RL-series circuits Math 2410 Spring 2011 Consider the RL-series circuit shown in the figure below, which contains a counterclockwise current I = I(t), a resistance R, and inductance L, and a generator that supplies a voltage V (t) when the switch is closed. Kirchhoff’s voltage law satates that, “the algebraic sum of all voltage R V (t) I = I(t) L drops around an electric circuit is zero.” Applied to this RL-series circuit, the statement translates to the fact that the current I = I(t) in the circuit satisfies the first-order linear differential equation LI˙ + RI = V (t), where I˙ denotes the derivative of I with respect to t. Substituting obtain our usual form of a first-order nonhomogeneous equation: dI dt for I˙ and dividing by L, we R 1 dI + I = V (t). dt L L For convenience, we present an index of units and symbols commonly used in this theory. Quantity Unit Symbol Resistance ohm R Inductance Capacitance henry farad L C Voltage volt V Current ampere I Charge coulomb Q Time seconds sec Table 1: Table of Units Battery Resistor Inductor Capacitor Switch Table 2: Symbols of Circuit Elements Exercises 1. Use the method of integration factors to calculate what the general solution is for this differential equation. Solution. The integrating factor is µ(t) = e I(t) = R R L 1 −Rt e L L dt Z R = e L t , so we have R V (t)e L t dt. In particular, if V (t) = V0 is some constant voltage, then we may evaluate the integral on the right directly to get Z R R R 1 −Rt 1 −Rt L V0 t t L L L L I(t) = e V0 e dt = e V0 e + c = + ce− L t . L L R R 2. In an RL-series circuit, L = 4 henries, R = 5 ohms, V = 8 volts, and I(0) = 0 amperes. Find the current at the end of 0.1 seconds. What will the current be after a very long time? Solution. Using the formula from the last solution, Z 5 5 1 5 1 5 32 5 t 8 I(t) = e− 4 t 8e 4 t dt = e− 4 t e 4 + c = + ce− 4 t . 4 4 5 5 Using the initial condition I(0) = 0, we get c = − 85 , so I(t) = Then I(0.1) ≈ 0.188 and I(t) → 8 5 5 8 (1 − e− 4 t ). 5 as t → ∞. 3. Assume that the voltage V (t) is given by V0 sin(ωt), where V0 and ω are given constants. Find the solution of the differential equation above subject to the initial condition I(0) = I0 . Solution. I’ll give the argument using the method of undetermined coefficient here, as it is different from the method done in class and it represents something new that you learned. So a solution to this DE is of the form yh (t) + yp (t). First, Ih (t) = e− R R L dt R = ke− L t . Then, to find the particular solution Ip (t), we try Ip (t) = α cos(ωt) + β sin(ωt). Substituting this into the DE, multiplying through by L to get rid of the fractions, and combining like terms, we get (βωL + Rα) cos(ωt) + (−αωL + Rβ) sin(ωt) = V0 sin(ωt). Thus we have the system Rα + βωL = 0 −αωL + Rβ = V0 . The first equation immediately gives α = −βωLR−1 . Then, plugging this into the second equation gives βωL ωL + Rβ = V0 R 2 2 ω L + Rβ = V0 β R β= Then α=− V0 R 2 ω L2 + Hence R I(t) = ke− L t − V0 V0 R = 2 2 . ω L + R2 +R ω 2 L2 R R2 · V0 ωL 2 ω L2 + R2 ωL V0 ωL =− 2 2 . R ω L + R2 cos(ωt) + V0 R 2 ω L2 + R2 sin(ωt). It remains to solve for k using the initial condition I(0) = I0 . We have I(0) = I0 = k − V0 ωL V0 ωL =⇒ k = I0 + 2 2 . ω 2 L2 + R2 ω L + R2 Thus I(t) = Ih (t) + Ip (t) = I0 + V0 ωL 2 ω L2 + R2 R e− L t − V0 ωL 2 ω L2 + R2 cos(ωt) + V0 R 2 ω L2 + R2 sin(ωt). 4. Given that L = 3 henries, R = 6 ohms, V (t) = 3 sin(t), and I(0) = 10 amperes, compute the value of the current at any time t. Solution. Using the equation from above, we have V0 ωL = 3 · 3 = 9, V0 R = 3 · 6 = 18, and ω 2 L2 + R2 = 12 32 + 62 = 45, so I(t) = 459 −2t 1 2 e − cos(t) + sin(t). 45 5 5 References [1] N. Finizio and G. Ladas, Ordinary Differential Equations with Modern Applications, 3rd edition, Simon & Schuster Custom Publishing, Boston, 1999.