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. 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 V-iR/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!
I could't work out in the editor how to show something raised to a power. Hope you can understand this easily. 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(1-e(-t/T))=34.3(1-e^(-0.05/0.05))=21.68A (2s.f.) But - 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 These figures don't agree ii) the time to develop maximum current. When t≥5T. Imax=V/R=24/15=1.6A
Another way is the same way calculators do it, 2.2E4 is . There is another way, I'm trying to figure it out, using LaTeX. Use {tex][/tex}with the number between them, and brackets [] around tex and /tex. A how to guide for math on this forum... http://forum.allaboutcircuits.com/showthread.php?t=21380
To find the time rearrange i=Imax(1-e(-t/T)) as to get t=-T*ln(-i/Imax+1) when i=Imax, ln(-1+1)=ln(0)=infinity Thus, the time to reach Imax is infinite. This the reason it is said that Imax is reached after 5T.
t=-T*ln(-i/Imax+1) Well. I had two values of i (not sure which one was correct) t=-0.05*ln(-1.008/1.6+1)=i get mathematical error or t=-0.05*ln(-21.68/1.6+1)=i get mathematical error
In the first case you shouldn't get a mathematical error however the value for i is wrong. It should be i=Imax. In the second case you will get a mathematical error because you have a negative number in the ln() which is not allowed.
if i=Imax then i=1.6A t=-0.05*ln(-1.6/1.6+1) but i'm confused cause they both have a minus sign in the ln(-1.008) and ln(-21.68) but which one of these values is also correct from the earlier workings? 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(1-e(-t/T))=34.3(1-e^(-0.05/0.05))=21.68A (2s.f.) But - 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 These figures don't agree
ln(-1.008/1.6+1)=ln(0.37) ln(-21.68/1.6+1)=ln(-12.55) this is wrong because 21.68 is wrong The correct value of current after T is 1.008.
hold on! Your earlier post: To find the time rearrange i=Imax(1-e(-t/T)) as to get t=-T*ln(-i/Imax+1) when i=Imax, ln(-1+1)=ln(0)=infinity Thus, the time to reach Imax is infinite. This the reason it is said that Imax is reached after 5T. So, the answer to this part of the question is infinity? It can't be -1s. It doesn't make any sense for the current to be maximum at -1s, the circuit isn't connected before time t=0.
If you calculate the time mathematically, then (t) is infinity. However, if you assume that (i) reaches Imax after 5T, then t=5*0.05=0.25.
ahh! I see. So the answer to my question is: (i) reaches Imax after 5T, then t=5*0.05=0.25. But, you have been showing me how to calculate the time mathematically, then (t) is infinity? But why is my answer -1s?
Yes, the exact time is infinity however we assume that after 5T it reaches Imax because (i) is very close to Imax. How you get -1?
How you get -1? It's a good question. I must have eaten a strange fruit this morning. I will use 0.25s then. Once again mik3, thanks for all your great support today. It is very appreciated.