When a voltage source is removed from a fully charged RC circuit, the capacitor, C will discharge back through the resistance, R
RC discharging circuits use the inherent RC time constant of the resisot-capacitor combination to discharge a cpacitor at an exponential rate of decay.
In the previous RC Charging Circuit tutorial, we saw how a Capacitor charges up through a resistor until it reaches an amount of time equal to 5 time constants known as 5T. It then remains fully charged as long as a constant supply is applied to it.
If this fully charged capacitor is now disconnected from its DC battery supply voltage, the stored energy built up during the charging process would stay indefinitely on its plates, (assuming an ideal capacitor and ignoring any internal losses), keeping the voltage stored across its connecting terminals at a constant value.
If the battery was replaced by a short circuit, when the switch is closed 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 we saw in the previous tutorial, 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 by 63% of its initial value which is 1 – 0.63 = 0.37 or 37% of its final value.
Thus the time constant of the circuit is given as the time taken for the capacitor to discharge down to within 63% of its fully charged value. So one time constant for an RC discharge circuit is given as the voltage across the plates representing 37% of its final value, with its final value being zero volts (fully discharged), and in our curve this is given as 0.37Vs.
As the capacitor discharges, it does not lose its charge at a constant rate. At the start of the discharging process, 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 at t = 0 across the capacitors plates is at its highest value, maximum discharge current therefore flows around the RC circuit.
When the switch is first closed, the capacitor starts to discharge as shown. The rate of decay of the RC discharging curve is steeper at the beginning because the discharging rate is fastest at the start, but then tapers off exponentially as the capacitor looses charge at a slower rate. As the discharge continues, VC reduces resulting in less discharging current.
We saw in the previous RC charging circuit that 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 and C in Farads.
Thus 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.
Time Constant |
RC Value | Percentage of Maximum | |
Voltage | Current | ||
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 | 36.8% | 63.2% |
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 decaying curve for a RC discharging circuit is exponential, for all practical purposes, after five time constants the voltage across the capacitor’s plates is much less than 1% of its inital starting value, so the capacitor is considered to be fully discharged.
So an RC circuit’s time constant is a measure of how quickly it either charges or discharges.
A capacitor is fully charged to 10 volts. Calculate the RC time constant, τ of the following RC discharging circuit when the switch is first closed.
The time constant, τ is found using the formula T = R*C in seconds.
Therefore the time constant τ is given as: T = R*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, (equal to 5 time constants)
1 time constant ( 1T ) = 2.2 seconds. Therefore, 5T = 5 x 2.2 = 11 Seconds
Good
Printed value in Capacitor C=1F. 5.5V
How much time is required for charging a capacitor when using resistance 1000ohm through 10V battery??
If capacitor has 5.5v then can I charge this capacitor through 10 v battery by adjusting resistance??
You’ll only learn if you do the math yourself, and not have others do your homework.
Clearly, if 1T = CR = 1000s, then 5T = 5000s. No, if the DC working voltage of the capacitor is 5.5V, then that is the maximum voltage which can be applied.
Charging of capacitor when Vs>VC
For complete charging of capacitor is it depends on source voltage??
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What value of τ will you choose when you are designing a DC power supply from a rectifier circuit usinga capacitor? Suppose, the supply frequency of the input of the rectifier circuit is 50 Hz.
Greater than 5T
)A 1000 μF capacitor wired in a simple series RC circuit is initially charged to 20 μC and then
discharged through a 10 kΩ resistor.
(a) What is the time constant of the circuit?
(b) What is the initial current that flows?
(c) How much charge is left on the capacitor after 6 time constant?
(d) What is the current after 6 time constant?
The discharge current in the table indicates it starts with 0 and goes to 100%. It is misinterpretation. Current will start at maximum of V/R and decay towards 0 as Capacitor discharges, but in the reverse direction.
If the current is 0% in the beginning, there is no possibility for discharging.
The exponential decay of the current through a capacitor in an RC circuit shows it will reach 100% of its decaying value after 5T. That is, it will be fully discharged after 5T = 5RC. Therefore the tutorial is correct as given.
Kindly refer to the graph that shows current changing from -V/R to 0 in 5T
Exactly, 100% discharged at 5T
Best explanation
Sir, thank nice tutorial notes please send me my email ID sir
2200 microF capacitor is charged through 2.2 k oms resistor with help 12v battery.plz tell me the voltage across the capacitor at 2nd time constant.i want to know how i find it.
Thank you
How did you get the value of Vc?
I have an old fused capacitor which the discharge through the resistor is 48- 19182-02 I’m trying to replace it with a conventional modern capacitor but I don’t know what the microfaring rating is or the UF rating is for a total overall capacity
give more heavy example
….
Brief Fall
If I have a square wave negative pulse, from a suitable device, eg a 555, will the (brief) pulse be delayed by the RC time value? Or will the RC to 0v simply remove the pulse?
In rc circuit I have connected resistor of value 1 megha ohm and cap value is 25v 100 micro farad, cap started charging slowly after some time the cap value is equal to battery voltage (9v dc).when I check the voltage accross cap with multimeter it shown voltage of 9v but constantly the charge reduced to 4.5 where it got stabilized .but when I kept another resistor to see what happens these time I have kept 200 ohm cap got charged to the battery voltage but when I checked the voltage is showing 9 v but these time voltage stayed at that position only .to check further I removed cap and removed wire from one end of resistor which is connected to the negative of battery ,the 200 ohm resistor it given a value of 9v and the 1 megha ohm showed the voltage of 4.5 can please any one explain these
Is it not obvious that the multimeter will have an input impedance of about 1 megaohm, so connecting it across the capacitor effectively connects the meter in series with the 1 megaohm resistor creating a voltage divider across the 9 volt supply with a center voltage of: 9V x (Rmeter/(1MΩ + Rmeter)) = 9*0.5 = 4.5V and it is this voltage that the capacitor charges too. Reducing the resistor value to 200 ohms gives: 9V x (Rmeter/(200Ω + Rmeter)) = 9*0.9998 = 8.99V.
I would like to know how long a 100000 uF cap. could hold it’s full charge with no load, jerrw
Depends on many factors including internal leakage, but if pure, indefinitely