Capacitor discharge time problem.

Thread Starter

quest271

Joined Apr 1, 2016
23
I am studying for CET certification. I am hung up on a practice problem envolving capacitor discharge time. Here is the question. A 0.09 microfarad capacitor is charged to 220 Volts. How long in milliseconds will it discharged to a level of 110 V if the discharged resistor has a resistance of 20, 000 ohms? The answer is 1.25 milliseconds. I can up with 9 milliseconds. This is what I did. t=rc 20k times 0.09 micro farads then I multiplied by 5 time constants.
 

Ian0

Joined Aug 7, 2020
3,517
Why five time constants?
You need to start with the equation for the discharge of a capacitor
\(
V=V_0e^{\frac{-t}{RC}}
\)
 

dl324

Joined Mar 30, 2015
13,320
Welcome to AAC!

Why did you choose 5 time constants? We use that number for calculating complete discharge time.
 

Thread Starter

quest271

Joined Apr 1, 2016
23
I am studying for CET certification. I am hung up on a practice problem envolving capacitor discharge time. Here is the question. A 0.09 microfarad capacitor is charged to 220 Volts. How long in milliseconds will it discharged to a level of 110 V if the discharged resistor has a resistance of 20, 000 ohms? The answer is 1.25 milliseconds. I can up with 9 milliseconds. This is what I did. t=rc 20k times 0.09 micro farads then I multiplied by 5 time constants.
I assumed five time constants were used.
 

Thread Starter

quest271

Joined Apr 1, 2016
23
I am studying for CET certification. I am hung up on a practice problem envolving capacitor discharge time. Here is the question. A 0.09 microfarad capacitor is charged to 220 Volts. How long in milliseconds will it discharged to a level of 110 V if the discharged resistor has a resistance of 20, 000 ohms? The answer is 1.25 milliseconds. I can up with 9 milliseconds. This is what I did. t=rc 20k times 0.09 micro farads then I multiplied by 5 time constants.
Do you have to solve that equation for the t in the numerator of -t/rc ?
 

Papabravo

Joined Feb 24, 2006
17,028
OK so the algebra looks like this

\( 110\;=\;220e^{\left(-\cfrac{t}{RC}\right)} \)
\( ln(0.5)\;=\;\cfrac{-t}{RC} \)
\( -0.693\;=\;-\cfrac{t}{RC} \)
\( t\;=\;0.693RC \)

On my calculator the right answer drops out. That is approximately 70% of ONE time constant
 
Last edited:

Thread Starter

quest271

Joined Apr 1, 2016
23
OK so the algebra looks like this

\( 110\;=\;220e^{\left(-\cfrac{t}{RC}\right)} \)
\( ln(0.5)\;=\;\cfrac{-t}{RC} \)
\( -0.693\;=\;-\cfrac{t}{RC} \)
\( t\;=\;0.693RC \)

On my calculator the right answer drops out. That is approximately 70% of ONE time constant
Thanks for your help. It makes sense now.
 

Thread Starter

quest271

Joined Apr 1, 2016
23
Since everyone is giving you equations...

You know that:
\(V_f=V_ie^{\frac{-t}{RC}}\)

Solving for t, you get:
\( t=-RCln(\frac{V_f}{V_i})\)
I know how to solve for t . I was unsure if I was on the right path to getting the answer. Thanks again.
 

MrAl

Joined Jun 17, 2014
8,504
I know how to solve for t . I was unsure if I was on the right path to getting the answer. Thanks again.
The only thing you did wrong was assumed that it took 5 full time constants to reach the voltage they gave as the final voltage (110v). After 5 time constants the voltage would reach a level much lower than that and the usual way of describing that is that after 5 time constants the voltage reaches a level below 1 percent of the starting voltage. 1 percent of 220v is 2.20v and that is much lower than 110v so it 'cant be 5 time constants it must be less. If you calculate the correct time you can express that in terms of the time itself in milliseconds or you could express it in terms of the time constants. 1 time constant takes the voltage down to about 37 percent of the starting voltage and since that would still be less than 110v then that would mean the time must be even less than 1 full time constant.

The above kind of reasoning is what we sometimes call a "sanity check" because it gives us a secondary way to calculate the answer in rough terms and that tells us if our more accurate calculation is within reason.
 

Papabravo

Joined Feb 24, 2006
17,028
The only thing you did wrong was assumed that it took 5 full time constants to reach the voltage they gave as the final voltage (110v). After 5 time constants the voltage would reach a level much lower than that and the usual way of describing that is that after 5 time constants the voltage reaches a level below 1 percent of the starting voltage. 1 percent of 220v is 2.20v and that is much lower than 110v so it 'cant be 5 time constants it must be less. If you calculate the correct time you can express that in terms of the time itself in milliseconds or you could express it in terms of the time constants. 1 time constant takes the voltage down to about 37 percent of the starting voltage and since that would still be less than 110v then that would mean the time must be even less than 1 full time constant.

The above kind of reasoning is what we sometimes call a "sanity check" because it gives us a secondary way to calculate the answer in rough terms and that tells us if our more accurate calculation is within reason.
It is hard to know those things for a sanity check if you have never solved an exponential equation. Useful for those members of the cognoscenti who have, useless otherwise.
 
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