Capacitors and Dialectric Constant

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SUMMARY

The discussion focuses on calculating the time constants for two different configurations of four identical capacitors connected with a resistor. In the first configuration, the time constant is established as 0.67 seconds. The calculations reveal that the time constant for the second configuration is three times that of the first, resulting in a time constant of 2.01 seconds. The key to solving the problem lies in accurately determining the equivalent capacitance for each circuit arrangement.

PREREQUISITES
  • Understanding of capacitor configurations (series and parallel)
  • Knowledge of the time constant formula (T=RC)
  • Ability to calculate equivalent capacitance for series and parallel combinations
  • Familiarity with basic circuit analysis principles
NEXT STEPS
  • Study the calculation of equivalent capacitance in series and parallel circuits
  • Learn about the implications of time constants in RC circuits
  • Explore practical applications of capacitors in electronic circuits
  • Investigate the effects of different resistor values on time constants
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Electrical engineering students, circuit designers, and anyone involved in analyzing or designing RC circuits will benefit from this discussion.

kris24tf
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Capacitors, Hard Problem

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 0.67 s. What is the time constant when they are connected with the same resistor as in part b?

The pictures included are http://edugen.wiley.com/edugen/courses/crs1000/art/qb/6e/ch20p_99.gif

I really have no idea what I'm doing here. My work is as follows:

T=RC
Ta=RCa, Tb=RCb
Ca=Cs1+Cs2
1/Cs=1/C+1/C
1/2/C=1/2C=2C
1/Cb=1/Cp+1/C
Cb=1/(1/Cp+1/C)
Cp=C+C=2C
Cb=1/(3/2C)=s/3C
Ta=R(2)(C)
Tb=R(2/3)(C)
Tb/Ta=R(2)(C)/R(2/3)(C)
Tb=3Ta

If anyone could help me with what I did wrong/what I need to fix, I'd appreciate it
 
Last edited:
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I didn't work through your equations, but basically you need to calculate the total capacitance in each situation to determine the RC time constant.

For the first circuit, the total combination of capacitance is C, right? Each leg is half of C, and the two parallel C/2 legs together gives you back a whole C, right? So the time constant for the first circuit is RC.

You need to then calculate the total combination of capacitance for the second circuit. The two caps in parallel are easy -- what is the combination of the two C caps in parallel? And then you end up with 3 caps in series -- the combo of the two parallel caps, plus the two C caps. What is the equation for combining caps in series? Just combine the equivalent capacitance from the two parallel C caps and the other two series C caps, and you will get an equivalent total capacitance. That will give you the relative time constant compared to the original RC from the first circuit. Makes sense?
 
I'm still confused. I tried working through what you said but I don't come up with anything...
 
You've got parts of it right in the work that you showed.

What is the equivalent capacitance of the caps in the first circuit arrangement? (parallel combo of two series C)

What is the equivalent capacitance of the caps in the second circuit arrangement? (two parallel C then in series with two other C)

What is the ratio between these two equivalent capacitance values? That will be the ratio of the time constants...
 

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