Electric circuit - time constant to charge

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Homework Help Overview

The discussion revolves around the time constant of an electric circuit involving four identical capacitors connected with a resistor in two different configurations. The original poster seeks to determine the time constant for the second configuration based on a given time constant for the first.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationships between the time constants and capacitances in different configurations, questioning the validity of their equations and assumptions. Some express uncertainty about the application of the charge equation and the natural logarithm in their calculations.

Discussion Status

Some participants have provided guidance on focusing directly on the time constants rather than the charge equations. There is an ongoing exploration of the relationships between the capacitors' configurations and their respective time constants, with no explicit consensus reached yet.

Contextual Notes

Participants note potential confusion regarding the configuration of the capacitors, with some misidentifying the arrangement as parallel. The original poster's calculations are challenged, and there is a suggestion to recalculate certain values.

seb26
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Homework Statement


Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawing, the time constant to charge up this circuit is 0.91 s. What is the time constant when they are connected with the same resistor as in part b?
http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c20/ch20p_99.gif

Homework Equations



Q=Q0*(1-e-t/RC)
Q0=CV
t=RC

The Attempt at a Solution


a) Q0=Q/(1-e-ta/RC) = 4C/V(1-e-ta/RC)
b) Q0=Q/(1-e-tb/RC) = 5C/2V(1-e-tb/RC)

4C/V(1-e-ta/RC) = 5C/2V(1-e-tb/RC)

4/(1-e-ta/(4ta)) = 5/2(1-e-tb/(5/2tb))

e-2/5 = 3/8 + 5/8 * e-1/4

I'm stuck there.. Not sure if what I did is right.. If it is, I can't find a way to use the natural logarithm
 
Last edited:
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You don't need the equation giving the charge at a certain time because the question doesn't ask you to find the charge at a certain time. Just write out an expression for the time constant for (a), then do the same for (b). Solve the two resulting equations.
 
I tried using the equation t=RC because Ra = Rb; Ca = 4C and Cb= (5/2)C
R= ta/4C = 5tb/2C
tb = (5/8)ta
Why doesn't this give me the right answer?

I can't find another equation to solve for tb.

4C/V(1-e-ta/RC) = 5C/2V(1-e-tb/RC)

4/(1-e-ta/RCa) = 5/2(1-e-tb/RCb)

e-2tb/5RC = 3/8 + 5/8 * e-ta/4RC

I'm stuck there again..
 
Recalculate Ca.

Again: you do not need the time dependence of charge. By the way, you use wrong formulas. The time constant is T=RC. Ta= RCa, Tb=RCb. Ta is given: 0.91 s. Tb/Ta=Cb/Ca.

ehild
 
When the capacitors are connected in parallel the total C is C1+C2+...+Cn which in case a) gives 4*C..
And i did use tb/ta = cb/ca
tb= taCb / Ca = ta(5/2)C / 4C = 5ta / 8

I used the equation with the charge because if there's a way to take the natural logarithm the charge would cancel.
 
The four capacitor are not connected in parallel in case "a".

Q(t)=Q0*(1-e-(t/RC) ) = Q0*(1-e-(t/T) ) is the formula for the time dependence of the charge on the capacitor. t is the time after connecting the capacitor to a voltage source. T is he time constant .

ehild
 
ah.. got it.. thanks.. that was easy...
 

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