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In this lab several passive elements of electrical circuits will be examined and their relationship to the field of differential equations explained. The elements of interest are the resistor, capacitor, and inductor . Resistors are a fundamental electrical unit and some form of them are found in almost every electrical circuit; they primarily serve the purpose of providing a desired voltage drop from some source voltage. Capacitors and inductors are designed to store energy in different forms, through an electric field and magnetic field respectively. They see wide use in frequency filter circuits, and specifically for inductors, in AC voltage transformers. The voltage drop across each component is known to be as follows:
Circuit Element Voltage Drop
Resistor RI(t)
Capacitor (1/C)Q(t)
Inductor LI'(t)
Table 1. Derivatives are with respect to time.
The values R,C, and L in Table 1 are component values of resistors, capacitors, and inductors respectively. It is also known that a current flows through a conductor when there is a difference in charge between two points. That is, a current function can be expressed as the rate of change of charge or
I(t)=Q^' (t) (1)
Kirchoff’s Voltage Law states, “The sum of the voltage drops across the elements in a simple loop of an electrical circuit is equal to the applied voltage.” The simple circuit mentioned in Kirchoff’s law will be shown as
Figure 1. A simple series RLC circuit.
Using the voltage drops from Table 1 we can build the equation
LI^' (t)+RI(t)+1/C Q(t)=E(t) (2)
Inserting the relationship from Eq. (1) into Eq. (2) yields
L〖Q'〗^' (t)+RQ'(t)+1/C Q(t)=E(t) (3)
By differentiating each function with respect to time and again using Eq. (1) we find
L〖I'〗^' (t)+RI'(t)+1/C I(t)=E'(t) (4)
Equations (3) and (4) are now representative of linear second order non-homogonous ordinary differential equations. As will be shown in later examples, the behavior of the circuit can be modeled using the characteristic equation for the homogonous part and the method of undetermined coefficients for the non-homogonous part.
Since the series RLC circuit can be expressed as a linear second order non-homogonous ordinary differential equation it useful for modeling other natural systems that are in the same form. Most notably, mechanical systems dependant on the oscillations of components can be modeled by cheap electrical components where it is not feasible to perform tests of the mechanical system design with the mechanical components themselves. This relationship between mechanical and electrical systems is an indicator of how important mathematical modeling can be to applied fields such as engineering.
Using the principle of superposition the solution to the current function can be represented by
I(t)=I_tr (t)+I_sp (t) (5)
Itr is the transient current of the circuit and Isp is the steady periodic current. Assuming all coefficients are positive (they are in physical circuits) the transient current will approach zero as time approaches infinity. When this happens the current will most resemble the steady periodic current function only. In general form, the transient current is the function formed from the homogonous solution to the ODE while the steady periodic current is found from the solution to the non-homogonous component.
The following six examples will examine the properties of circuits containing only a voltage source, resistor, and inductor. They will use the following circuit with voltage loop equation LI’+RI=E(t).
Figure 2. Series RL circuit
Example 1
L= 5 H, R= 25 Ω, E(t) = 100 V Assume switch 1 has been closed for a long time such that a steady 4 Amps flows through the circuit. At t = 0 seconds switch 1 opens and switch 2 closes simultaneously so I(0) = 4 A and E = 0 for t ≥ 0.
5I^'+25I=0
This is a first order ODE with characteristic equation
5r+25=0
This yields the root r = -5.
The general form of the solution is
I(t)=c_1 e^(r_1 t)
Plugging in our root of -5 we find
I(t)=c_1 e^(-5t)
Given our initial condition I(0) = 4 A we can find the particular solution
I(0)=4=c_1
Finally the current in this circuit can be expressed as
I(t)=4e^(-5t)
Example 2
Use same values for R and L as Example 1. Assume switch 1 and 2 are open. At t = 0 switch 1 closes. I(0) = 0 and E(t) = 100 for t ≥ 0. Find I(t) and show I(t) = 4 A as t → +∞.
This time the voltage loop equation is
5I^'+25I=100
This non-homogonous ODE will have general solution
I(t)=I_tr (t)+I_sp (t)
From the previous example we know the homogonous solution, or transient current, is
I_tr (t)=-4e^(-5t)
We can find the steady periodic current using the method of undetermined coefficients. We guess
I_sp=A
〖I'〗_sp=0
Plugging these into our voltage loop equation gives us
25A=100
A=4
Finally, the current can be expressed as
I(t)=-4e^(-5t)+4
As t → +∞ the transient solution will approach zero. The solution will resemble the steady periodic current which is equal to the constant 4. This agrees with the idea in circuit brown townysis that as t → +∞ that inductors will start acting like short circuits.
Example 3
Replace the direct current voltage source used in Example 2 with an alternation current source with voltage E(t) = 100cos(60t). Find I(t).
Voltage loop equation
5I^'+25I=100cos(60t)
I_tr (t)=c_1 e^(-5t)
I_sp=Acos(60t)+Bsin(60t)
〖I'〗_sp=-60Asin(60t)+60Bcos(60t)
Plugging in our guesses for Isp into the voltage loop equation yields
-300Asin(60t)+300Bcos(60t)+25Bsin(60t)+25Acos(60t)=100cos(60t)
Coefficients A and B can be found by splitting the above equation into a system and solving.
(-300A+25B) sin(60t)=0sin(60t)
(25A+300B) cos(60t)=100cos(60t)
A = 4/145 B = 48/145
We now plug our found values of A and B into Isp
I_sp=4/145 cos(60t)+48/145 sin(60t)
Remember the general solution to the ODE is
I(t)=I_tr (t)+I_sp (t)
I(t)=c_1 e^(-5t)+4/145 cos(60t)+48/145 sin(60t)
We are given initial condition I(0) = 0 so
I(0)=0=c_1+4/145
c_1=-4/145
I(t)=(-4)/145 e^(-5t)+4/145 cos(60t)+48/145 sin(60t)
Example 4
L = 2, R = 40, E(t) = 100e-10t, I(0) = 0. Find the maximum current in the circuit for t ≥ 0 with switch 1 closed and switch 2 open.
First we will want to find the particular solution for I(t) with our given initial condition. This is straight forward as with the previous examples.
Voltage loop equation
2I^'+40I=100e^(-10t)
Character equation of homogonous solution and roots
2r+40=0,r=-20
Transient current
I_tr (t)=〖c_1 e〗^(-20t)
Steady periodic current
I_sp=Ae^(-10t)
〖I'〗_sp=-10Ae^(-10t)
-20Ae^(-10t)+40Ae^(-10t)=100e^(-10t)
20Ae^(-10t)=100e^(-10t)
A=5
I_sp=5e^(-10t)
General solution
I(t)=〖c_1 e〗^(-20t)+5e^(-10t)
Particular solution
I(0)=0=c_1+5,c_1=-5
I(t)=〖-5e〗^(-20t)+5e^(-10t)
To find the maximum current in this circuit we must recognize that our particular solution is bounded by two equilibrium solutions. To find them we first need to find the derivative of our current function and find at what values of time it is equal to zero.
I^' (t)=100e^(-20t)-50e^(-10t)
Equilibrium solutions at times where I’(t) = 0. From the above equation we can find one equilibrium solution by noticing as t → +∞ the function will approach zero.
100e^(-20t)-50e^(-10t)=0
2e^(-20t)=e^(-10t)
2=e^10t
10t=ln2
t=.0593 s
Given that zero is our lower limit we can plug this value of t into our I(t) to find the maximum current that will flow through the circuit.
I_max=I(.0593)=〖c_1 e〗^(-20(.0593))+5e^(-10(.0593))
I_max=1.25 A
Example 5
Switch 1 is closed and switch 2 opened. E(t) = 100e-10tcos(60t), R = 20, L = 2, I(0) = 0. Find I(t).
Voltage loop equation
2I^'+20I=100e^(-10t) cos(60t)
Characteristic equation with roots
2r+20=0,r=-10
Transient current
I_tr (t)=〖c_1 e〗^(-10t)
Steady periodic current guess
I_sp=e^(-10t) A cos〖(60t)+e^(-10t) B sin〖(60t)〗 〗
〖I'〗_sp=-10e^(-10t) Acos〖(60t)-10e^(-10t) B sin〖(60t)-60e^(-10t) A sin〖(60t)+60e^(-10t) B cos〖(60t)〗 〗 〗 〗
2〖I'〗_sp+20I_sp=100e^(-10t) cos(60t)
The above equation simplifies to
120Be^(-10t) cos〖(60t)-120Ae^(-10t) sin〖(60t)=100e^(-10t) cos(60t) 〗 〗
120B=100
-120A=0
A = 0 B = 5/6
I_sp=5/6 e^(-10t) sin〖(60t)〗
General solution
I(t)=〖c_1 e〗^(-10t)+5/6 e^(-10t) sin〖(60t)〗
Particular solution
I(0)=0=c_1
I(t)=5/6 e^(-10t) sin〖(60t)〗
Example 6
L = 1, R = 10, E(t) = 30cos(60t)+40sin(60t). Switch 1 is closed and switch 2 is open. Find the steady periodic and express it in the form Ccos(ωt-α).
We are only asked to find the steady periodic current so we do not need to utilize the characteristic equation. The transient current typically dies out quickly (meaning it only affects and is affected by initial behavior) so the steady periodic current is of greater interest for long term circuit behavior.
Voltage loop equation
I^'+10I=30cos〖(60t)+40 sin〖(60t)〗 〗
We are told the function to use for our Isp guess
I_sp=A cos〖(60t)+B sin(60t) 〗
〖I'〗_sp=-60A sin〖(60t)+60B cos〖(60t)〗 〗
〖I'〗_sp+10I_sp=30 cos〖(60t)+40 sin〖(60t)〗 〗
-60A sin〖(60t)+60B cos〖(60t)〗 〗+10A cos〖(60t)+10B sin〖(60t)=30 cos〖(60t)+40 sin(60t) 〗 〗 〗
-60A+10B=40
10A+60B=30
A=(-21)/37,B=22/37
I_sp=(-21)/37 cos〖(60t)+22/37 sin(60t) 〗
Since both trigonometric functions share the same frequency they can be easily combined into a single function. First we will determine the phase shift α.
α=tan^(-1)〖(B/A)=tan^(-1)〖(22/(-21))〗 〗=-0.8086
Since the coefficient in front of our sine function is positive and negative in front of the cosine function our phase shift angle is in the second quadrant, so α=π-0.8086=2.33 radians.
I_sp=(5√37)/37 〖(cos〗〖60t-2.33)=〗 5/√37 〖(cos〗〖60t-2.33) A〗
The next four examples will be based off a circuit with a capacitor in series with a voltage source and resistor. The voltage loop for the following circuit can be shown as RQ’ + (1/C)Q = E(t).
Example 7
