Solving RC Circuit Equation with Internal Resistance

In summary, the equation V = EMF (1-e^-t/RC) in a RC circuit becomes V = EMF ( r / R+r) (1-e^-((R+r)t/RrC)) when taking into account the internal resistance r. Kirchoff's current law still applies, but you need to use another equation to get the voltage across the capacitor.
  • #1
quietrain
655
2

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!
 
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  • #2
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.)
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
Last edited:
  • #6
Yes, [itex]V_r=V_c[/itex], but [itex]I\ne EMF/R[/itex].
 
  • #7
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!
 
  • #8
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?
 
  • #9
The time constant is whatever divides t in the exponential, without the minus sign.
 
  • #10
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
 
  • #11
Yes, that's the time constant. Try looking at [itex]1/\tau[/itex].
 
  • #12
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!
 

FAQ: Solving RC Circuit Equation with Internal Resistance

What is an RC circuit equation?

The RC circuit equation is a mathematical representation of the behavior of a resistor-capacitor (RC) circuit. It describes the relationship between the voltage across the capacitor, the current flowing through the circuit, and the values of the resistor and capacitor in the circuit.

How is internal resistance accounted for in the RC circuit equation?

Internal resistance is taken into account by including it as a separate term in the RC circuit equation. This resistance is typically represented by the letter "r" and is added to the value of the resistor in the equation.

What is the significance of solving the RC circuit equation with internal resistance?

Solving the RC circuit equation with internal resistance allows us to accurately predict and analyze the behavior of real-world circuits. Internal resistance is present in most circuits and can significantly affect the performance and efficiency of the circuit. By including it in the equation, we can better understand and optimize the circuit's behavior.

What are the key components needed to solve the RC circuit equation with internal resistance?

To solve the RC circuit equation with internal resistance, you will need the values of the resistor and capacitor in the circuit, as well as the internal resistance value. Additionally, you will also need to know the initial voltage or current values in the circuit, as these will affect the solution of the equation.

Are there any simplifications that can be made to the RC circuit equation with internal resistance?

Yes, there are some simplifications that can be made to the RC circuit equation with internal resistance, depending on the specific circuit and its components. For example, if the internal resistance is significantly smaller than the resistance and capacitance values in the circuit, it can be neglected in the equation. Additionally, if the circuit is in a steady-state, certain terms in the equation may also be simplified or eliminated.

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