42","slug":"the-rlc-series-circuit","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/the-rlc-series-circuit\/","title":{"raw":"The\u00a0RLC\u00a0Series Circuit","rendered":"The\u00a0RLC\u00a0Series Circuit"},"content":{"raw":"<div data-type=\"note\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3 style=\"text-align: center;\">Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Solve a second-order differential equation representing charge and current in an RLC series circuit.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170571071163\">Consider an electrical circuit containing a resistor, an inductor, and a capacitor, as shown in <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/simple-harmonic-motion\/\">Simple Harmonic Motion<\/a> Figure 9. Such a circuit is called an\u00a0<strong><span id=\"1f3e94ba-eee2-44b2-881a-5e5f8fe37136_term321\" data-type=\"term\"><em data-effect=\"italics\">RLC<\/em>\u00a0series circuit<\/span>.<\/strong>\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0circuits are used in many electronic systems, most notably as tuners in AM\/FM radios. The tuning knob varies the capacitance of the capacitor, which in turn tunes the radio. Such circuits can be modeled by second-order, constant-coefficient differential equations.<\/p>\r\n<p id=\"fs-id1170571448902\">Let [latex]I(t)[\/latex] denote the current in the\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0circuit and [latex]q(t)[\/latex] denote the charge on the capacitor. Furthermore, let [latex]L[\/latex] denote inductance in henrys ([latex]H[\/latex]), [latex]R[\/latex] denote resistance in ohms ([latex]\\Omega[\/latex]), and [latex]C[\/latex] denote capacitance in farads ([latex]F[\/latex]). Last, let [latex]E(t)[\/latex] denote electric potential in volts ([latex]V[\/latex]).<\/p>\r\n<p id=\"fs-id1170573581916\">Kirchhoff\u2019s voltage rule states that the sum of the voltage drops around any closed loop must be zero. So, we need to consider the voltage drops across the inductor (denoted [latex]E_L[\/latex]), the resistor (denoted [latex]E_R[\/latex]), and the capacitor (denoted [latex]E_C[\/latex]). Because the\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0circuit shown in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/simple-harmonic-motion\/\">Simple Harmonic Motion<\/a> Figure 9\u00a0includes a voltage source, [latex]E(t)[\/latex], which adds voltage to the circuit, we have [latex]E_L+E_R+E_C=E(t)[\/latex].<\/p>\r\n<p id=\"fs-id1170573515970\">We present the formulas below without further development. Those of you interested in the derivation of these formulas should consult a physics text. Using Faraday\u2019s law and Lenz\u2019s law, the voltage drop across an inductor can be shown to be proportional to the instantaneous rate of change of current, with proportionality constant [latex]L[\/latex]. Thus,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{E_L=L\\frac{dI}{dt}}[\/latex].<\/p>\r\n<p id=\"fs-id1170571131649\">Next, according to Ohm\u2019s law, the voltage drop across a resistor is proportional to the current passing through the resistor, with proportionality constant [latex]R[\/latex]. Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{E_R=RI}[\/latex].<\/p>\r\n<p id=\"fs-id1170571087077\">Last, the voltage drop across a capacitor is proportional to the charge, [latex]q[\/latex], on the capacitor, with proportionality constant [latex]1\/C[\/latex]. Thus,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{E_C=\\frac{1}{C}q}[\/latex].<\/p>\r\n<p id=\"fs-id1170571260221\">Adding these terms together, we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{L\\frac{dI}{dt}+RI+\\frac1Cq=E(t)}[\/latex].<\/p>\r\n<p id=\"fs-id1170571042904\">Noting that [latex]I=(dq)\/(dt)[\/latex], this becomes<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{L\\frac{d^2q}{dt^2}+R\\frac{dq}{dt}+\\frac1Cq=E(t)}[\/latex].<\/p>\r\n<p id=\"fs-id1170571293694\">Mathematically, this system is analogous to the spring-mass systems we have been examining in this section.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n\r\n[caption id=\"attachment_3293\" align=\"aligncenter\" width=\"325\"]<img class=\"wp-image-3293 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18022941\/7-3-9.jpeg\" alt=\"This figure is a diagram of a circuit. It has broken lines at the bottom labeled C. On the left side there is an open circle labeled E. The top has diagonal lines labeled R. The right side has little bumps labeled L.\" width=\"325\" height=\"232\" \/> Figure 1. An RLC series circuit can be modeled by the same differential equation as a mass-spring system.[\/caption]\r\n\r\n<\/div>\r\n<div data-type=\"note\"><\/div>\r\n<div data-type=\"note\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: the <em>rlc<\/em> series circuit<\/h3>\r\nFind the charge on the capacitor in an\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0series circuit where [latex]L=5\/3[\/latex] H, [latex]R=10\\Omega[\/latex], [latex]C=1\/30[\/latex] F, and [latex]E(t)=300[\/latex] V. Assume the initial charge on the capacitor is\u00a0[latex]0[\/latex] [latex]C[\/latex] and the initial current is\u00a0[latex]9[\/latex] [latex]A[\/latex]. What happens to the charge on the capacitor over time?\r\n\r\n[reveal-answer q=\"729861165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"729861165\"]\r\n<p id=\"fs-id1170570995082\">We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nL\\frac{d^2q}{dt^2}+RL\\frac{dq}{dt}+\\frac1Cq&amp;=E(t) \\\\\r\n\\frac53\\frac{d^2q}{dt^2}+10\\frac{dq}{dt}+30q&amp;=300 \\\\\r\n\\frac{d^2q}{dt^2}+6\\frac{dq}{dt}+18q&amp;=180\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170571483434\">The general solution to the complementary equation is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{e^{-3t}(c_1\\cos(3t)+c_2\\sin(3t))}[\/latex].<\/p>\r\n<p id=\"fs-id1170573515927\">Assume a particular solution of the form [latex]q_p=A[\/latex], where [latex]A[\/latex] is a constant. Using the method of undetermined coefficients, we find [latex]A=10[\/latex]. So,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{q(t)=e^{-3t}(c_1\\cos(3t)+c_2\\sin(3t))+10}[\/latex].<\/p>\r\n<p id=\"fs-id1170571230479\">Applying the initial conditions [latex]q(0)=0[\/latex] and [latex]i(0)=((dq)\/(dt))(0)=9[\/latex], we find [latex]c_1=-10[\/latex] and [latex]c_2=-7[\/latex]. So the charge on the capacitor is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{q(t)=-10e^{-3t}\\cos(3t)-7e^{-3t}\\sin(3t)+10}[\/latex].<\/p>\r\n<p id=\"fs-id1170571171130\">Looking closely at this function, we see the first two terms will decay over time (as a result of the negative exponent in the exponential function). Therefore, the capacitor eventually approaches a steady-state charge of [latex]10 \\ C[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the charge on the capacitor in an\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0series circuit where [latex]L=1\/5 [\/latex] H, [latex]R=2\/5\\Omega[\/latex], [latex]C=1\/2[\/latex] F, and [latex]E(t)=50[\/latex] V. Assume the initial charge on the capacitor is 0 [latex]C[\/latex] and the initial current is [latex]4 \\ A[\/latex].\r\n\r\n[reveal-answer q=\"981734122\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"981734122\"]\r\n\r\n[latex]q(t)=-25e^{-t}\\cos(3t)-7e^{-t}\\sin(3t)+25[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250346&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=7FbR492dU2g&amp;video_target=tpm-plugin-9g88d5wl-7FbR492dU2g\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.21_transcript.html\">transcript for \u201cCP 7.21\u201d here (opens in new window).<\/a><\/center><\/div>","rendered":"<div data-type=\"note\">\n<div class=\"textbox learning-objectives\">\n<h3 style=\"text-align: center;\">Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Solve a second-order differential equation representing charge and current in an RLC series circuit.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170571071163\">Consider an electrical circuit containing a resistor, an inductor, and a capacitor, as shown in <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/simple-harmonic-motion\/\">Simple Harmonic Motion<\/a> Figure 9. Such a circuit is called an\u00a0<strong><span id=\"1f3e94ba-eee2-44b2-881a-5e5f8fe37136_term321\" data-type=\"term\"><em data-effect=\"italics\">RLC<\/em>\u00a0series circuit<\/span>.<\/strong>\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0circuits are used in many electronic systems, most notably as tuners in AM\/FM radios. The tuning knob varies the capacitance of the capacitor, which in turn tunes the radio. Such circuits can be modeled by second-order, constant-coefficient differential equations.<\/p>\n<p id=\"fs-id1170571448902\">Let [latex]I(t)[\/latex] denote the current in the\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0circuit and [latex]q(t)[\/latex] denote the charge on the capacitor. Furthermore, let [latex]L[\/latex] denote inductance in henrys ([latex]H[\/latex]), [latex]R[\/latex] denote resistance in ohms ([latex]\\Omega[\/latex]), and [latex]C[\/latex] denote capacitance in farads ([latex]F[\/latex]). Last, let [latex]E(t)[\/latex] denote electric potential in volts ([latex]V[\/latex]).<\/p>\n<p id=\"fs-id1170573581916\">Kirchhoff\u2019s voltage rule states that the sum of the voltage drops around any closed loop must be zero. So, we need to consider the voltage drops across the inductor (denoted [latex]E_L[\/latex]), the resistor (denoted [latex]E_R[\/latex]), and the capacitor (denoted [latex]E_C[\/latex]). Because the\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0circuit shown in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/simple-harmonic-motion\/\">Simple Harmonic Motion<\/a> Figure 9\u00a0includes a voltage source, [latex]E(t)[\/latex], which adds voltage to the circuit, we have [latex]E_L+E_R+E_C=E(t)[\/latex].<\/p>\n<p id=\"fs-id1170573515970\">We present the formulas below without further development. Those of you interested in the derivation of these formulas should consult a physics text. Using Faraday\u2019s law and Lenz\u2019s law, the voltage drop across an inductor can be shown to be proportional to the instantaneous rate of change of current, with proportionality constant [latex]L[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{E_L=L\\frac{dI}{dt}}[\/latex].<\/p>\n<p id=\"fs-id1170571131649\">Next, according to Ohm\u2019s law, the voltage drop across a resistor is proportional to the current passing through the resistor, with proportionality constant [latex]R[\/latex]. Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{E_R=RI}[\/latex].<\/p>\n<p id=\"fs-id1170571087077\">Last, the voltage drop across a capacitor is proportional to the charge, [latex]q[\/latex], on the capacitor, with proportionality constant [latex]1\/C[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{E_C=\\frac{1}{C}q}[\/latex].<\/p>\n<p id=\"fs-id1170571260221\">Adding these terms together, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\large{L\\frac{dI}{dt}+RI+\\frac1Cq=E(t)}[\/latex].<\/p>\n<p id=\"fs-id1170571042904\">Noting that [latex]I=(dq)\/(dt)[\/latex], this becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\large{L\\frac{d^2q}{dt^2}+R\\frac{dq}{dt}+\\frac1Cq=E(t)}[\/latex].<\/p>\n<p id=\"fs-id1170571293694\">Mathematically, this system is analogous to the spring-mass systems we have been examining in this section.<\/p>\n<\/div>\n<div data-type=\"note\">\n<div id=\"attachment_3293\" style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3293\" class=\"wp-image-3293 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18022941\/7-3-9.jpeg\" alt=\"This figure is a diagram of a circuit. It has broken lines at the bottom labeled C. On the left side there is an open circle labeled E. The top has diagonal lines labeled R. The right side has little bumps labeled L.\" width=\"325\" height=\"232\" \/><\/p>\n<p id=\"caption-attachment-3293\" class=\"wp-caption-text\">Figure 1. An RLC series circuit can be modeled by the same differential equation as a mass-spring system.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"note\"><\/div>\n<div data-type=\"note\">\n<div class=\"textbox exercises\">\n<h3>Example: the <em>rlc<\/em> series circuit<\/h3>\n<p>Find the charge on the capacitor in an\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0series circuit where [latex]L=5\/3[\/latex] H, [latex]R=10\\Omega[\/latex], [latex]C=1\/30[\/latex] F, and [latex]E(t)=300[\/latex] V. Assume the initial charge on the capacitor is\u00a0[latex]0[\/latex] [latex]C[\/latex] and the initial current is\u00a0[latex]9[\/latex] [latex]A[\/latex]. What happens to the charge on the capacitor over time?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q729861165\">Show Solution<\/span><\/p>\n<div id=\"q729861165\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170570995082\">We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  L\\frac{d^2q}{dt^2}+RL\\frac{dq}{dt}+\\frac1Cq&=E(t) \\\\  \\frac53\\frac{d^2q}{dt^2}+10\\frac{dq}{dt}+30q&=300 \\\\  \\frac{d^2q}{dt^2}+6\\frac{dq}{dt}+18q&=180  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170571483434\">The general solution to the complementary equation is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{e^{-3t}(c_1\\cos(3t)+c_2\\sin(3t))}[\/latex].<\/p>\n<p id=\"fs-id1170573515927\">Assume a particular solution of the form [latex]q_p=A[\/latex], where [latex]A[\/latex] is a constant. Using the method of undetermined coefficients, we find [latex]A=10[\/latex]. So,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{q(t)=e^{-3t}(c_1\\cos(3t)+c_2\\sin(3t))+10}[\/latex].<\/p>\n<p id=\"fs-id1170571230479\">Applying the initial conditions [latex]q(0)=0[\/latex] and [latex]i(0)=((dq)\/(dt))(0)=9[\/latex], we find [latex]c_1=-10[\/latex] and [latex]c_2=-7[\/latex]. So the charge on the capacitor is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{q(t)=-10e^{-3t}\\cos(3t)-7e^{-3t}\\sin(3t)+10}[\/latex].<\/p>\n<p id=\"fs-id1170571171130\">Looking closely at this function, we see the first two terms will decay over time (as a result of the negative exponent in the exponential function). Therefore, the capacitor eventually approaches a steady-state charge of [latex]10 \\ C[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the charge on the capacitor in an\u00a0<em data-effect=\"italics\">RLC<\/em>\u00a0series circuit where [latex]L=1\/5[\/latex] H, [latex]R=2\/5\\Omega[\/latex], [latex]C=1\/2[\/latex] F, and [latex]E(t)=50[\/latex] V. Assume the initial charge on the capacitor is 0 [latex]C[\/latex] and the initial current is [latex]4 \\ A[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q981734122\">Show Solution<\/span><\/p>\n<div id=\"q981734122\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]q(t)=-25e^{-t}\\cos(3t)-7e^{-t}\\sin(3t)+25[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250346&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=7FbR492dU2g&amp;video_target=tpm-plugin-9g88d5wl-7FbR492dU2g\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.21_transcript.html\">transcript for \u201cCP 7.21\u201d here (opens in new window).<\/a><\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1332\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 7.21. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":428269,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 7.21\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1332","chapter","type-chapter","status-publish","hentry"],"part":25,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1332","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/428269"}],"version-history":[{"count":36,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1332\/revisions"}],"predecessor-version":[{"id":6188,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1332\/revisions\/6188"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/25"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1332\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1332"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1332"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1332"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1332"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}