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Registered Member #63
Joined: Thu Feb 09 2006, 06:18AM
Location:
Posts: 1425
Hi all, I'm looking for some help on a fairly primitive question that is confusing me (shamefully).
I'm not sure how many typos are present in the question or the reference material I've been provided (in fact, I'd probably bet money that's not supposed to be a "tau" but a "t" in the formula?)...
I am presented with a voltage vs time graph of an inductor L= 120mH with the target of my calculations at t = 10us.
So, how exactly DO I calculate the current at t = 10us using the 'current in an inductor' formula?
What do I use for t0 and i(t0)?
I've done some scouring of Wikipedia and hyperphysics, but all their examples seem to have graphs with lines with slopes. :-/
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
First, since there is no information about the current in the inductor at time t = 0, assume I = 0
Using I= (1/L) x Integral (V.dt) =Io
Think of it as (change in current) = (integral of V.dt)/L During the first 7 us (Integral of V.dt) = 500V x 7 us = 3500 uV.s = 3.5 mV.s So the change in current is 3.5 mV.s / 120 mH = 0.029166666666 A = 29.166666666 mA
From 7 to 11 us the current in the inductor will change in the opposite direction by (-1750V x 4) us / 120 mH = -7mV.s / 120 mH = -0.05833333 A = -58.33333 mA so the current in the inductor will now be 29.16666 - 58.33333 = -29.16666666 mA
From 11 to 18 uS the current will change by (7 us x 500 V) / 120 mH = 29.166666 mA So at 18 us the current in the inductor will be -29.1666666 + 29.16666 mA = 0 mA
You could have calculated it by adding all the (integral of v.dt) together then dividing by inductance which would be [(7 x 500) - (1750 x 4) + (7 x 500)] = 0 uV.s so the overall change in current would be 0 uV.s / 120 mH = 0 A But then you would have no appreciation of the actual inductor current during the sequence so you would not be able to design/specify a suitable inductor. Doing it step by step you know that the peak current achieved during the cycle was 29.1666666 mA
I'll leave the calculation of current at 10 us for you to solve, it should be easy now!
Registered Member #63
Joined: Thu Feb 09 2006, 06:18AM
Location:
Posts: 1425
Thanks indeed Sulaiman! Where else would I get a prompt and thorough response -- you rock! We must have started typing our posts at the same time -- I did sit at this for an hour (an hour too long?) and mull it out eventually, and here's how I put the solution into words:
1) I hand-integrate the voltage curve in a piecemeal fashion until I get to Tx, take that value, and multiply it by 1/L, which gives me the current. Solved.
2) If I multiply the calculated current by the voltage off the graph at Tx, I get the power. Solved.
3) I use the formula 0.5 * L * I^2, taking the current figure from part 1, to find the stored energy. Solved.
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calculating current in an inductor
