RC Circuit with alternating voltage source

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The discussion focuses on solving for the current in an RC circuit with an alternating voltage source represented as V=V0ejwt. The equation I=C*dv/dt is clarified, indicating that 'v' refers to the voltage across the capacitor. Kirchhoff's Voltage Law (KVL) is applicable in this scenario, and the user is guided to consider the real part of the phasor representation, specifically V0cos(ωt). The conversation emphasizes the importance of understanding phasors for accurate analysis of the circuit. Overall, the thread provides insights into applying fundamental circuit laws to solve the problem effectively.
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Homework Statement


I'm supposed to find the current in a circuit with a voltage source, capacitor, and resistor in series. The voltage source is described by V=V0ejwt.
Here, j is the complex number j2=-1, and i is the current

Homework Equations


I=C*dv/dt

The Attempt at a Solution


I have several sources of confusion with this question. First, in the equation I=C*dv/dt, does that v refer to the voltage source, or specifically the voltage drop across the capacitor? Can I use Kirchoff's voltage law with this type of circuit? I tried this and got:

V0ejwt - iR - q/c = 0
Then taking the time derivative of the equation:

V0jwejwt - R*di/dt - i/c = 0
I then have no idea how to solve this differential equation. Even rewriting with euler's identity:
V0jw[cos(wt)+jsin(wt)]=R*di/dt+i/c

Any help would be greatly appreciated.
 
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I suspect that the ##V = V_o e^{j \omega t}## is meant to be a phasor representation for the input. Have you covered phasors yet? If so you're meant to deal with the real portion of the resulting voltage, which in this case would be ##V_o cos(\omega t)##.

Yes, the v in I = C*dv/dt represents the voltage across the capacitor.

Yes, Kirchhoff's Voltage Law (KVL) can be applied as you have done.
 
We haven't covered phasors yet but I read ahead and it makes sense now.
Thank you for the response.
 
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