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

• mike88si
In summary, the initial rate of discharge for a 3.9 microF capacitor charged to 10V through a DMM with a 10M ohms internal resistance is 1uA, or 1 coulomb per second. If this rate is maintained for 3 seconds, the capacitor will be discharged by 7.6% of its initial charge. The relationship between amps and coulombs per second is that one amp is equal to one coulomb per second. The equation used to calculate this is I = dQ/dt.
mike88si

## 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!

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.

## 1. How do I calculate the initial rate of discharge of a capacitor?

To calculate the initial rate of discharge of a capacitor, you will need to know the capacitance (C) of the capacitor, the voltage (V) across the capacitor, and the resistance (R) in the circuit. The formula for initial discharge rate is given by V/R*C.

## 2. Can I calculate the initial rate of discharge without knowing the capacitance?

No, the capacitance is a crucial component in calculating the initial rate of discharge. It determines how much charge the capacitor can hold and how quickly it will discharge.

## 3. How does the initial rate of discharge affect the overall discharge of a capacitor?

The initial rate of discharge determines how quickly the capacitor will discharge its stored energy. A higher initial discharge rate means the capacitor will discharge more quickly, while a lower initial discharge rate means it will take longer to discharge.

## 4. Is the initial rate of discharge affected by the voltage or resistance in the circuit?

Yes, the initial rate of discharge is affected by both the voltage and resistance in the circuit. A higher voltage will result in a higher initial discharge rate, while a higher resistance will result in a lower initial discharge rate.

## 5. What units are used to measure the initial rate of discharge?

The units for initial rate of discharge are amperes (A) or coulombs per second (C/s). This value represents the amount of charge flowing through the circuit per unit of time during the initial discharge of the capacitor.

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