RC Circuit Differential Equation

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SUMMARY

The forum discussion focuses on solving the differential equation for an RC circuit, specifically the equation L \frac{d^2I}{dt^2} + R \frac{dI}{dt} + \frac{I}{C} = \frac{dV}{dt} with L set to 0 and V defined as V_0 cos(\omega t). The general solution provided 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}. The discussion emphasizes the approach of assuming a solution of the form Acos(ωt) + Bsin(ωt) to derive the final result.

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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!
 
Last edited:

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