How to calculate the initial rate of dischrage of a capacitor.

AI Thread Summary
To calculate the initial rate of discharge of a 9.9 microF capacitor charged to 10 V through a DMM with 10M ohms internal resistance, the current (I) is found using Ohm's law, yielding 1 microampere. This current translates to a discharge rate of 1 micro-Coulomb per second. If this rate is maintained for 3 seconds, the capacitor would discharge 3 micro-Coulombs, which is approximately 7.6% of its total charge. The calculations confirm the relationship between current, charge, and time, leading to a clearer understanding of capacitor discharge dynamics. The discussion emphasizes the importance of correctly identifying resistance values and applying relevant equations.
mike88si
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Homework Statement



A) A DMM with its 10M(ohms symbol) internal resistance is connected to a 9.9 microF capacitor that has been charged to 10 V. Calculate the intial rate of discharge of the capacitor through the DMM in micro-Coulombs per second. Hint: You do not need exponentials for this calculations.

B) Assuming this initial rate of discharge in the question above is maintained for 3 seconds, by what percentage would the capacitor be discharged?


Homework Equations



How do I calculate the initial rate of discharge? My lab did not go over it but I Found this formula online but it only goes over time.

T = ( C * V ) / I


The Attempt at a Solution



I = V/R
= (10 / 10
= 1

T = ( C * V ) / I
= (3.9 * 10) / 1
= 39 seconds

So the rate is .1 microF a second?
And if it were discharged for 3 seconds it would be .3 which is 7.6% of the charge discharged.


I am lost! Any help is greatly appreciated, thanks!
 
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mike88si said:

Homework Statement



A) A DMM with its 10M(ohms symbol) internal resistance is connected to a 9.9 microF capacitor that has been charged to 10 V. Calculate the intial rate of discharge of the capacitor through the DMM in micro-Coulombs per second. Hint: You do not need exponentials for this calculations.

B) Assuming this initial rate of discharge in the question above is maintained for 3 seconds, by what percentage would the capacitor be discharged?


Homework Equations



How do I calculate the initial rate of discharge? My lab did not go over it but I Found this formula online but it only goes over time.

T = ( C * V ) / I


The Attempt at a Solution



I = V/R
= (10 / 10
= 1

T = ( C * V ) / I
= (3.9 * 10) / 1
= 39 seconds

So the rate is .1 microF a second?
And if it were discharged for 3 seconds it would be .3 which is 7.6% of the charge discharged.


I am lost! Any help is greatly appreciated, thanks!

You are close when you say I = 10V / 10Meg Ohms (you wrote 10 Ohms, BTW).

After fixing the missing "Meg", you would get I = 1uA

Now you need to convert that into coulombs per second. What is the relationship between Amps and Coulombs per second?
 
mike88si said:

Homework Statement



A) A DMM with its 10M(ohms symbol) internal resistance is connected to a 3.9 microF capacitor that has been charged to 10 V. Calculate the intial rate of discharge of the capacitor through the DMM in micro-Coulombs per second. Hint: You do not need exponentials for this calculations.

B) Assuming this initial rate of discharge in the question above is maintained for 3 seconds, by what percentage would the capacitor be discharged?


Homework Equations



How do I calculate the initial rate of discharge? My lab did not go over it but I Found this formula online but it only goes over time.

T = ( C * V ) / I


The Attempt at a Solution



I = V/R
= (10 / 10
= 1

T = ( C * V ) / I
= (3.9 * 10) / 1
= 39 seconds

So the rate is .1 microF a second?
And if it were discharged for 3 seconds it would be .3 which is 7.6% of the charge discharged.


I am lost! Any help is greatly appreciated, thanks!

berkeman said:
You are close when you say I = 10V / 10Meg Ohms (you wrote 10 Ohms, BTW).

After fixing the missing "Meg", you would get I = 1uA

Now you need to convert that into coulombs per second. What is the relationship between Amps and Coulombs per second?


The relationship between coulombs per second and amps is that one Amp is one coulomb per second. I think I need this equation:

Q= I * t

t = Q / I also I = Q/t

So to find Q

C = Q/V so we get Q = CV

Q = ( 3.9 * 10^-6 * 10 V ) I = 10/10*10^6
= 3.9*10^-5 couloumbs = 1*10^-6

t = Q / I
= 3.9*10^-5 couloumbs / 1 * 10^-6 couloumbs/second
= 39 seconds

I just realized that what I just did is pretty much the same as above.
 
Better is I = \frac{dQ}{dt}
 
thanks. i think i got it. turned it in yesterday.
 
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