Differential Equations and Circuits

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SUMMARY

This discussion focuses on solving LR circuits using differential equations, specifically the equation L(dI/dt) + RI = V. Participants explore the integration process, particularly the arrangement of variables during integration. The integration of both sides leads to ∫(dI/(V-IR)) = ∫(dt/L), highlighting the necessity of keeping variable terms together. The conversation clarifies that while L can be rearranged, V must remain with IR to maintain the integrity of the equation.

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with electrical circuit concepts (LR, RC, LC circuits)
  • Knowledge of integration techniques
  • Basic principles of circuit analysis
NEXT STEPS
  • Study the derivation of differential equations for RC and LC circuits
  • Learn about the Laplace transform for circuit analysis
  • Explore the concept of time constants in RL and RC circuits
  • Investigate the impact of variable resistance and inductance on circuit behavior
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in the mathematical modeling of electrical circuits using differential equations.

nikki__10234
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So, I'm learning how to solve LR, RC, LC etc. types of circuits using differential equations. I understand how to do the math with differential equations, but I am confused as to why the variables are split in the way they are.

For example, for an LR circuit you have the equation
L(dI/dt)+RI=V

and then the book integrates both sides:
∫(dI/(V-IR))=∫(dt/L)
and so on...

It is justified to group the I term with the dI, but I don't understand why the L, V, and R terms are placed as they are? Could you get the same result if they were not arranged in this way since the integral is not being taken in terms of V, R or L?
 
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welcome to pf!

hi nikki! welcome to pf! :smile:
nikki__10234 said:
L(dI/dt)+RI=V

and then the book integrates both sides:
∫(dI/(V-IR))=∫(dt/L)
and so on...

It is justified to group the I term with the dI, but I don't understand why the L, V, and R terms are placed as they are? Could you get the same result if they were not arranged in this way since the integral is not being taken in terms of V, R or L?

the V has to stay with the IR …

can you see any way of getting the IR over onto the LHS (with the dI), without the V coming with it?​

but the L could go either side
 
nikki__10234 said:
So, I'm learning how to solve LR, RC, LC etc. types of circuits using differential equations. I understand how to do the math with differential equations, but I am confused as to why the variables are split in the way they are.

For example, for an LR circuit you have the equation
L(dI/dt)+RI=V

and then the book integrates both sides:
∫(dI/(V-IR))=∫(dt/L)
and so on...

It is justified to group the I term with the dI, but I don't understand why the L, V, and R terms are placed as they are? Could you get the same result if they were not arranged in this way since the integral is not being taken in terms of V, R or L?
The only requirement is that any variables which actually vary in the range of integration stay inside an integral, but it will help if it varies as a function of the variable of integration with which it is placed.
E.g. if L = L(t) then it would best be placed in the integral .dt as above, whereas if R is a constant then you could just as easily write
∫(dI/(V/R-I))=R∫(dt/L)
OTOH, if V = V(t) it's going to get tricky ;-).
 

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