- #1
xconwing
- 20
- 0
How can you get from the first expression to the second. I've done calculus, just need a fresher. I don't know how to deal with the dt/dt
thanks
How to Simplify Inductor Expressions Using Calculus?
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SUMMARY
This discussion focuses on simplifying inductor expressions using calculus, specifically addressing the integral of current over time. The key expression derived is L∫_{−∞}^t i(τ)(di/dτ)dτ = (1/2)L(i^2(t)), assuming that the limit of current i(τ) approaches zero as τ approaches infinity. The distinction between treating "Li" as a constant times a function versus a single function is crucial for accurate simplification.
PREREQUISITES- Understanding of calculus, particularly integration techniques.
- Familiarity with electrical engineering concepts, specifically inductance.
- Knowledge of current as a function of time, denoted as i(t).
- Basic grasp of limits in calculus.
- Study the application of integration in electrical circuits, focusing on inductors.
- Learn about the relationship between current and inductance in RL circuits.
- Explore advanced calculus techniques for solving differential equations in electrical engineering.
- Review the concept of limits and their application in calculus for better understanding of function behavior.
Electrical engineers, physics students, and anyone involved in circuit analysis or studying inductor behavior in electrical systems.
How can you get from the first expression to the second. I've done calculus, just need a fresher. I don't know how to deal with the dt/dt
thanks
- #2
HallsofIvy
Science Advisor
Homework Helper
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- 983
What you have written is not clear. Is "Li" a single function or is this the constant, L, times the function i?
If the latter this is easy. [itex]L\int_{-\infty}^t i(\tau) (di/d\tau)d\tau= L\int_{-\infty}^t i di = L\left[(1/2)i^2(\tau)\right]_{\infty}^t[/itex] which, if [itex]\lim_{\tau\to\infty} i(\tau)= 0[/itex] is [itex](1/2)i^2(\tau)[/itex]
If Li is a single function, then it is NOT always true.
If the latter this is easy. [itex]L\int_{-\infty}^t i(\tau) (di/d\tau)d\tau= L\int_{-\infty}^t i di = L\left[(1/2)i^2(\tau)\right]_{\infty}^t[/itex] which, if [itex]\lim_{\tau\to\infty} i(\tau)= 0[/itex] is [itex](1/2)i^2(\tau)[/itex]
If Li is a single function, then it is NOT always true.
- #3
berkeman
Admin
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- 25,111
Yeah, L is the constant value of the inductance. i(t) is the current as a function of time.
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