In the previous RC Charging Circuit tutorial, we saw how a Capacitor, C charges up through the resistor until it reaches an amount of time equal to 5 time constants or 5T and then remains fully charged.
If this fully charged capacitor was now disconnected from its DC battery supply voltage it would store its energy built up during the charging process indefinitely (assuming an ideal capacitor and ignoring any internal losses), keeping the voltage across its terminals constant.
If the battery was now removed and replaced by a short circuit, when the switch was closed again the capacitor would discharge itself back through the resistor, R as we now have a RC discharging circuit. As the capacitor discharges its current through the series resistor the stored energy inside the capacitor is extracted with the voltage Vc across the capacitor decaying to zero as shown below.
As with the previous RC charging circuit, in a RC Discharging Circuit, the time constant ( τ ) is still equal to the value of 63%. Then for a RC discharging circuit that is initially fully charged, the voltage across the capacitor after one time constant, 1T, has dropped to 63% of its initial value which is 1 – 0.63 = 0.37 or 37% of its final value.
So now this is given as the time taken for the capacitor to discharge down to within 37% of its fully discharged value which will be zero volts (fully discharged), and in our curve this is given as 0.37Vc.
As the capacitor discharges, it loses its charge at a declining rate. At the start of discharge the initial conditions of the circuit, are t = 0, i = 0 and q = Q. The voltage across the capacitors plates is equal to the supply voltage and Vc = Vs. As the voltage across the plates is at its highest value maximum discharge current flows around the circuit.
With the switch closed, the capacitor now starts to discharge as shown, with the decay in the RC discharging curve steeper at the beginning because the discharging rate is fastest at the start and then tapers off as the capacitor looses charge at a slower rate. As the discharge continues, Vc goes down and there is less discharge current.
As with the previous charging circuit the voltage across the capacitor, C is equal to 0.5Vc at 0.7T with the steady state fully discharged value being finally reached at 5T.
For a RC discharging circuit, the voltage across the capacitor ( Vc ) as a function of time during the discharge period is defined as:
Just like the previous RC Charging circuit, we can say that in a RC Discharging Circuit the time required for a capacitor to discharge itself down to one time constant is given as:
Where, R is in Ω’s and C in Farads.
Then we can show in the following table the percentage voltage and current values for the capacitor in a RC discharging circuit for a given time constant.
|RC Value||Percentage of Maximum|
|0.5 time constant||0.5T = 0.5RC||60.7%||39.3%|
|0.7 time constant||0.7T = 0.7RC||49.7%||50.3%|
|1.0 time constant||1T = 1RC||36.6%||63.4%|
|2.0 time constants||2T = 2RC||13.5%||86.5%|
|3.0 time constants||3T = 3RC||5.0%||95.0%|
|4.0 time constants||4T = 4RC||1.8%||98.2%|
|5.0 time constants||5T = 5RC||0.7%||99.3%|
Note that as the discharging curve for a RC discharging circuit is exponential, for all practical purposes, after five time constants a capacitor is considered to be fully discharged.
So a RC circuit’s time constant is a measure of how quickly it either charges or discharges.
Calculate the RC time constant, τ of the following RC discharging circuit.
The time constant, τ is found using the formula T = R x C in seconds.
Therefore the time constant τ is given as: T = R x C = 100k x 22uF = 2.2 Seconds
a) What value will be the voltage across the capacitor at 0.7 time constants?
At 0.7 time constants ( 0.7T ) Vc = 0.5Vc. Therefore, Vc = 0.5 x 10V = 5V
b) What value will be the voltage across the capacitor after 1 time constant?
At 1 time constant ( 1T ) Vc = 0.37Vc. Therefore, Vc = 0.37 x 10V = 3.7V
c) How long will it take for the capacitor to “fully discharge” itself, (equals 5 time constants)
1 time constant ( 1T ) = 2.2 seconds. Therefore, 5T = 5 x 2.2 = 11 Seconds