# graph of time constant rc network

Discussion in 'Electronics Homework Help' started by bhuvanesh, Apr 20, 2014.

1. ### bhuvanesh

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Aug 29, 2013
in the below image u see that time constant is 0.37 of initial value and 63% of final value.if that so see the second graph in that image the point should be on 0.37 (from initial value) but actually the time constant point is on 0.63.
this is messing with my i dont know what i misunderstood.

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2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
In both cases, after 1 RC time interval the voltage has reached initial + 0.63*(final - initial)

5,165
1,087
Dec 18, 2013
Yes and that is because e^-1 = 0.36 which is 36% when -Tau/RC =-1
I thinks that's right Steve?

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
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Jan 21, 2010
Yeah, the whole point is if you compare conditions 1 RC time interval apart, the second one will be 63% of the way to the steady state condition compared to the first (assuming conditions were constant during that period)

If you turn one of those graphs upside down you will find it matches precisely with the other.

In fact it gets even better. If you take either graph and make a copy where the Y axis is scaled, you will find a place where the 2 graphs can be overlapped for an exact match.

Last edited: Apr 21, 2014

5,165
1,087
Dec 18, 2013
Nice one

6. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Your first misunderstanding is in the definition of "time constant".

The time constant of an RC circuit is defined like this:

t = R C

In other words, t, the time constant of an RC circuit is equal to the resistance (in ohms) multiplied by the capacitance (in farads). Its unit is seconds.

Whether the capacitor is being discharged (as in the first graph), or charged up (as in the second graph), the time constant is the same. The time constant depends only on the resistor and the capacitor.

The reason why we talk about time constants is that any combination of resistor and capacitor values that has the same time constant behaves exactly the same (assuming there are no other resistances or capacitances in the circuit, i.e. pretending that the measuring equipment does not load the circuit in any way).

For example, you could make up three RC circuits.

RC circuit 1: R = 100 ohms; C = 100 µF.
RC circuit 2: R = 5k; C = 2 µF.
RC circuit 3: R = 20k; C = 500 nF.

Use your calculator to calculate the time constants of these three circuits using the formula t = R C. Remember to handle the units properly.

You should find that all three circuits have time constants of 0.01 seconds (10 ms). This means that they will produce IDENTICAL charge and discharge waveforms; the amount of time taken to reach each point on the charge or discharge graph will be the same for all three circuits.

So the reason for describing an RC circuit by its time constant is that it gives you a single number that tells you the most important thing about that combination of resistance and capacitance.

Now, the numbers 63% and 37% relate to points on those graphs. Let's take them one at a time.

Say we have a resistor and a capacitor connected in series. Their time constant t is 10 ms, so they could be any of the three circuits from before, or any other combination of R and C that produce a result of 0.01 when multiplied together.

Initially, the capacitor is completely discharged, i.e. there is 0V across it. But we can close a switch and connect a voltage source, VS, to the series RC circuit.

(That image is from http://people.sinclair.edu/nickreeder/eet155/mod02.htm which looks like quite a good article for you to read.)

VS is a 100V DC power source. When we close the switch, current will flow from the battery, through the switch, through the resistor, through the capacitor, and back to the battery.

This current flow will cause the capacitor to charge up. In other words, the voltage across the capacitor will increase.

You can see how the voltage across the capacitor (VC) increases over time in the second graph in your post #1 on this thread. To start with, it increases quite quickly (the steep line at the left side of the graph), but as VC gets higher, its rate of rise decreases, in other words, it rises more and more gradually.

After 10 ms, VC will have reached about 63V.

The shape of that graph is the same for ALL RC circuits - that is, for ANY values of R and C. The only difference is the time scale. If R and C are small, e.g. 100 ohms and 100 pF, Vc will reach almost 100V in a very short time - less than 1 µs (one millionth of a second). But if R and C are large, e.g. 100k and 470 µF, the time taken for Vc to reach nearly 100V will be much longer - several minutes, in fact.

Remember, we can simplify R and C into a single number, t, measured in seconds, which is equal to R multiplied by C. We can also relate this value, t, to the graph.

So, when the switch is closed, VC starts to increase towards 100V. After t seconds, VC will be 63V, i.e. 63% of the applied voltage VS.

So that is what t means. It is a time period, measured in seconds, equal to R multiplied by C, and when a voltage VS is applied to the RC circuit with C initially discharged, t is the amount of time that it will take for VC to reach 63% of VS.

The left hand graph in your post shows the discharge of the capacitor. Initially, the capacitor is charged - i.e. VC is some value - let's use 100V again. Then at the left side of the graph, the resistor is connected across the charged capacitor, and VC starts to drop from 100V towards 0V. It drops quite quickly at first, but as it drops, its rate of fall decreases, in other words it falls more and more gradually. And after a certain time, VC will have fallen to 37% of its initial value; this time interval is t, which can be calculated by multiplying R and C together.

You can see that the two graphs are just upside down versions of the same shape. That's just because the right one shows the capacitor charging up from 0V towards the applied voltage, and the left one shows the capacitor discharging from an initial voltage towards 0V.

In fact, both graphs will look exactly the same if you graph the voltage across the resistor, instead of across the capacitor. They will look like the left hand graph. In both cases (charge and discharge of the capacitor), the voltage across the resistor starts at the high value, and starts dropping quite quickly, and its rate of fall becomes more gradual as the voltage drops.

Actually, the rate of fall (the angle of the line) is directly proportional to the voltage, at every point of the curve. This makes sense when you realise that this angle is proportional to the current (from the formula dV/dT = I/C) and the current through a resistor is proportional to the voltage across it.

I can go into more detail if you want. See if you can understand all of that first. Read it over until you can follow what I'm getting at. If you get stuck at some point, let me know what you don't understand, and I'll try to explain it better.

Last edited: Apr 22, 2014