Circuit with two resistances and inductance

Click For Summary

Homework Help Overview

The discussion revolves around a circuit containing two resistors and an inductor, where the original poster seeks to determine the currents through the resistors. They present a system of equations related to the circuit's behavior and express uncertainty about its correctness.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to solve a differential equation for the circuit and consider leveraging known properties of first-order circuits. There are inquiries about the correctness of the original equations and suggestions for writing expressions for the currents based on potential at the junction of the components.

Discussion Status

Some participants have offered guidance on deriving a differential equation from the existing equations and suggested alternative approaches, such as using nodal analysis or simplifying the circuit with a Norton equivalent. The discussion reflects a collaborative effort to clarify the setup and explore different methods without reaching a definitive conclusion.

Contextual Notes

The original poster expresses difficulty in formulating the correct equations for the system, indicating a potential gap in understanding the relationships between the currents and circuit elements.

bznm
Messages
181
Reaction score
0

Homework Statement


http://img534.imageshack.us/img534/5788/gimpjl.png

I have this circuit and I want to know [itex]i_1[/itex] and [itex]i_2[/itex] (currents in [itex]R_1[/itex] and [itex]R_2[/itex]).
[itex]\varepsilon[/itex] is electromotive force and L is inductance.

I'd like to know if the following system of equations is correct and if I can get [itex]i_1[/itex] and [itex]i_2[/itex].

2. The attempt at a solution

\begin{cases} \varepsilon=R_1(i_1+i_2)+R_2i_2 \\ \varepsilon-L\frac{d}{dt}i_1=R_1(i_1+i_2) \end{cases}

I think I'm wrong, because i1 is the current in L, not in R1, but I must get current in R1!
How can I correct the system of equations?
 
Last edited by a moderator:
Physics news on Phys.org
Do you have to solve the differential equation for the circuit, or can you take advantage of the known properties of first order circuits (initial conditions at t = 0+; steady state conditions; the forms of their transient solutions) to write the solution?

As a general hint/suggestion, note that if you happened to have an expression for the potential where R1 and R2 (and L) meet, then it would be a simple matter to write expressions for the two currents that you're looking for.
 
gneill said:
Do you have to solve the differential equation for the circuit, or can you take advantage of the known properties of first order circuits (initial conditions at t = 0+; steady state conditions; the forms of their transient solutions) to write the solution?

As a general hint/suggestion, note that if you happened to have an expression for the potential where R1 and R2 (and L) meet, then it would be a simple matter to write expressions for the two currents that you're looking for.


Thank you for your answer! My main problem is writing the equations of the system. Can you help me?
 
The set of equations that you've already written will allow you to find a differential equation for the current through the inductor (i1 in your equations). If you solve for this current you can then go back and find i2, and then find the current through R1 since its the sum of the other two. The algebra may get tedious.

I think that if I were to do this problem I'd consider starting with a nodal equation and find the potential at the one independent node (so only one equation to deal with). This would involve an integral equation rather than a differential equation, but a quick differentiation would reduce it to familiar form. Solving it, having the expression for the node voltage would let me easily find the two currents. Also, I think I'd simplify the circuit first: a Norton equivalent would get rid of one of the resistances.
 
thank you so much! :)
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
7
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K