Calculating Voltage Across a Capacitor in a RC Circuit with Two Resistors

[ProjectENIS]

Homework Statement

Given a circuit with two resistors, R and r, and a capacitor of C, and EMF of V₀ as shown in the diagram, find the voltage across the capacitor during charging. Prove that this voltage, V is given by V = V₀ (r/(R+r)) (1-e^{-((R+r)t)/(RrC)})

Homework Equations

N.A.

The Attempt at a Solution

This is what I have

Loop A (with C): V₀ + RI + q/c = 0 ==> I = (V₀ - V_C)/R
Loop B (with r): V₀ + RI + rI_r = 0
I = I_r + I_c ==> I = V/r + C(dv/dt)

From first and third,
(V₀ - V_C)/R = V/r + C(dv/dt)

Simplify to get,
V₀ - RC(dv/dt) = V(1+R/r)
V = (r/[R+r])(V₀ - RC(dV/dt))

The shape of the equation is getting there (I hope), but what do I do next? To get the given equation, RC(dV/dt) must be V₀e^{-((R+r)t)/(RrC)}.

RC (dV/dt) = V, solving this differential equation to get ln (V) = -t/(RC) + k, hence, RC (dV/dt) = e^{-t/(RC) + k}. And I am totally stuck.

Did I do something wrong somewhere? I can't think of anyway to get the RrC term in e^{-((R+r)t)/(RrC)}, not to mention the V₀ and the (R+r) terms, unless k is like Rt/r. But that still does not give me a V₀?

 

[ehild]

The differential equation is correct. You need not solve it to prove that the given V is solution, just plug in V and dV/dt. What is the derivative of the given V(t) function? ehild

 

[ProjectENIS]

Hi ehild,

Thanks for your reply.

For the differential equation, V = RC (dV/dt), solved for V = e^{-t/(RC) + k},

e^{-t/(RC) + k} = RC (dV/dt)

dV/dt = e^{-t/(RC) + k}/RC

Is this what you meant?

 

[ehild]

Hi ehild,

Thanks for your reply.

For the differential equation, V = RC (dV/dt), solved for V = e^{-t/(RC) + k},

e^{-t/(RC) + k} = RC (dV/dt)

dV/dt = e^{-t/(RC) + k}/RC

Is this what you meant?

NO.

Your differential equation is not V = RC (dV/dt), but V = (r/[R+r])(V0 - RC(dV/dt)).


ehild

 

[ProjectENIS]

Sorry for the mistake,

That will get me

(V - (r/[R+r])V₀)^-1dV = [-(r/[R+r])RC)]^-1dt

Which gets me ln [V - (r/[R+r])V₀)] = t/(RrC/[R+r])
V - (r/[R+r])V₀) = e^{(t/(RrC/[R+r])]}
And a little rearranging gets me V = e^{(t/(RrC/[R+r])]} + (r/[R+r])V₀)

Hmm, did i do something wrong? Even if I subsitute V = RC(dV/dt) into the equation, i will get

V = (r/[R+r])(V₀ - e^{(t/(RrC/[R+r])]} - (r/[R+r])V₀))

which is not equal to the answer..

 

[ehild]

Sorry for the mistake,

That will get me

(V - (r/[R+r])V₀)^-1dV = [-(r/[R+r])RC)]^-1dt

Which gets me ln [V - (r/[R+r])V₀)] = t/(RrC/[R+r])

You missed to add the arbitrary constant. Your result will be is the same as the given function with the condition that the initial voltage on the capacitor is zero: V(0)=0.

Hmm, did i do something wrong? Even if I subsitute V = RC(dV/dt) into the equation, i will get

V = (r/[R+r])(V₀ - e^{(t/(RrC/[R+r])]} - (r/[R+r])V₀))

which is not equal to the answer..

Why do you substitute V=RC(dV/dt)? It is valid for a simple RC circuit only.

ehild

 

[ProjectENIS]

Hi again ehild, I added in the arbitrary constant and got the equation! Thank you so much!

Regarding the V = RC (dV/dt), I have misunderstood the derivation, and assumed that it is a general formula for all RC circuits. Thanks for pointing that out, I'll read up on it tonight.

Once again, thanks for your help!

 

[ehild]

You are welcome

ehild