Second Order Differential Equations
3.9 Application: RLC Electrical Circuits
In Section 2.5F, we explored first-order differential equations for electrical circuits consisting of a voltage source with either a resistor and inductor (RL) or a resistor and capacitor (RC). Now, equipped with the knowledge of solving second-order differential equations, we are ready to delve into the analysis of more complex RLC circuits, which incorporate a resistor, inductor, and capacitor.
Previously, we established that:
- Ohm’s law dictates that the voltage drop [asciimath]E_R[/asciimath] across a resistor is proportional to the current I flowing through it, expressed as [asciimath]E_R=RI[/asciimath], where [asciimath]R[/asciimath] is the resistance.
- Faraday’s law, complemented by Lenz’s law, describes that the voltage drop [asciimath]E_L[/asciimath] across an inductor is proportional to the rate of change of current, given as [asciimath]E_L=L(dI)/(dt)[/asciimath], where [asciimath]L[/asciimath] is the inductance.
- The voltage drop [asciimath]E_C[/asciimath] across a capacitor is proportional to the electric charge [asciimath]q[/asciimath] stored on it, represented as [asciimath]E_C=1/Cq[/asciimath], with [asciimath]C[/asciimath] being the capacitance
Figure 3.9.1 Schematic of an RLC series circuit
With these foundations, consider [asciimath]E(t)[/asciimath] as the external voltage supplied to the RLC series circuit in Fig. 3.9.1. By applying Kirchhoff’s voltage law, we have
[asciimath]E_L+E_R+E_c=E(t)[/asciimath]
Substituting [asciimath]E_R=RI[/asciimath] ,[asciimath]E_L=L (dI)/dt[/asciimath] , and [asciimath]E_C=1/Cq[/asciimath] into this equation yields
[asciimath]L (dI)/dt+RI+1/Cq=E(t)[/asciimath](3.9.1)
Differentiating this equation with respect to time and substituting [asciimath]I=(dq)/dt[/asciimath] transform it into a second-order differential equation.
[asciimath]L(d^2I)/(dt^2)+R (dI)/dt+1/CI=(dE)/dt[/asciimath](3.9.2)
Alternatively, Equation 3.9.1 can be expressed in terms of charge [asciimath]q(t)[/asciimath].
[asciimath]L (d^2q)/(dt^2)+R(dq)/(dt)+1/Cq=E(t)[/asciimath](3.9.3)
Given [asciimath]E(t)[/asciimath] and an initial condition, such as initial current[asciimath]I(0)[/asciimath] and initial charge [asciimath]q(0)[/asciimath], we can solve the equation for [asciimath]I(t)[/asciimath] using techniques discussed in previous sections, such as the method of undetermined coefficients. Once [asciimath]I(t)[/asciimath] is determined, the voltage across different components of the circuit can be calculated.
Consider an RLC series circuit with a resistor of [asciimath]0.06\ Omega[/asciimath] and an inductor of [asciimath]0.01\ "H"[/asciimath], and a capacitor of [asciimath]50/89\ "F"[/asciimath] powered by a voltage source [asciimath]E(t)=0.1sin(10t)\ "V"[/asciimath]. Initially, the current and charge on the capacitor are zero. Determine the current in the circuit as a function of time.
Show/Hide Solution
Given information:
- Resistor: [asciimath]R=0.06\ Omega[/asciimath]
- Inductor: [asciimath]L=0.01 \ "H"[/asciimath]
- Capacitor: [asciimath]C=50/89\ "F"[/asciimath]
- Voltage source: [asciimath]E(t)=0.1sin(10t)\ "V"[/asciimath]
- Initial current on capacitor: [asciimath]I(0)=0 \ "A"[/asciimath]
- Initial charge on capacitor: [asciimath]q(0)=I'(0)=0 \ C[/asciimath]
The differential equation for an RLC series circuit is given by Equation 3.9.1.
[asciimath]L(d^2I)/(dt^2)+R (dI)/dt+1/CI=(dE)/dt[/asciimath]
The initial value problem is then
[asciimath]0.01(d^2I)/(dt^2)+0.06 (dI)/dt+89/50I=cos(10t),[/asciimath] [asciimath]I(0)=0, \ I'(0)=0[/asciimath]
Multiplying the equation by 100, we get
[asciimath](d^2I)/(dt^2)+6 (dI)/dt+178I=100cos(10t),[/asciimath] [asciimath]I(0)=0, \ I'(0)=0[/asciimath]
Given the characteristic equation has complex conjugates [asciimath]r_(1,2)=-3+-13i[/asciimath], the complementary solution is
[asciimath]I_c(t)=e^(-3t)(c_1cos(13t)+c_2sin(13t))[/asciimath]
Finding the particular solution:
To find the particular solution, we use undetermined coefficients. Given the forcing cosine function, we guess the form of the particular solution to be
[asciimath]I_p=Acos(10t)+Bsin(10t)[/asciimath]
The derivatives are
[asciimath]I'_p=-10Asin(10t)+10Bcos(10t)[/asciimath]
[asciimath]I''_p=-100Acos(10t)-100Bsin(10t)[/asciimath]
Substituting [asciimath]I_p[/asciimath] and its derivatives into the differential equation yields
[asciimath]-100Acos(10t)-100Bsin(10t) +6(-10Asin(10t)+10Bcos(10t) )[/asciimath] [asciimath]+178(Acos(10t)+Bsin(10t))=[/asciimath] [asciimath]100cos(10t)[/asciimath]
Simplifying gives
[asciimath](78A+60B)cos(10t)+(-60A+78B)sin(10t)=100cos(10t)[/asciimath]
By matching coefficients of sine and cosine terms and solving the system of two equations in unknowns [asciimath]A[/asciimath] and [asciimath]B[/asciimath], we get
[asciimath]A=650/807,\ B=500/807[/asciimath]
Therefore, the particular solution is
[asciimath]I_p=650/807cos(10t)+500/807sin(10t)[/asciimath]
Combining the particular and complementary solutions gives the general solution
[asciimath]I(t)=650/807cos(10t)+500/807sin(10t) +[/asciimath] [asciimath]e^(-3t)(c_1cos(13t)+c_2sin(13t))[/asciimath]
Applying the initial conditions:
[asciimath]I_0(0)=0 \ ->[/asciimath] [asciimath]c_1=-650/807[/asciimath]
[asciimath]I'_0(0)=0\ ->[/asciimath] [asciimath]c_2=-6950/10491[/asciimath]
The equation of the object’s displacement is then
[asciimath]I(t)=650/807cos(10t)+500/807sin(10t) +[/asciimath] [asciimath]e^(-3t)(-650/807 cos(13t)-6950/10491 sin(13t))[/asciimath]
As with forced mechanical vibration scenarios, the current in an RLC circuit is composed of two distinct parts: the transient current, represented by the complementary solution that diminishes to zero as time progresses to infinity, and the steady-state current, described by the particular solution which is sinusoidal and persists over time.
Try an Example
Section 3.9 Exercises
- Consider an RLC circuit with a [asciimath]17/50\Omega[/asciimath] resistor, a [asciimath]1/100\H[/asciimath] inductor, and a [asciimath]100/93\F[/asciimath] capacitor driven by the voltage [asciimath]E(t) = 0.06sin(3t)V[/asciimath]. a) Write the differential equation associated with this circuit in terms of current [asciimath]I[/asciimath] . b) If the initial charge and initial current on the capacitor are both zero, find the current [asciimath]I[/asciimath] and the voltages across the resistor [asciimath]E_R[/asciimath] in terms of time [asciimath]t[/asciimath].
Show/Hide Answer
a) [asciimath]I''+34I'+93I = 18cos(3t)[/asciimath]
b) [asciimath]I(t)=0.0205 e^(-31t)-0.1071 e^(-3t)+[/asciimath] [asciimath]0.1052sin(3t)+0.08660cos(3t)[/asciimath]
c) [asciimath]E_R(t)=0.34(0.0205 e^(-31t)-0.1071 e^(-3t)+[/asciimath] [asciimath]0.1052sin(3t)+0.08660cos(3t) )[/asciimath]
- Consider an RLC circuit with a [asciimath]1/50\Omega[/asciimath] resistor, a [asciimath]1/100\H[/asciimath] inductor, and a [asciimath]50/61\F[/asciimath] capacitor driven by the voltage [asciimath]E(t) = 0.09t^2\ V[/asciimath]. a) Write the differential equation associated with this circuit in terms of current [asciimath]I[/asciimath]. b) If the initial charge and initial current on the capacitor are both zero, find the current [asciimath]I[/asciimath].
Show/Hide Answer
a) [asciimath]I''+2I'+122I = 18t[/asciimath]
b) [asciimath]I(t)=e^(-t)(9/3721cos(11t)-540/40931sin(11t))+9/61t-9/3721[/asciimath]
