Differential equation to represent circuit

[TigerBrah]

**edit** This problem has been solved.For the circuit shown in the figure, find a differential equation the relates the input current i₁(t) to the output voltage e₀(t). Note that you cannot take a KVL across a current source because you cannot figure out the voltage drop across it.

the circuit:

http://www.flickr.com/photos/70948910@N08/6412941773

my work so far:

http://www.flickr.com/photos/70948910@N08/6413318181/Am I on the right track? We never had any problems with a current source during the lectures, and I don't know if I can set the voltages of parallel branches equal like that.

Any ideas will be greatly appreciated! It has been very tough getting in touch with my professor over Thanksgiving break and this assignment is due tomorrow.

 

[gneill]

Hi TigerBrah, welcome to Physics Forums.

You're doing pretty well with the problem, but I think you'll want to keep R2 in the equations (rather than substituting e_o), since R2 influences the current in its branch. If you solve for the differential equation for the branch current, multiplying it through by R2 will give you the equation for the output voltage.

Another approach would be to use mesh currents. The first loop's mesh current is set by the input current, so you only have one loop to deal with.

Yet another approach would be to convert i_i(t) and R1 into a Thevenin equivalent voltage source and series resistance. Then you'd have a simple voltage divider to solve!

 

[TigerBrah]

Thanks gneill!

Using Mesh Currents is definitely the fastest way. I reworked the problem and got almost the same answer as above, but I also caught a mistake. My last term should be divided by R1.

I was also able to check my work using a secondary method we learned in class using Laplace transformations.

I appreciate your help!

 

[gneill]

Hey, glad to help

 

[DeeNos]

I would like to provide a response that is based on the principles of circuit analysis and differential equations.

Firstly, it is important to note that the circuit shown in the figure is a series-parallel circuit, where the current i1(t) splits into two parallel branches, one containing a capacitor and the other containing a resistor and an inductor.

To find the differential equation that relates the input current i1(t) to the output voltage e0(t), we can use Kirchhoff's voltage law (KVL) to analyze the circuit. Since the voltage drop across a current source cannot be determined, we can apply KVL to the two parallel branches separately.

For the branch containing the capacitor, KVL gives us:

i1(t)R + e0(t) = 0

Where R is the resistance in the branch.

For the branch containing the resistor and inductor, KVL gives us:

i1(t)R + L(di2/dt) + e0(t) = 0

Where i2 is the current in this branch.

Now, since the two branches are in parallel, the current i1(t) is the same in both branches. Therefore, we can set the two equations equal to each other:

i1(t)R + e0(t) = i1(t)R + L(di2/dt) + e0(t)

Rearranging this equation, we get:

L(di2/dt) = 0

This differential equation relates the input current i1(t) to the output voltage e0(t) of the circuit.

In summary, to represent this circuit using a differential equation, we can use Kirchhoff's voltage law to analyze the circuit and then set the equations for the two parallel branches equal to each other. This will give us a differential equation that relates the input current to the output voltage.