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. Kirchhoffs voltage law satates that, the algebraic sum of all voltage
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 dI
dt for I and dividing by L, we
obtain our usual form of a first-order nonhomogeneous equation:
dI R 1
+ 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 henry L
Capacitance farad 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.
R R
R
dt
Solution. The integrating factor is (t) = e L = e L t , so we have
Z
1 Rt R
I(t) = e L V (t)e L t dt.
L
In particular, if V (t) = V0 is some constant voltage, then we may evaluate the integral on
the right directly to get
Z
1 Rt R 1 Rt L R V0 R
I(t) = e L L t
V0 e dt = e L t
V0 e + c =
L + 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
1 5 5 1 5 32 5 t 8 5
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
8 5
I(t) = (1 e 4 t ).
5
8
Then I(0.1) 0.188 and I(t) 5 as t .
3. Assume that the voltage V (t) is given by V0 sin(t), where V0 and are given constants.
Find the solution of the differential equation above subject to the initial condition I(0) = I0 .
Solution. Ill 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,
R R
R
Ih (t) = e L dt
= ke L t .
Then, to find the particular solution Ip (t), we try Ip (t) = cos(t) + sin(t). Substituting
this into the DE, multiplying through by L to get rid of the fractions, and combining like
terms, we get
(L + R) cos(t) + (L + R) sin(t) = V0 sin(t).
Thus we have the system
R + L = 0
L + R = V0 .
The first equation immediately gives = LR1 . Then, plugging this into the second
equation gives
L
L + R = V0
R
2 2
L
+ R = V0
R
V0 V0 R
= 2 L2
= 2 2 .
R +R L + R2
Then
V0 R L V0 L
= = 2 2 .
L2 +
2 R2 R L + R2
Hence
R V0 L V0 R
I(t) = ke L t cos(t) + sin(t).
L2 + R2
2 L2 +
2 R2
It remains to solve for k using the initial condition I(0) = I0 . We have
V0 L V0 L
I(0) = I0 = k = k = I0 + 2 2 .
2 L2 + R2 L + R2
Thus
V0 L R V0 L V0 R
I(t) = Ih (t) + Ip (t) = I0 + e L t cos(t) + sin(t).
L2 + R2
2 L2 + R2
2 L2 +
2 R2
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 L = 3 3 = 9, V0 R = 3 6 = 18, and
2 L2 + R2 = 12 32 + 62 = 45, so
459 2t 1 2
I(t) = 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.
