RC circuit with AC - finding current

[Akhilleus]

Homework Statement

I am given an RC circuit with an alternating current. The circuit contains a capacitor and a resistor in parallel. Part (a) says "Use KCL to find a differential equation for I in terms of V." Part (b) says "For an applied voltage V = V_oexp(iwt), find the current I."

Homework Equations

I_Total = I₁ + I₂
Z = impedance
Z_R = R
Z_C = (1/wiC)
V = IZ

The Attempt at a Solution

I understand how KCL works, but I'm not sure how to get it in a differential equation. Ignoring this and using KCL anyway, I get:

V - I₁Z_R = V - I₁R = 0
V - I₂Z_C = V - I₂(1/iwC) = 0
I₁ = ViwC
I₂ = (V/R)
I_Total = ViwC + (V/R)

Am I on the right track, or does the fact that I need a differential equation mean I need to do something completely different?

 

[Delphi51]

Looks like you are supposed to use the differential equation method (which is more general than the impedance method you started).
The current through a capacitor is related to the voltage across it by i=C*dV/dt. Apply KCL to the node where the current splits, part through R and part through C. You should have only the combined i, the common V and of course R and C in your equation. And a derivative.

 

[Akhilleus]

I've gotten as far as:

i = C*dV/dt + V/R

Is it mathematically correct to say that V is common to both terms on the right, and therefore:

i = V(C*d/dt + 1/R) ?

If so, I would plug in the V value given in step (b):

i = V_oe^iwt(C*d/dt + 1/R)

How would I go about integrating this (or should I have integrated before plugging in V)?

 

[Delphi51]

The first formula looks great, but you can't factor the V out of the derivative. For the (b) part you'll have to find the derivative of the V function. After that, likely you'll be able to take out a common factor because the derivative of an exponential function is the same function multiplied by some constants.

 

[Akhilleus]

Is the i in the voltage function V = V_oe^iwt the current I want? If so, I've been spending MUCH longer on this than I needed to... haha

In that case, it'd just be:

V = V_oe^iwt

dV/di = V_o(iwt)e^iwt

 

[Delphi51]

dV/dt = iwVo*e^(iwt)

 

[Akhilleus]

RIGHT that's what I meant.

Is that my answer (solving for i) or do I set that equal to I/C and solve for I?

I guess I'm getting hung up on the fact that there's an instantaneous current (i) and an average/overall current (I), and I don't know which to solve for.

 

[Delphi51]

You can't solve it for i.
The answer is your i = C*dV/dt + V/R with that expression for dV/dt subbed in. Oh, and replace V with its given value. Maybe you can simplify it. Factoring out the exponential would be nice.

The average current will always be zero with AC.

 

[Akhilleus]

Ah, that makes sense! Thank you!

Just to double check, I got for an answer:

i = V_oe^iwt (iwC + 1/R)

Does it matter that there's also an i in the exponent?

 

[Delphi51]

Looks good!

 

[Akhilleus]

Excellent! Thanks for your help.