Okay so the image i uploaded only because of the cirscuit never mind the text to it. So as some of you may have noticed in the past month i regularly have been posting questions and that's because i am not actualy learning this yet in university i havent yet reached that part, but im learning it anyway. And even if i was learning it, im not good at math and i dont understand things mathematicaly verry well. I like to have a good physical grasp on what is going on with the circuit. That being said, i dont mean to insult any mathematicians, but for this one i need a good physical explanation on why is it that way.
So the time constant of this circuit is Requivalent x C. Through calculations using Thevenin's equivalent circuit method i have arrived at this outcome - If you add a resistor in parallel to the capacitor will decrease the Time constant. And now i ask Why and some maybe will say " Because with resistor in parallel the equivalent resistance is now less than it was before with only one resistor'' but that is not an answer atleast for me.
The way i think is that the current through the capacitor deposits the charge on it so if the current through it decreases it should take longer to charge up to a specific voltage. This is why increasing the resistance increases the time constant. However, from that Thevenin equivalent circuit method it doesnt seem that way. If you have two resistors in series and you add another branch in parallel with the bottom resistor, you would expect that now the current through the bottom resistor will decrease abit, maybe in half if the resistance of the new branch is equal. And thats pretty much what would happen in a circuit such as the one above in the picture after a time when the capacitor no longer behaves like a short circuit. So what im saying is because of the 1k Ohm resistor the current through the capacitor should be less than if there was no 1kOhm therefore it would take more time for the capacitor to charge, but again according to the Thevenin equivalent circuit method thats not the way and it is infact the opposite. So what is happening, why adding resistors in parallel with the capacitor decreases the Time constant?

Thanks in advance

 

Without the 1k in the circuit, the voltage across the capacitor will rise more quickly but the voltage it is charging to is higher as well.

 

Oh im starting to grasp it. What you said is actualy what i thought of first and then later dismissed since raising or decreasing voltage cant possibly alter the time constant since increasing the voltage proportionaly increases the current. But in a voltage divider we are not raising or lowering the source voltage. Instead the capacitor would want to charge up to the source voltage however when it reaches the output voltage of the voltage divider current through the capacitor will cease to exist since there will be no more potential difference and the voltage divider is effectively fixing the voltage at this particular value. Is this the answer? parallel resistors aka Voltage divider formation decreases the time constant because it ''cuts'' the voltage on the capacitor before it reachest the source?

 

Thevenin works no matter what you connect to the output. So remove the capacitor and look back into terminals 'a' and 'b'. The open-circuit voltage is 10V * 1k/1.5k (the output potential divider) = 6.66V. And this is the final voltage the capacitor can charge to. Now short circuit the voltage source (mentally - not in fact!) and you have the two resistors in parallel. which equal 366 ohms. So the Thevenin model is 6.66V in series with 366 ohms so the capacitor will finally charge to 6.66 v with a time constant of 366 *1u.

 

Although this relies a bit more on the math, also consider the following.

No matter what resistance you put in parallel with the capacitor, you don't change the INITIAL current (rate of charge) because, initially, the capacitor has zero volts across it and so, initially, no current flows in any of the paths parallel to the resistor. So, in this circuit, the initial current is simply 10 V / 500 Ω = 20 mA.

Here is where the math comes in. In a first order capacitor circuit, the initial current is such that, if that current could be maintained throughout the charge (or discharge) process, the capacitor would go from its initial voltage to its final voltage in one time constant. Hence, if you know the initial current and the voltage difference between the initial and final states (and the capacitance), you have what you need to find the time constant. You simply use the current and the capacitance to determine the initial charge rate (volts/sec) and divide that into the voltage difference.

By putting additional resistors in parallel, what you change is the voltage span over which the capacitor charges. You make that smaller, so it doesn't take as long to reach it, so the time constant goes down.

While I definitely applaud your desire to understand what is happening on an intuitive/physical level (which more people made that effort), I also need to point out that math is the language of engineering in general and that applies in spades to electrical engineering. So if that is a weakness for you, you would be well served to put in particular efforts to strengthen those skills.

 

Sorry, hit the wrong key before - equivalent resistance is 333 ohms so with Thevenin equivalent voltage of 6.66 V.you get 20 mA initial current. But I disagree with WBahn - the time constant is for the capacitor to charge to 63.2% of its final voltage (see Wikipedia). In theory, the capacitor never reaches the final voltage, but it it within 2% of the final value after 4 time constants.

 

Sorry, hit the wrong key before - equivalent resistance is 333 ohms so with Thevenin equivalent voltage of 6.66 V.you get 20 mA initial current. But I disagree with WBahn - the time constant is for the capacitor to charge to 63.2% of its final voltage (see Wikipedia). In theory, the capacitor never reaches the final voltage, but it it within 2% of the final value after 4 time constants.

You can disagree all you want, that doesn't make what I said incorrect.

Th proof is in the pudding, as they say, so let's see how this works out.

In the circuit presented the initial current is 20 mA (regardless of the value of the parallel resistance, provided it is non-zero) and, with a 1 μF capacitor, that means that the initial charge rate is 1 μF / 20 mA which is 20 V/ms.

Now let's find the time constants for that circuit under three conditions:

1) No parallel resistor: Final voltage is 10 V so τ = 10 V / (20 V/ms) = 500 μs.
2) The circuit as shown: Final voltage is 6.67 V, so τ = 6.67 V / (20 V/ms) = 333 μs.
3) The circuit with an additional 1 kΩ resistor put in parallel: Final voltage is 5 V, so τ = 5 V / (20 V/ms) = 250 μs.

Now, analyze the circuit however you want and find the time constants for each of these circuits.

 

Sorry, hit the wrong key before - equivalent resistance is 333 ohms so with Thevenin equivalent voltage of 6.66 V.you get 20 mA initial current. But I disagree with WBahn - the time constant is for the capacitor to charge to 63.2% of its final voltage (see Wikipedia). In theory, the capacitor never reaches the final voltage, but it it within 2% of the final value after 4 time constants.

WBhan states that the capacitor will fully charge in one time constant IF the initial charging current can be maintained/constant throughout the entire charging period.

Yes, in most applications (at least those that I have done little research on) the current is not maintained when charging a capacitor. In this case, yes, the time to charge/discharge a capacitor takes about 5 time constants.

I think that he wrote that initially in order to help the OP gain a better qualitative understanding of the matter in question, but I could be wrong.

 

Putting a resistor in parallel reduces the thevenin resistance reducing the time constant. It's been a long time but I remember using an exponential function that describes charging and discharging of a capacitor.

 

Putting a resistor in parallel reduces the thevenin resistance reducing the time constant.

Yes, it does. I don't think the TS is questioning the validity of the math, but just asking for a non-mathematical explanation of why it is valid because, to him, that result seemed non-intuitive and contradictory.

It's been a long time but I remember using an exponential function that describes charging and discharging of a capacitor.

For a first-order circuit that is correct. The capacitor voltage and current change from an initial state to a final state. The rate of change exhibits a decaying exponential behavior.

 

See

 

neweare, you are correct. In the circuit shown, the capacitor voltage is given by:
Vc = Vs(1 - exp(-t/RC))
where, Vs is the supply voltage.By definition, the time constant is the product of Resistance x Capacitance. So in this case we have the equivalent resistance of 333 ohms and a time constant of 333 us. WBhan is taking the rather unusual situation of a constant current supply, rather than the constant voltage as in the diagram, and yes, in that situation, (thanks BearSub for pointing that out) he is correct - I read his port too quickly!

 

Tau=(500Ohm/q)||1kOm*1u=0.5ms

 

neweare, you are correct. In the circuit shown, the capacitor voltage is given by:
Vc = Vs(1 - exp(-t/RC))
where, Vs is the supply voltage.By definition, the time constant is the product of Resistance x Capacitance. So in this case we have the equivalent resistance of 333 ohms and a time constant of 333 us. WBhan is taking the rather unusual situation of a constant current supply, rather than the constant voltage as in the diagram, and yes, in that situation, (thanks BearSub for pointing that out) he is correct - I read his port too quickly!

No, I'm NOT taking the situation of a constant current source (though that would have that result).

I'm making an observation about the relationship between the time constant and the INITIAL capacitor current. That's ALL!

For a first-order RC circuit you have the capacitor voltage being

\(
V_c \; = \; V_{fin} \; + \; \( V_{ini} \; - \; V_{fin} \) e^{-\frac{t}{\tau}}
\)

There is NOTHING in this about what time of circuit is driving the system. Put it in a black box and use the Thevenin equivalent of it.

What is the rate of change of the capacitor voltage?

\(
\frac{dV_c}{dt} \; = \; \frac{\( V_{fin} \; - \; V_{ini} \)}{\tau} e^{-\frac{t}{tau}}
\)

What is the current in the capacitor?

The defining relation for a capacitor is Q = CV. Differentiating this we get

\(
i_c(t) \; = \; C \frac{dV_c}{dt}
\)

which means that

\(
i_c(t) \; = \; C\frac{\( V_{fin} \; - \; V_{ini} \)}{\tau} e^{-\frac{t}{tau}}
\)

What do we get when we evaluate this at t = 0?

\(
I_{c0} \; = \; i_c(t=0) \; = \; C\frac{\( V_{fin} \; - \; V_{ini} \)}{\tau}
\)

What do we get when we solve this for the time constant?

\(
\tau \; = \; C\frac{\( V_{fin} \; - \; V_{ini} \)}{I_{c0}}
\)

Now, how does that compare to what I said, which was

Hence, if you know the initial current and the voltage difference between the initial and final states (and the capacitance), you have what you need to find the time constant. You simply use the current and the capacitance to determine the initial charge rate (volts/sec) and divide that into the voltage difference.

The relevance to the TS's query (albeit it more mathematical than he was probably looking for, but still a different perspective that might make a bell or two ring) is that changing the parallel resistance has no effect on the INITIAL capacitor current, but it DOES reduce the difference between initial and final voltages. Thus, shunting current away from the capacitor actually decreases the time constant.

From a practical standpoint, it can allow you to find (or verify) a time constant without having to do as much analysis because finding only the INITIAL current and the INITIAL and FINAL voltages are often easier than finding the Thevenin resistance.

EDIT: Corrected typo -- changed "parallel capacitance" to "parallel resistance" in the next to last paragraph.

 

Hi Bordodynov
I simulated it directly in LTSpice with a dc source of 10V BUT you need to open - Edit the Simulation Command - and tick the box to set all power supplies to 0v.

 

Hi Bordodynov
I simulated it directly in LTSpice with a dc source of 10V BUT you need to open - Edit the Simulation Command - and tick the box to set all power supplies to 0v.

If you carefully looked at my scheme, you should have noticed: .tran 3m uic
It's enough.

 

You have my thanks!

 

The relevance to the TS's query (albeit it more mathematical than he was probably looking for, but still a different perspective that might make a bell or two ring) is that changing the parallel capacitance has no effect on the INITIAL capacitor current, but it DOES reduce the difference between initial and final voltages. Thus, shunting current away from the capacitor actually decreases the time constant.

For the avoidance of confusion: "changing the parallel capacitance" should read "changing the parallel resistance"

 

I did look at you scheme - it's just an alternative.

 

I guess the point of the thread is to get a intuitive "feel" for whats going on. Not always easy when your just learning about a subject thats not as concrete as say mechanics. I went to school for EE but I suck at circuits and never had the natural insight but was quick to understand software. So I understand where the OP is coming from.