# RC discharging and ohms law

#### Drmario5237

Joined Oct 14, 2018
65
Hello I understand that t=R*C as to get the time constant of how long it takes a capsitor to drain though a resistor with voltage and current lowering exponentially over that period of time but do I still use ohms law (I=V/R) to get the intial current that the capsitor is draining though the resistor where V equals the supply voltage or what the capsitor is charged to the r represents the resistance of the resistor and the I represents the amount of current I'm getting from the capsitor and resistor keeping in mind the voltage is changing on me as the capsitor discharges and the capsitance of the capsitor is the amount of charge stored up that that it is expontially discharging from. Thanks.

#### shteii01

Joined Feb 19, 2010
4,647

#### shteii01

Joined Feb 19, 2010
4,647
Last edited:

#### MrAl

Joined Jun 17, 2014
6,605
Hello I understand that t=R*C as to get the time constant of how long it takes a capsitor to drain though a resistor with voltage and current lowering exponentially over that period of time but do I still use ohms law (I=V/R) to get the intial current that the capsitor is draining though the resistor where V equals the supply voltage or what the capsitor is charged to the r represents the resistance of the resistor and the I represents the amount of current I'm getting from the capsitor and resistor keeping in mind the voltage is changing on me as the capsitor discharges and the capsitance of the capsitor is the amount of charge stored up that that it is expontially discharging from. Thanks.
Hi,

The short answer is yes, if you need that current that is.

The voltage goes down to about 0.36787944 times the initial voltage after 1 time constant.
The voltage then goes down by another 0.36787944 after 2 time constants.
The voltage then goes down by another 0.36787944 after 3 time constants.
Etc., etc.
So for each additional time constant, multiply the previous result by 0.36787944 to get the voltage at the next time constant.

#### MrChips

Joined Oct 2, 2009
19,379
The answer is yes, Ohm's Law still applies for the resistor.
Note that the voltage across the resistor is used.
Hence if the voltage Vc is across the capacitor then the instantaneous discharge current I = Vc / R.

For a resistor in series between the supply voltage Vs and the capacitor being charged, the instantaneous charging current would be I = (Vs - Vc) / R.

#### Drmario5237

Joined Oct 14, 2018
65
Hi,

The short answer is yes, if you need that current that is.

The voltage goes down to about 0.36787944 times the initial voltage after 1 time constant.
The voltage then goes down by another 0.36787944 after 2 time constants.
The voltage then goes down by another 0.36787944 after 3 time constants.
Etc., etc.
So for each additional time constant, multiply the previous result by 0.36787944 to get the voltage at the next time constant.
Is the same drop measurement the same for current (0.36787944) and how did you arrive at that number?

#### shteii01

Joined Feb 19, 2010
4,647
You can use Ohm's Law for the value at that specific instant.
Is the same drop measurement the same for current (0.36787944) and how did you arrive at that number?
"RC time constant. It is the time required to charge the capacitor, through the resistor, from an initial charge voltage of zero to approximately 63.2% of the value of an applied DC voltage, or to discharge the capacitor through the same resistor to approximately 36.8% of its initial charge voltage."
https://en.wikipedia.org/wiki/RC_time_constant

36.8% is 0.368 which is what 0.36787944 is when you round it up to 3 decimal places.

#### MrAl

Joined Jun 17, 2014
6,605
Is the same drop measurement the same for current (0.36787944) and how did you arrive at that number?
It is simply:
e^(-1)

or of course:
1/e

where 'e' is the base of the natural log system.
It is approximately: 2.7182818

#### AnalogKid

Joined Aug 1, 2013
8,137
Hello I understand that t=R*C as to get the time constant of how long it takes a capsitor to drain though a resistor with voltage and current lowering exponentially over that period of time but do I still use ohms law (I=V/R) to get the intial current that the capsitor is draining though the resistor
Yes. At any instant in time, the current through the resistor is equal to the instantaneous voltage divided by the resistance. As the capacitor continues to discharge, the voltage is decreasing continuously. But instant to instant, Ohm's law applies.

ak