Solving RC Circuit Equation with Internal Resistance

[quietrain]

Homework Statement

ok erm, i know how to get the equation V = EMF (1 - e^(-t/RC)) in a RC circuit
but what if i had to take into account the internal resistance r of the voltmeter measuring the voltage across the capacitor? meaning now, the circuit contains a resistor R in series with, an internal resistance r parallel with capacitor C.

so how does the equation becomes V = EMF ( r / R+r) ( 1 - e^-((R+r)t/RrC))

so by kirchhoffs law, EMF = voltage across resistance R + voltage across the [parallel capacitor C and internal resistance r]
EMF = RI + q/C , here's the problem, the current through the parallel component will split up into r and C. so how do i form the new equation?

also will the voltage drop across the capacitor still be the same as when there is no r? so, will it still be q/C? or something else?

please help thanks!

 

[vela]

Label the two currents that go through the internal resistance and the capacitor. You can get another equation applying Kirchoff's current law to one of the nodes in the circuit. If you go around two loops in the circuit, applying Kirchoff's voltage law, you'll get two more equations. (You already have one above.)

 

[quietrain]

so my 2nd equation should be V = RI + rI_r ? where I_r is the current through the internal resistor

so i just equate the 2 equation's V together?

 

[vela]

The V in your equation is the EMF in your original post, right? You can set the two equations equal to each other, but whether you should is debatable. You have three equations and three unknowns. You should do whatever you usually do to solve such a system of equations.

 

[quietrain]

oh i think i mixed up the Vs and EMFs
it shoudl be

EMF = RI + q/C
EMF = RI + rI_r
I = I_r + I_C ===> EMF/R = V_r / r + C dv/dt , btw V_r = Voltage across capacitor right? parallel? so = V

so i am trying to find V, which is the voltage across the capacitor . the EMF is EMF.

this is confusing@@

so to get the V = EMF ( r / R+r) ( 1 - e^-((R+r)t/RrC)), i should eliminate I from my equations right?

i still can't get the equation V = EMF ( r / R+r) ( 1 - e^-((R+r)t/RrC)) , from the 3 equations above. help?

 

[vela]

Yes, $V_r=V_c$, but $I\ne EMF/R$.

 

[quietrain]

OMG I FINALLY SOLVED IT! SO HAPPPY>>>>><<<<

oh, ic... I is not = to EMF / R . so i just need to equate my first equation iwth my 3rd equation , eliminating I to get the equation!

WOW>>>><<< 6.5 hours to solve this part of my question! physics is great...


THANKS A LOT VELA!

 

[quietrain]

OMG ... now i have another problem...

from that equation, i have to plot a straight line graph of either time constant = RC ,against R or vice versa... to find the internal resistance and capacitance from the graph.

but how am i suppose to do that? i tried manipulating it from the intermediate step

-(R+r)t/RrC = ln |(V(R+r) - Er)/(-Er)| but i can't single out the R term in the ln function...

any ideas?

or am i suppose to use some other formula? i only know time constant (tau) = RC. but if i include internal resistance r, will it become (tau) = (R+r)C?

 

[vela]

The time constant is whatever divides t in the exponential, without the minus sign.

 

[quietrain]

so from the e^-(R+r)t/RrC, my time constant will be RrC / (R+r) ?

so tau = RCr / (R+r) ?

but how do i plot a graph of R against RC(tau) or vice versa to get a straight line graph from this?

since i have a R at the denominator too

 

[vela]

Yes, that's the time constant. Try looking at $1/\tau$.

 

[quietrain]

AH isee... so i have to plot 1/(tau) against 1/R

and my gradient is 1/C and y intercept is 1/Cr?

THANK YOU SOOO MUCH!