Second Order Circuit Analysis - Mesh - Issue with DE Setup

In summary, vela was trying to solve for the current through an inductor but made a mistake in the second equation. She solved for the current through a capacitor instead. Her mesh equations were correct, but her initial conditions were not.
  • #1
ghoti
5
0
Hi,

I am sure I have mistaken my DE setup or my initial conditions assumption.

Homework Statement


[URL]http://ivila.net/E8.png[/URL]
R1 will be the middle Resistor
R2 the top one.

Homework Equations


i1' + R1/L i1 = R1/L i2
i2(R1+R2)*1/(R1) + 1/(R1*C)*integral(i2)=i1

The Attempt at a Solution


My initial conditions are i1(0) = 1A
Vl = L di1/dt
at (0+) VL = 0 therefore L di1/dt = 0 therefore di1/dt = 0
i1'(0) = 0
Substitute for i1 I end up with sqrt(2)exp(-2)cos(t+45')
 
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  • #2
Your first initial condition, i1(0) = 1A, is fine, but your second one, vL(0+)=0, is wrong. The current through an inductor must be continuous, but the voltage across it doesn't have to be.

The opposite is true for capacitors. The voltage must be continuous, but the current doesn't have to be. So the second initial condition would be vC(0) = 1 V.
 
  • #3
Thanks vela, I have changed my working however I am still tragically stuck.

for the mesh, is = i1

initial conditions
i1(0) = 1A
Vc(0) = 1V

Mesh Equations
[tex]i_1 = (\frac{R_1+R_2}{R_1})i_2 + (\frac{1}{R_1C}) \int i_2[/tex]
[tex]i_2 = (\frac{L}{R_1})\frac{di_1}{dt} + i_1[/tex]

Sub (2) into (1)
[tex]i_1 = (\frac{R_1+R_2}{R_1})((\frac{L}{R_1})\frac{di_1}{dt} + i_1) + (\frac{1}{R_1C}) \int ((\frac{L}{R_1})\frac{di_1}{dt} + i_1)[/tex]

Expand
[tex]i_1 = ((\frac{R_1+R_2}{R_1})(\frac{L}{R_1})\frac{di_1}{dt} + (\frac{R_1+R_2}{R_1})i_1) + (\frac{1}{R_1C})(\frac{L}{R_1}) (\int \frac{di_1}{dt} + \int i_1)[/tex]

Simplify
[tex]i_1 = (\frac{L(R_1+R_2)}{R_1^2}\frac{di_1}{dt} + (\frac{R_1+R_2}{R_1})i_1) + (\frac{L}{R_1^2C})i_1 + (\frac{L}{R_1^2C})\int i_1[/tex]

[tex]i_1 = \frac{L(R_1+R_2)}{R_1^2}\frac{di_1}{dt} + (\frac{(R_1^2C + R_1CR_2 + L)}{R_1^2C})i_1 + (\frac{L}{R_1^2C})\int i_1[/tex]

[tex]\frac{L(R_1+R_2)}{R_1^2}\frac{d^2i_1}{dt} + (\frac{(R_1^2C + R_1CR_2 + L)}{R_1^2C} -1)\frac{di_1}{dt} + (\frac{L}{R_1^2C})i_1 = 0[/tex]

Check if DE is dimensionally correct.
[tex]\frac{d^2i_1}{dt} + (\frac{R_1CR_2 + L}{CL(R_1+R_2)})\frac{di_1}{dt} + (\frac{1}{C(R_1+R_2)})i_1 = 0[/tex]

I believe this to be incorrect due to the i1' term with an R/L coefficient, this is not dimensionally correct. CR^2/CLR gives a R/L term (no good) and L/CLR gives a 1/RC term, also no good. My understanding is that this co-ef should be some mixture of [tex]\tau[/tex]

We will continue just to check

Put in a few numbers,
L = 0.5
C = 0.5
R1 = 1
R2 = 1

[tex]\frac{d^2i_1}{dt} - 0.5\frac{di_1}{dt} + i_1 = 0[/tex]

[tex]i_1(0) = 1A[/tex]
[tex]V_c(0) = 1V [/tex] therefore [tex] V_r(0) = 1V[/tex]
[tex]V_r(0) = V_l(0) = 1V[/tex]
[tex]V_l = L\frac{di_1}{dt}[/tex]

[tex]\frac{di_1}{dt} = 2[/tex]

Putting the DE into MatLab (to save some time)
pretty(simplify(dsolve(' D2y-0.5*Dy+y=0' , ' Dy(0)=2' , ' y(0)=1 ' )))
results in,

1/15*exp(1/4*t)*(7*sin(1/4*15^(1/2)*t)*15^(1/2)+15*cos(1/4*15^(1/2)*t))

As predicted, miles off, I am expecting a pair of sine and cosine terms with a common coefficient of 1
 
  • #4
After you substitute (2) into (1), differentiate the result to get rid of the integral, and let [itex]\tau_C = R_1C[/itex] and [itex]\tau_L = L/R_1[/itex], both of which have units of time. Then you'll have

[tex]i_1' = \frac{R_1+R_2}{R_1}(\tau_L i_1'' + i_1') + \frac{1}{\tau_C} (\tau_L i_1' + i_1)[/tex]

Collect terms to get

[tex]\frac{R_1+R_2}{R_1}\tau_L i_1'' + \left(\frac{R_1+R_2}{R_1}+\frac{\tau_L}{\tau_C}-1\right)i_1' + \frac{1}{\tau_C} i_1 = 0[/tex]

It's pretty easy to see that result is dimensionally correct.
 
  • #5
Working with [tex] \tau [/tex] is a much simpler way to work the Algebra. I never thought of carrying around the time-constants are pairs but it makes complete sense. thankyou! should save me a lot of paper if nothing else.

I re-worked my solution from my initial mesh equations to your final result but I am confused about the final DE.

Does the [tex] \frac{\tau_L}{\tau_C} [/tex] term not reduce to dimensionless?
 
Last edited:
  • #6
Yes, that combination is dimensionless as is (R1+R2)/R1 and 1.

You also made a mistake deriving di1/dt. Check the signs in your equations.
 

1. What is a second order circuit analysis?

A second order circuit analysis is a method used in electrical engineering to analyze and understand the behavior of circuits with two energy storage elements, such as capacitors and inductors. It involves solving differential equations to determine the voltages and currents in the circuit.

2. What is a mesh in circuit analysis?

In circuit analysis, a mesh is a loop in the circuit that contains only one energy storage element, such as a capacitor or inductor. The analysis of a circuit using mesh analysis involves applying Kirchhoff's voltage law to each mesh to determine the voltages and currents in the circuit.

3. What are some common issues with setting up differential equations in second order circuit analysis?

Some common issues with setting up differential equations in second order circuit analysis include incorrect application of Kirchhoff's laws, incorrect identification of energy storage elements, and missing or incorrect initial conditions. It is important to carefully analyze the circuit and understand the behavior of the components before setting up the differential equations.

4. How can I solve for the unknown variables in a second order circuit analysis?

The unknown variables in a second order circuit analysis, such as the voltages and currents, can be solved using various techniques such as Laplace transforms, differential equations, or numerical methods. The appropriate method to use will depend on the complexity of the circuit and the available tools and resources.

5. What are some applications of second order circuit analysis?

Second order circuit analysis has many practical applications, such as in the design and analysis of electronic circuits, control systems, and communication systems. It is also used in fields such as signal processing, power systems, and electromagnetics. Understanding second order circuit analysis is essential for any electrical engineer or scientist working with complex circuits.

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