How Is Current Derived in a Series RC Circuit with an AC Source?

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Homework Help Overview

The discussion revolves around deriving an expression for current in a series RC circuit with an AC source, specifically using the example of a source voltage described by 50cos(100t). Participants are exploring the differences in approach between AC and DC circuit analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Kirchhoff's Voltage Law (KVL) and the need for a two-part solution involving homogeneous and particular solutions. Questions arise regarding the correct formulation of differential equations and the relationship between current and voltage in capacitors.

Discussion Status

The conversation is ongoing, with participants providing insights and suggestions for rewriting equations. There is a mix of interpretations regarding the derivation process, and some participants express confusion about the steps involved in replacing current expressions with voltage. No consensus has been reached yet.

Contextual Notes

Initial conditions and the specific form of the source voltage are noted, with participants questioning how these factors influence the derivation of current in the circuit.

canicon25
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can anyone show me the derivation of an expression for current in a series RC circuit with an AC source? I have seen many for DC sources simply using KVL is it the same for AC?

I would imagine that a two part solution is required: 1) homogeneous solution 2)using undetermined coefficients as the source is equal to 50cos(100t)

Is the derivation the same as for a DC source?
 
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canicon25 said:
can anyone show me the derivation of an expression for current in a series RC circuit with an AC source? I have seen many for DC sources simply using KVL is it the same for AC?

I would imagine that a two part solution is required: 1) homogeneous solution 2)using undetermined coefficients as the source is equal to 50cos(100t)

Is the derivation the same as for a DC source?

Welcome to the PF. Are you familiar with the relationship between current and voltage for a capacitor? That is where the DiffEq aspect of the circuit comes in. Write the KVL equation for the loop, using the simple V=IR for the resistive drop, and the differential for the voltage drop across the capacitor. Then solve the DiffEq...
 
i am familiar. i have seen a lot of examples for DC ccts, but not AC. following your suggestions i would make this:

E=VR+VC
E=iR+q/C

E=dq/dt R + q*1/C

or

d/dt E= di/dt R + I/C

where E=50cos(100t).

would i solve the first order homogeneous ( could probably just separate) first then add on the solution from undetermined coefficients?

on a second note the initial condition i have is q(0)=0 but the question wants an expression for loop current. can i solve for q(t) then divide through by C and then divide through by Z=R-jX?
 
canicon25 said:
i am familiar. i have seen a lot of examples for DC ccts, but not AC. following your suggestions i would make this:

E=VR+VC
E=iR+q/C

E=dq/dt R + q*1/C

or

d/dt E= di/dt R + I/C

where E=50cos(100t).

That's not quite the equation I meant. For a capacitor, the following equation expresses the relationship between voltage and current:

[tex]i(t) = C \frac{dV(t)}{dt}[/tex]

Try re-writing the KVL with that for the capacitor current and voltage...
 
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d/dt E= di/dt R + 1/C int[ i(t) dt ] ?
 
canicon25 said:
d/dt E= di/dt R + 1/C int[ i(t) dt ] ?

Almost, but you mixed a couple steps together...

It looks like you wrote the correct equation for the voltages in the series circuit, and then went to differentiate to get rid of the integral term on the right. But you only show the differentiation of the left hand side (LHS) and the first term on the RHS, but you haven't differentiated the integral yet to get your final DiffEq...
 
i'm confused why i want to replace a current expression with voltage? integrating the expression you gave me will produce an expression for capacitor current i believe.
 
canicon25 said:
i'm confused why i want to replace a current expression with voltage? integrating the expression you gave me will produce an expression for capacitor current i believe.

(keep in mind that I could be wrong, but...) Do you see why I think your equation is not correct yet? Why did you differentiate the LHS term and the first RHS term, but not the 2nd RHS term?

And as for solving for current in a series RC circuit, I thought that's what you were asking to do?

canicon25 said:
can anyone show me the derivation of an expression for current in a series RC circuit with an AC source?
 

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