Learning Objectives
- Solve a second-order differential equation representing charge and current in an RLC series circuit.
Consider an electrical circuit containing a resistor, an inductor, and a capacitor, as shown in Simple Harmonic Motion Figure 9. Such a circuit is called an RLC series circuit. 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. Such circuits can be modeled by second-order, constant-coefficient differential equations.
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]).
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 shown in Simple Harmonic Motion Figure 9 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].
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’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,
[latex]\large{E_L=L\frac{dI}{dt}}[/latex].
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,
[latex]\large{E_R=RI}[/latex].
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,
[latex]\large{E_C=\frac{1}{C}q}[/latex].
Adding these terms together, we get
[latex]\large{L\frac{dI}{dt}+RI+\frac1Cq=E(t)}[/latex].
Noting that [latex]I=(dq)/(dt)[/latex], this becomes
[latex]\large{L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac1Cq=E(t)}[/latex].
Mathematically, this system is analogous to the spring-mass systems we have been examining in this section.
Figure 1. An RLC series circuit can be modeled by the same differential equation as a mass-spring system.
Example: the rlc series circuit
Find the charge on the capacitor in an RLC series 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 [latex]0[/latex] [latex]C[/latex] and the initial current is [latex]9[/latex] [latex]A[/latex]. What happens to the charge on the capacitor over time?
try it
Find the charge on the capacitor in an RLC series 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].
Watch the following video to see the worked solution to the above Try It
