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

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SUMMARY

The initial rate of discharge of a capacitor through a Digital Multimeter (DMM) with a 10 MΩ internal resistance and a 9.9 µF capacitor charged to 10 V is calculated using Ohm's Law and the relationship between current and charge. The current (I) is determined to be 1 µA, leading to a discharge rate of 0.1 µF per second. If this rate is maintained for 3 seconds, the capacitor discharges approximately 7.6% of its charge.

PREREQUISITES
  • Understanding of Ohm's Law (I = V/R)
  • Knowledge of capacitor charge calculations (Q = CV)
  • Familiarity with the relationship between current and charge (1 Amp = 1 Coulomb/second)
  • Basic concepts of electrical discharge in capacitors
NEXT STEPS
  • Learn about capacitor discharge time constants and exponential decay.
  • Explore the implications of internal resistance in measuring circuits.
  • Study the effects of varying capacitance and voltage on discharge rates.
  • Investigate practical applications of capacitors in electronic circuits.
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Students in electrical engineering, electronics enthusiasts, and anyone involved in circuit design or analysis who seeks to understand capacitor discharge behavior.

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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