RC Circuit Differential Equation

AI Thread Summary
The discussion focuses on solving the differential equation for an RC circuit, specifically L d²I/dt² + R dI/dt + I/C = dV/dt, with L set to zero and V defined as V_0 cos(ωt). The user is unsure how to derive the general solution, which the textbook provides as I = Ae^(-t/RC) - (V_0 ω C (sin(ωt) - ω R C cos(ωt))) / (1 + ω² R² C²). A participant suggests assuming a solution of the form A cos(ωt) + B sin(ωt) to reach the correct answer. The user expresses gratitude for the clarification and acknowledges a mistake in copying the equation. This exchange highlights the importance of understanding differential equations in circuit analysis.
sritter27
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Homework Statement


Find the general solution of L \frac{d^2I}{dt^2} + R \frac{dI}{dt} + \frac{I}{C} = \frac{dV}{dt} given L = 0 and V = V_0 cos(\omega t).

Homework Equations


The Attempt at a Solution


So the equation basically turns into a first-order RC circuit equation R \frac{dI}{dt} + \frac{I}{C} = \frac{dV}{dt}, but I'm not sure how to approach it to find a general solution.

The answer the book gives is I = Ae^{-\frac{t}{RC}} - \frac{V_0 \omega C (sin(\omega t) - \omega R C cos(\omega t))}{1 + \omega^2 R^2 C^2} and I'm not sure how they came to that conclusion, so any help or nudge in the right direction would be greatly appreciated.
 
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sritter27 said:
R \frac{dI}{dt} + \frac{I}{C} = \frac{dV}{dt}, but I'm not sure how to approach it to find a general solution.

The answer the book gives is I = Ae^{-\frac{t}{RC}} - \frac{V_0 \omega C (sin(\omega t) - \omega R C cos(\omega t))}{1 + \omega^2 R^2 C^2} and I'm not sure how they came to that conclusion, so any help or nudge in the right direction would be greatly appreciated.

Hi sritter27! Welcome to PF! :smile:

To find the general solution of R dI/dt + I/C = -ωV0sinωt,

assume it's of the form Acosωt + Bsinωt, and you get the given result,

except that you've copied it wrong … it's V_0 \omega C\frac{ (sin(\omega t) - \omega R C cos(\omega t))}{1 + \omega^2 R^2 C^2} :wink:
 
Oh wow I should have been able to see that. My many thanks for the help!
 
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