So, using the phasor approach, what is the solution to this circuit problem?

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Discussion Overview

The discussion revolves around solving a circuit problem using the phasor approach, specifically addressing an integrodifferential equation related to current in a circuit. The focus is on the conversion of terms into phasor form and the implications of using phasors in the context of differential equations.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant presents an integrodifferential equation for current and seeks guidance on converting the left side into phasor form.
  • Another participant suggests that by expressing current as the derivative of charge, a second-order differential equation can be derived, implying a straightforward solution process.
  • A different participant notes that the integral of current can be represented as charge, indicating a relationship between current and charge in the context of the problem.
  • Another contribution emphasizes the use of phasor format, providing derivatives and integrals of the exponential function in phasor representation, which may aid in solving the equation.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as participants offer different methods and perspectives on how to approach the problem without resolving which method is preferable or correct.

Contextual Notes

Participants do not clarify certain assumptions regarding the initial conditions or the specific definitions of terms used in the equations, which may affect the interpretation of the problem.

hazim
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[SOLVED] phasor approach question

using phasor approach, determine the current i(t) in a circuit described by the integrodifferential equation: 4i + 8integral(idt) - 3di/dt = 50cos(2t + 75*)
the right side is so easy to convert to phasor form, but how can i convert the left side or how to solve such a problem?
 
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dear it is simple.
just write I=dq/dt.you will get a second order differential equation.
you can solve it easily.
 
also integral(Idt)=q.
 
I think the point is to use the phasor format, and realize the for the phasor [tex]Ae^{i\omega{t}}[/tex], that

[tex]d/dt(e^{i\omega{t}})\,=\,i\omega{e^{i\omega{t}}}[/tex], and

[tex]\int(e^{i\omega{t}})dt\,=\,\frac{1}{i\omega}{e^{i\omega{t}}}\,=\,\frac{-i}{\omega}{e^{i\omega{t}}}[/tex]
 

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