{"id":1578,"date":"2025-10-17T17:21:25","date_gmt":"2025-10-17T17:21:25","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1578"},"modified":"2025-11-11T17:40:08","modified_gmt":"2025-11-11T17:40:08","slug":"applications-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/applications-learn-it-4\/","title":{"raw":"Applications: Learn It 4","rendered":"Applications: Learn It 4"},"content":{"raw":"

The\u00a0RLC\u00a0Series Circuit<\/h2>\r\nConsider an electrical circuit containing a resistor, an inductor, and a capacitor, as shown in Figure 1. Such a circuit is called an RLC series circuit<\/strong>. RLC circuits 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.\r\n\r\n[caption id=\"attachment_3293\" align=\"aligncenter\" width=\"325\"]\"This Figure 1. An RLC series circuit can be modeled by the same differential equation as a mass-spring system.[\/caption]\r\n\r\nSuch circuits can be modeled by second-order, constant-coefficient differential equations.\r\n\r\nLet [latex]I(t)[\/latex] denote the current in the RLC circuit 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]).\r\n
\r\n

Kirchhoff's Voltage Rule<\/h3>\r\n

Kirchhoff's 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 RLC circuit includes 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

Using Faraday's law and Lenz's 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

[latex]E_L=L\\frac{dI}{dt}[\/latex].<\/p>\r\n

Next, according to Ohm's 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

[latex]E_R=RI[\/latex].<\/p>\r\n

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

[latex]E_C=\\frac{1}{C}q[\/latex].<\/p>\r\n

Adding these terms together, we get:<\/p>\r\n

[latex]L\\frac{dI}{dt}+RI+\\frac1Cq=E(t)[\/latex].<\/p>\r\n

Noting that [latex]I=(dq)\/(dt)[\/latex], this becomes:<\/p>\r\n

[latex]L\\frac{d^2q}{dt^2}+R\\frac{dq}{dt}+\\frac1Cq=E(t)[\/latex].<\/p>\r\n

Mathematically, this system is analogous to the spring-mass systems we have been examining in this section.<\/p>\r\n\r\n

Notice the similarity between the RLC circuit equation and the spring-mass system equation:\r\n
    \r\n \t
  • Spring-mass<\/strong>: [latex]mx^{\\prime\\prime}+bx^\\prime+kx=f(t)[\/latex]<\/li>\r\n \t
  • RLC circuit<\/strong>: [latex]L\\frac{d^2q}{dt^2}+R\\frac{dq}{dt}+\\frac{1}{C}q=E(t)[\/latex]<\/li>\r\n<\/ul>\r\n

    The inductance [latex]L[\/latex] corresponds to mass, resistance [latex]R[\/latex] corresponds to damping, and the reciprocal of capacitance [latex]1\/C[\/latex] corresponds to the spring constant.<\/p>\r\n\r\n<\/section><\/div>\r\n

    Find the charge on the capacitor in an\u00a0RLC<\/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?[reveal-answer q=\"729861165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"729861165\"]\r\n

    We have<\/p>\r\n

    [latex]\\begin{aligned}\r\nL\\frac{d^2q}{dt^2}+RL\\frac{dq}{dt}+\\frac1Cq&=E(t) \\\\\r\n\\frac53\\frac{d^2q}{dt^2}+10\\frac{dq}{dt}+30q&=300 \\\\\r\n\\frac{d^2q}{dt^2}+6\\frac{dq}{dt}+18q&=180\r\n\\end{aligned}[\/latex].<\/p>\r\n

    The general solution to the complementary equation is<\/p>\r\n

    [latex]\\large{e^{-3t}(c_1\\cos(3t)+c_2\\sin(3t))}[\/latex].<\/p>\r\n

    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

    [latex]\\large{q(t)=e^{-3t}(c_1\\cos(3t)+c_2\\sin(3t))+10}[\/latex].<\/p>\r\n

    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

    [latex]\\large{q(t)=-10e^{-3t}\\cos(3t)-7e^{-3t}\\sin(3t)+10}[\/latex].<\/p>\r\n

    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<\/section><\/div>","rendered":"

    The\u00a0RLC\u00a0Series Circuit<\/h2>\n

    Consider an electrical circuit containing a resistor, an inductor, and a capacitor, as shown in Figure 1. Such a circuit is called an RLC series circuit<\/strong>. RLC circuits 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.<\/p>\n

    \"This
    Figure 1. An RLC series circuit can be modeled by the same differential equation as a mass-spring system.<\/figcaption><\/figure>\n

    Such circuits can be modeled by second-order, constant-coefficient differential equations.<\/p>\n

    Let [latex]I(t)[\/latex] denote the current in the RLC circuit 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

    \n

    Kirchhoff’s Voltage Rule<\/h3>\n

    Kirchhoff’s 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 RLC circuit includes 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

    Using Faraday’s law and Lenz’s 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

    [latex]E_L=L\\frac{dI}{dt}[\/latex].<\/p>\n

    Next, according to Ohm’s 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

    [latex]E_R=RI[\/latex].<\/p>\n

    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

    [latex]E_C=\\frac{1}{C}q[\/latex].<\/p>\n

    Adding these terms together, we get:<\/p>\n

    [latex]L\\frac{dI}{dt}+RI+\\frac1Cq=E(t)[\/latex].<\/p>\n

    Noting that [latex]I=(dq)\/(dt)[\/latex], this becomes:<\/p>\n

    [latex]L\\frac{d^2q}{dt^2}+R\\frac{dq}{dt}+\\frac1Cq=E(t)[\/latex].<\/p>\n

    Mathematically, this system is analogous to the spring-mass systems we have been examining in this section.<\/p>\n

    Notice the similarity between the RLC circuit equation and the spring-mass system equation:<\/p>\n
      \n
    • Spring-mass<\/strong>: [latex]mx^{\\prime\\prime}+bx^\\prime+kx=f(t)[\/latex]<\/li>\n
    • RLC circuit<\/strong>: [latex]L\\frac{d^2q}{dt^2}+R\\frac{dq}{dt}+\\frac{1}{C}q=E(t)[\/latex]<\/li>\n<\/ul>\n

      The inductance [latex]L[\/latex] corresponds to mass, resistance [latex]R[\/latex] corresponds to damping, and the reciprocal of capacitance [latex]1\/C[\/latex] corresponds to the spring constant.<\/p>\n<\/section>\n<\/div>\n

      \n
      Find the charge on the capacitor in an\u00a0RLC<\/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