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A circuit comprises an inductor of 700mH of negligible resistance 
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#1
Feb710, 05:31 PM

P: 201

1. The problem statement, all variables and given/known data
A circuit comprises an inductor of 700mH of negligible resistance connected in series with a 15Ω resistor and a 24V d.c. source. Calculate: a) i) the initial rate of change of current ii)the time constant (T) for the circuit b) the value of the current after 1T seconds, and ii) the time to develop maximum current. 2. Relevant equations V=L x d_{i}/d_{t}+iR ( i think) 3. The attempt at a solution The inductor has negligible resistance so we can use V=IR to calculate the current through the circuit = 24/15 = 1.6A ViR/L = 24(1.6x15)/(700x10^{3})=0 So, I got zero. So, I got nothing, right? What am I doing wrong? I suck at this! 


#2
Feb810, 04:04 PM

P: 29

You're on the right track with the differential equation.
You need to solve the differential equation for the current in the circuit with a 24V step input. This is easiest to do using Laplace transforms. See my sketch below showing the circuit, the Laplace transform of the differential equation, as well as the Laplace transform of the input: http://img687.imageshack.us/img687/4498/0208001346.jpg I'll leave the math up to you but the result you should get for the current once you apply an inverse laplace transform is: [tex] I(t) =\frac{24V}{R} [1e^{\frac{R}{L}t}][/tex] Most circuit texts will present the result above for RL circuits, good ones will show the derivation. Once you have I(t) you should be able to find the all the answers you need with some basic calculus and plotting. Keep in mind that R/L is the time constant. 


#3
Feb810, 06:01 PM

P: 201

Hi:
Thanks for the reply. It's a bit over my head though. I just started electronics recently I_{s}=24/15+(700x10^{3})=1.5A (2s.f.) If I use the equation you gave in your drawing I get the above. Though I don't really know what I'm doing here, to be honest. This is an ac circuit, right? So, the initial rate of change of current is the switch from positive to negative that induces an emf? It says the inductor has negligible resistance so why is it added to the resistance in the calculation above? And what exactly is a time constant T anyway? I know it is used in a capacitor circuit as well which is equal to the product or R by C. 


#4
Feb810, 06:41 PM

P: 29

A circuit comprises an inductor of 700mH of negligible resistance
I'll try to give you some physical insight into what is happening in the circuit.
1) The circuit starts at time t=0 with no current flowing. 2) Right at time t=0 the 24v voltage source is connected to the circuit. The voltage source is a DC signal and tries to force a current of [tex]24V/15\Omega[/tex] through the circuit. This is because as you pointed out the inductor has no intrinsic resistance. 3) When the circuit is connected the inductor does produce an EMF as you mentioned. This EMF prevents current from flowing within the circuit. The inductor isn't acting as a resistance, but it is limiting the amount current within the circuit. 4) Over time the energy in the inductor will decrease allowing more current to flow. 5) At time [tex]t=\infty[/tex] the emf in the inductor is zero and the circuit will operate as a voltage source in series with a resistor. To understand this better with the math look at the time dependent current equation I gave you in two terms. The first term: [tex]\frac{24V}{15\Omega}[/tex] accounts for the current that would be flowing in the circuit at [tex]t=\infty[/tex]. The second term: [tex]\frac{24V}{15\Omega}e^{\frac{R}{L}t}[/tex] accounts for the current the inductor is preventing from flowing. When I think of inductors I always think of a device that inhibits any change in the current of a circuit. Since the circuit is switching from 0A to 24/15 A the inductor is generating an EMF to prevent this change. To answer you question on time constant T, any exponential function of the form [tex]e^{\alpha t}[/tex] has a time constant [tex]\alpha[/tex] the time constant just determines how fast the function decreases if the time constant is negative, or increases if the time constant is positive. There are many applications for time constants from RC and RL circuits to population growth. Hope this helps 


#5
Feb910, 05:01 PM

P: 201

a) i) the initial rate of change of current
I have found this in my book: Quote: "At the instant of closing the switch, the rate of change of current is such that it induces and e.m.f. in the inductance which is equal and oppposite to V, hence V=V_{L}+0, i.e. V_{L}=V. Because V=V_{L}+V_{R}, then V_{R}=0 and i=0." This is saying my initial rate of change of current is 0, isn't it? But if we are saying "an initial 'rate of change' of current", then it has to have changed from 0A. I found this also: When t=0, i=0, therefore, V_{R}=0, therefore, V=L x di/dt+0, so di/dt=V/L If this is correct we get 24V/(700x10^{3})=34.3A/s (3s.f.) ii)the time constant (T) for the circuit This is given in the same way as the time constant for a series connected CR circuit T=LR Therefore T=(700x10^{3})x15Ω=10.5s b) the value of the current after 1T seconds Ok. Maybe here we have entered into current decay, as the induced emf is not restricting the current. So, i=Ie^{(t/T)}=is this 34.3e^{(10.5/10.5)}=12.6 (3s.f.) or 34.3e^{(1/10.5)}=31.2 (3s.f.) or, When i=max, di/dt=0, so i=I=V/R=24/15=1.6A after 1T, i ≈ 0.63 x 1.6A = 1.008A ii) the time to develop maximum current. when t≥5T. This is a given approximation I found in the book. How'd I do 


#6
Feb910, 05:29 PM

P: 29

The initial rate of change of the current is correct, another way to find it would have been to evaluate the derivative of I(t) at t=0.
The time constant is incorrect, the time constant of an RL circuit is R/L. If exponential decay is of the form e^at where a is the time constant then the time constant in our equation for I(t) is R/L. Check the EM units of R and L and you will see you only get units of time with R/L. Redo the rest of your answers with the correct time constant and you should be alright, never forget the power of dimensional analysis when you don't know an equation. 


#7
Feb1110, 05:25 PM

P: 201

klouchis
The time constant is incorrect, the time constant of an RL circuit is R/L. Shouldn't that be L/R? 


#8
Feb1110, 06:07 PM

P: 29

Oh my, yes you are correct, I mistyped. The time constant is L/R so that way the ratio or (R/L)*t is unitless. Look back at the last post I had and reverse the R and L values.
Henries are equivalent units to Ohm*Seconds, only way to get a time constant it to ratio henries/ohm, or L/R. My mistake, I hope I didn't cause any confusion. 


#9
Feb1310, 07:35 AM

P: 201

No. No harm done.
a)i) So, the initial rate of change of current is when di/dt=V/L Which is 24V/(700x10^{3})=34.3A/s (3s.f.) ii) the time constant (T) for the circuit This is not given in the same way as the time constant for a series connected CR. It is, instead, circuit T=L/R Therefore T=(700x10^{3})/15Ω=0.0467s or 0.05s (2s.f.) b) the value of the current after 1T seconds Well there are two equations here  one for growth of current and one for decay. I think it is growth because the question says 'initial rate' Growth=I(1e^{(t/T)})=34.3(1e^{(0.05/0.05)})=21.68A (2s.f.) But  Currents will be approximately 63% or their final values after 1T when growing. I_{max}=V/R=24/15=1.6A So, at 1T, I=0.63x1.6=1.008A These figures don't agree ii) the time to develop maximum current. When t≥5T. I_{max}=V/R=24/15=1.6A 


#10
Feb1310, 01:22 PM

P: 29

[tex]I(t)=\frac{V}{R} (1e^{\frac{R}{L} t})[/tex] All of you questions can be answered by this one equation. For the initial rate of change I': [tex]I'=\frac{d I(t)}{dt}\right_{t=0}[/tex] Just take the derivative and the initial rate of change is the derivative at t=0. For the value of the current at and time t just substitute that value into the equation of I(t). For t=L/R: [tex]I(T)=\frac{V}{R}(1e^{1})[/tex] 


#11
Feb1310, 02:04 PM

P: 201

a)i) So, the initial rate of change of current is when di/dt=V/L
Which is 24V/(700x10^3)=34.3A/s (3s.f.) ii) the time constant (T) for the circuit This is not given in the same way as the time constant for a series connected CR. It is, instead, circuit T=L/R Therefore T=(700x10^3)/15Ω=0.0467s or 0.05s (2s.f.) b) the value of the current after 1T seconds Currents will be approximately 63% or their final values after 1T when growing. Imax=V/R=24/15=1.6A So, at 1T, I=0.63x1.6=1.008A ii) the time to develop maximum current. When t≥5T. Imax=V/R=24/15=1.6A ln[(i/Imax)+1] 1.008/1.6 + 1=0.37 ln(0.37)=0.9942.... 1s (2s.f.) 


#12
Feb1310, 02:23 PM

P: 29

I'm still unsure exactly what you are trying to find in part b part ii. The time it takes to reach maximum current is infinity. Usually you want to know when the current is within a certain percentage of the maximum current. An example being, what time will the current be 99% of the maximum current. It doesn't make any sense for the current to be maximum at 1s, the circuit isn't connected before time t=0. So what exactly are you trying to calculate in part ii of part b?



#13
Feb1310, 02:38 PM

P: 201

t=T*ln(i/Imax+1)
when i=Imax, ln(1+1)=ln(0)=infinity So the answer to this part of the question is just infinity? 


#14
Feb1310, 03:24 PM

P: 29

That would be the answer assuming that's the question being asked. I have a feeling that a different question is being asked because knowing that I=1.6A when t is infinity doesn't seem like a practical problem. What exactly is the question for this part?



#15
Feb1310, 03:39 PM

P: 201

question b)ii)
1. The problem statement, all variables and given/known data A circuit comprises an inductor of 700mH of negligible resistance connected in series with a 15Ω resistor and a 24V d.c. source. Calculate: a) i) the initial rate of change of current ii)the time constant (T) for the circuit b) the value of the current after 1T seconds, and ii) the time to develop maximum current. 


#16
Feb1310, 03:41 PM

P: 201

Hmmm!
If you calculate the time mathematically, then (t) is infinity, right? However, if you assume that (i) reaches Imax after 5T, then t=5*0.05=0.25? 


#17
Feb1310, 03:51 PM

P: 29

Well if that's the only information you are given in the problem then the correct theoretical answer is infinity. No circuit will ever be left running until infinity though, it is important to know when the current is say 95% or 99% of the max value, so it would be a good idea to calculate those times as well.



#18
Feb1310, 04:51 PM

P: 201

The exact time is infinity however we assume that after 5T it reaches Imax because (i) is very close to Imax.
So, my answer is 0.25s 


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