Time constant

[sunny1982]

A 100 µF capacitor is connected in series with an 8000Ω resistor. Determine the time constant of the circuit.
If the circuit is suddenly connected to a 100 dc supply find

The initial rise of P.D across the capacitor, (rate of voltage rise per volt to the first time constant)

T= CR= 100µF * 10¯⁶ *8000Ω = 8 seconds

63.2% of 100v = 63.2v

V/T =63.2v/8S = 7.9v rise per volt to first time constant

The initial charging current, (at zero time)

I = V/R = 100v/8000Ω = 0.0125a

The ultimate charge in the capacitor

Q = C V = 100µF * 10¯⁶ * 100v = 0.01 coulomb

The ultimate energy stored in the capacitor

W = 1/2 C V² = (100µF * 10¯⁶ * 100v²)/2 = 0.5 joules

Hi Mr Al this is what that completed task looks like that you kindly helped me with and I am very thankful for. If you could just check it one last time before I hand it in and see if anything needs amending I would be very grateful.

 

[ericgibbs]

hi sunny,
Recheck the 8 Secs..??

100*10^-6 * 8 *10^3 = ?????

 

[sunny1982]

A 100 µF capacitor is connected in series with an 8000Ω resistor. Determine the time constant of the circuit.
If the circuit is suddenly connected to a 100 dc supply find

The initial rise of P.D across the capacitor, (rate of voltage rise per volt to the first time constant)

T= CR= 100µF * 10¯⁶ *8000Ω = 0.8 seconds

63.2% of 100v = 63.2v

V/T =63.2v/0.8S = 79v rise per volt to first time constant

The initial charging current, (at zero time)

I = V/R = 100v/8000Ω = 0.0125a

The ultimate charge in the capacitor

Q = C V = 100µF * 10¯⁶ * 100v = 0.01 coulomb

The ultimate energy stored in the capacitor

W = 1/2 C V² = (100µF * 10¯⁶ * 100v²)/2 = 0.5 joules

Yes sorry can you see if that is correct now as it was meant to be 0.8 seconds lol

 

[MrAl]

A 100 µF capacitor is connected in series with an 8000Ω resistor. Determine the time constant of the circuit.
If the circuit is suddenly connected to a 100 dc supply find

The initial rise of P.D across the capacitor, (rate of voltage rise per volt to the first time constant)

T= CR= 100µF * 10¯⁶ *8000Ω = 0.8 seconds

63.2% of 100v = 63.2v

V/T =63.2v/0.8S = 79v rise per volt to first time constant

The initial charging current, (at zero time)

I = V/R = 100v/8000Ω = 0.0125a

The ultimate charge in the capacitor

Q = C V = 100µF * 10¯⁶ * 100v = 0.01 coulomb

The ultimate energy stored in the capacitor

W = 1/2 C V² = (100µF * 10¯⁶ * 100v²)/2 = 0.5 joules

Yes sorry can you see if that is correct now as it was meant to be 0.8 seconds lol

Hello,


"The initial rise of P.D across the capacitor, (rate of voltage rise per volt to the first time constant)"

You do mean the rate of rise of voltage per second right? Because rate of voltage rise per volt means something else. Clarify this

Also, you either write:
100uF*100v^2

or you write:
100*10^-6*100v^2

but you dont write:
100uF*10^-6*100v^2

because that makes the capacitor very small. This is somewhat important or your reader wont know which you really mean.

Also, dont you want to calculate the energy stored in the cap after one time constant too, and the charge after 1 time constant also?
Or do you really want the values after infinite time has passed?

Using the calculation of Q=C*V is also a valid method of calculating the charge. I should have mentioned that.

So far:
Your calculation of the time constant is now correct.
Your calculation of the initial current is correct.
Your other two calculations are correct if you want the values after an infinite time has passed.

 

[sunny1982]

Hi MR AL
What that means is the rate of voltage rise per volt to the first time constant and the other two calculations are for the ultimate time past or infinite time passed.
So what is rate of voltage rise per volt? to the first time constant?

 

[sunny1982]

Hi MR AL
What that means is the rate of voltage rise per volt to the first time constant and the other two calculations are for the ultimate time past or infinite time passed.
So what is rate of voltage rise per volt? to the first time constant?

So would this be correct? 100/0.8 = 125volt per volt to first time constant?

 

[MrAl]

Hello,

Im asking this for a second time now, are you sure the question was for the "rate of voltage rise per volt" and not "rate of voltage rise per second"?

There is credence to both of these but the first one would be more unlikely. The second would be much more common.

The rate of voltage rise per second is simply the voltage rise (to the first time constant time) divided by the time constant: 63.2/0.8
The rate of voltage rise per volt would be interpreted as the rate of voltage rise per volt of input amplitude (to the first time constant) which would be: 0.632/0.8

The two interpretations are different in that one requires just knowing the initial slope of the voltage rise, while the other considers multiple slopes from various amplitude input voltage steps.

So you have to find a way to verify which one of these are required to know on the test. It could be that there was just a mistake in the writing and they really meant the rise of voltage per second to the first time constant, but you really should verify this is correct.

 

[sunny1982]

Thankyou very much mr Al, well now i know both the calculations so when I hand it in and the per volts calculation is wrong i'l just change it to volts per second that is minor but credit to you for helping me understand it. So I decided to give the next task ago myself. I just need you to see if I have done this correctly and you can see how well your teaching methods are lol. This question is about RL circuits.

A coil of 10H is connected to a 100Ω resistor across a 300v d.c supply instantaneously.
Calculate:

The time constant of the circuit

L/R = 10/100 = 0.1s

The current after 2ms

I = V/R = 300v/100Ω = 3a i = I (1-e¯ t/T) i = 3a (1-e¯ 2ms/0.1s) = 20 i = 3a (1-e¯²⁰) = 2.99a

The voltage drop across the resistor after 5ms

VR = V (1-e¯ t/T) VR = 300v (1- e¯ 5ms/0.1s) = 50 VR = 300v (1-e¯⁵⁰) = 300v

The time it takes for the current to settle down

5*0.1= 0.5s

The final current in the circuit

I = V/R = I = 300/100 = 3a

0.632*3a= 1.89a

 

[MrAl]

Hello again,

You have the final current calculation correct, as well as the time it takes for the current to settle.

But your calculation for the current at 2ms (0.002 seconds) is not correct yet.
Also your calculation for the voltage drop across the resistor at 5ms (0.005 seconds) is not correct yet either.
Try again being more careful with the math.

It would also help to use the symbol ^ for exponents, such as e^(-1) rather than whatever character you are using because it is unreadable in my browser as well as another browser i tried. Another example: A*e^(-t/T), which should be readable in any browser.

 

[sunny1982]

4) A coil of 10H is connected to a 100Ω resistor across a 300v d.c supply instantaneously.
Calculate:

The current after 2ms

I = V/R = 300v/100Ω = 3a i = I (1-e^- t/T) i = 3a (1-e^- 2ms/100ms) = 0.02 i = 3a (1-e^-0.02) = 0.059a

The voltage drop across the resistor after 5ms

VR = V (1-e^- t/T) VR = 300v (1- e^- 5ms/100ms) = 0.05 VR = 300v (1-e^-0.05) = 14.96v

Thanks again Mr Al, I have went through my maths again and this is what I have come up with.

 

[MrAl]

Hi again,

Your calculation of the current at 2ms looks better now.
Your calculation of the voltage at 5ms looks so close i hate to say anything, but i still think you should do that one over again. It looks like you may have got the answer correct but then copied some of the digits wrong. Check that once more and see what you can come up with.

Much better now overall though

 

[sunny1982]

Hi Mr Al
Don't you get sick of me lol I did have a look at 5ms again, 5/100ms =0.05, 300*(1-e^-0.05) = 14.63. Hey Mr Al where you from?, because if I ever came across you I would buy you a beer lol. Now I'm stuck on the last question of this paper because its not same principle as the last question wants me to calculate when the switch is immediately switched on and when the switch is immediately switched off.

So here is the question.

1. A coil of inductance 4H and resistance of 80Ω is connected in parallel with a 200Ω of resistor of negligible inductance across a 200v dc supply. The switch connecting these to the supply is then opened; the coil and resistor remain connected together. State in each case one for immediately before and one for immediately after opening the switch;

a) The current Through the resistor
b) The current through the coil
c) The e.m.f induced in the coil
d) The voltage across the coil

What is negligible inductance?

 

[WTP Pepper]

What is negligible inductance?

It's assuming all component inductance and resistance are accounted for in the given values. The resistance of the inductor is given so will form part of the solution. The inductance of the 200ohm resistance isn't so you can assume it's zero and does not form part of the solution (hence negligible inductance) as it couldn't be easily calculated in this simple example.

There are many good books (remember books) that explain this so you learn. Much better that requesting fast responses from a forum to avoid the hard slog of actually reading. Probably the way most members learned before the web existed.

The Art of Electronics by Horowitz & Hill is a book for life.

 

[sunny1982]

It's assuming all component inductance and resistance are accounted for in the given values. The resistance of the inductor is given so will form part of the solution. The inductance of the 200ohm resistance isn't so you can assume it's zero and does not form part of the solution (hence negligible inductance) as it couldn't be easily calculated in this simple example.

There are many good books (remember books) that explain this so you learn. Much better that requesting fast responses from a forum to avoid the hard slog of actually reading. Probably the way most members learned before the web existed.

The Art of Electronics by Horowitz & Hill is a book for life.

Thankyou very much for your help appreciate it and yes I will get a hold of that book thanks

 

[MrAl]

Hi Mr Al
Don't you get sick of me lol I did have a look at 5ms again, 5/100ms =0.05, 300*(1-e^-0.05) = 14.63. Hey Mr Al where you from?, because if I ever came across you I would buy you a beer lol. Now I'm stuck on the last question of this paper because its not same principle as the last question wants me to calculate when the switch is immediately switched on and when the switch is immediately switched off.

So here is the question.

1. A coil of inductance 4H and resistance of 80Ω is connected in parallel with a 200Ω of resistor of negligible inductance across a 200v dc supply. The switch connecting these to the supply is then opened; the coil and resistor remain connected together. State in each case one for immediately before and one for immediately after opening the switch;

a) The current Through the resistor
b) The current through the coil
c) The e.m.f induced in the coil
d) The voltage across the coil

What is negligible inductance?

Hello again,


Yes as WTP Pepper mentioned, negligible inductance means that the ever present inductance of that particular device is considered to be zero. There is always some inductance in almost everything that conducts current, but often it is so small that we just do not have to consider it there at all in order to obtain reasonable results about the circuit. In fact, doing two analyses one with the tiny inductance and one without, yields the same results to several significant figures, so we consider it moot in many cases.

I assume you have the formulas for your other questions on hand already?

Yes a good book will help a lot. However, you'll probably end up wanting more than one book because one book never seems to have everything you want to understand. Get used to it and hunt around on the web first to avoid spending what adds up to hundreds of dollars of literature and taking up quite a bit of space. I must have spent upwards of something like $5000.00 USD on my library and still dont have everything i would like to have. Also a visit to the local library might turn up some good information.

 

[sunny1982]

Wow $5000 USD, I think I'll stick to the internet lol , I have learning difficulties, (personal problem) but I give it ago just like Ive tried to learn off you guys and so far what I have learnt I have tried to tackle the problems myself. I feel I learn better when someone explains and breaks down information for me it sticks with me, I'm trying to get into engineering its just this one unit that is engineering science something I haven't done in high school but having to do in college.
I have got some formula's that need checking and would be very grateful if you can check them for me;-)

A coil of inductance 4H and resistance of 80Ω is connected in parallel with a 200Ω of resistor of negligible inductance across a 200v dc supply. The switch connecting these to the supply is then opened; the coil and resistor remain connected together. State in each case one for immediately before and one for immediately after opening the switch;

The current Through the resistor

Switch closed
200v/200Ω =1a
Switch open
200v/80Ω =2.5a

The current through the coil

Switch Closed
200v/80Ω = 2.5a
Switch open
200v/(200ohm +80Ω) = 0.714a


The e.m.f induced in the coil

Switch closed
(2.5a*80Ω)/200v = 0v
Switch Open
2.5a *(200Ω + 80Ω) = 700v

The voltage across the coil

Switch closed
200v/1a = 200v
Switch open
2.5a * 200Ω = 500v

 

[MrAl]

Hi again,

Not bad, but a few things to look at here...

When you state that:

The current through the coil:

Switch Closed
200v/80Ω = 2.5a

Switch open
200v/(200ohm +80Ω) = 0.714a [This is not correct sorry to say]

In this last line where are you getting that 200v from? The 200v supply goes away when the switch is opened. Recheck that one and note the behavior of the inductor. And inductors operate more on current than on voltage. See what you can come up with.

Small point here:

The e.m.f induced in the coil

Switch closed
(2.5a*80Ω)/200v = 0v [i think you meant to subtract here right?
Switch Open
2.5a *(200Ω + 80Ω) = 700v

Small point here too:
The voltage across the coil

Switch closed
200v/1a = 200v [Volts divided by amps comes out to units of Ohms. So what did you really mean to do here?]
Switch open
2.5a * 200Ω = 500v


Take a minute to recheck those lines indicated and see what you can find.

 

[WTP Pepper]

Wow $5000 USD

OK. Send me $4000 USD and I will send you the book discounted.
Or look here https://tinyurl.com/bt5zl3w

 

[sunny1982]

OK. Send me $4000 USD and I will send you the book discounted.
Or look here https://tinyurl.com/bt5zl3w

Thankyou very much appreciated, book looks really interesting, I'm going to start building a book collection, I wish I had $4000USD, Some of you are top engineers on here so you can afford where me I've just started out but I will get there one day, I'll never stop learning.

MR AL, I'm going to see what I can come up with, I was so confident that they were correct I had a smug grin, when I posted them lol

 

[MrAl]

Hi again,


I have no doubt you'll figure this out very soon seeing as how you picked up on the other stuff so fast
Dont let it make you a big head however or you'll start overlooking things (over confidence) always double check your work

We all make mistakes, but part of engineering is learning to minimize your errors. As well as learning what the errors are, try to figure out how they occurred in the first place and figure out what you can do in the future to minimize whatever happens to be causing that. The best bet is to have a back up plan...a second method for calculating the results...if the second method produces a value that is not the same as the first, something went wrong. It doesnt matter if the first method or the second method went wrong because they should both be perfect.