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Note: The 10 Ohm res. here has replaced the 10 Ohm which was tucked in as the relay contact resistance.
The aim here is to gain a further understanding of the amount of current flowing at any chosen time, say at 1ms after t=0 and i=0 start, from closing a switch in a simple RL circuit.(Opening the switch is a different scenario which is not discussed here).
Let's start, with each (numbered) step by step ...
1) total V = inductor V + resistor V
That description is in the form of a linear first order differential equation, and can be written as:
2) E(t) = L(di/dt) + iR
( note: v is replaced as v = 'i x R' for the resistor v,
and v = 'change in i/ change in time' x L for the inductor v, E is the battery source v which is held constant here throughout our time t )
Schematic example:
t = 1ms
E = 9v
L = 30mH ... E.C. ideal inductor
R = 10 Ohm
3) Thus: 9 = 0.03(di/dt) + 10i
Now put di/dt on its own: divide throughout by 0.03 :
Thus: 300 = di/dt + 333.333i
Let's turn that round to read as:
4) di/dt + 333.333i =300
5) Now multiply both sides by the integrating factor: (e^ [∫333.333 dt])
which is also: e^(333.333t)
being e^(coefficient of i)
so, from our step 4) where di/dt + 333.333i =300 :
Let's do that multiplication:
6) e^[333.333t] di/dt + 333.333ie^[333.333t] = 300e^[333.333t]
That step 6) can be re-arranged and written equivalently as the derivative of the product of the 'integrating factor e^[333.333t], and the function i' :
7) d (ie^[333.333t]) / dt = 300e^[333.333t]
8) Then integrate both sides:
∫(d/dt)( ie^[333.333t] ) dt = ∫(300e^[333.333t] )dt
Which gives:
9) ie^[333.333t] = 300/333.333 (e^[333.333t]) +C
10) Now divide throughout by e^[333.333t]
Which gives our main useful equation:
11) i = 0.9 + C(e^[-333.333t])
Then find C at i(t=0)
12) 0=0.9 + C .... where t=0 and i=0 and e^0 =1
So:
C= -0.9
13) Now use this known value of C where t=0.001 sec. as per original question above. use step 11) :
i(t=0.001) = 0.9 - 0.9(e^[-333.333 x 0.001])
So:
14) i(t=0.001) = 0.255 amp
This is what we set out to find.... as per the first paragraph above.
See the approx 0.255 amp reading on the scope.
15) check:
i(t)=(V/R) - (V/R)[e^(-Rt/L)] **
i(t)=9/10 - (9/10)[e^(-0.01/0.03)]
i(t)=0.9 - 0.9e^-0.333
i(t)= 0.255. amp. where t = 0.001sec
16) Thus we have derived this formula to obtain i(t) ... generalised from step 11) **
but now with a deeper understanding about what is happening when the current changes.
17) That handy formula can be re-arranged as:
i(t)= (V/R)(1 - e^[-Rt/L])
which is the often used formula found in texts for RL d.c. analysis.
Summary:
If you have followed all of the above, then you now understand in detail why a current reads as it does at any time after closing a switch, right from the first principle shown at step 1) above, which was:
Step 1): total V = inductor V + resistor V
Your now well-understood usable formula is:
Step17): i(t)= (V/R)(1 - e^[-Rt/L])
If you are stuck on any step then ask for help.
Extra notes :
Vₗ = Vₛ x e^[-Rt/L]
where Vₗ = induced EMF, Vₛ= full supply V
Also:
P = i^2R + Li(di/dt) .. total Power for the RL circuit
18) Note also that V/R is the max i (seen here as 0.9 amp) which occurs at about '5 x tau' seconds for t in step 17), called the 'transient time' ... where the 'time constant' tau= L/R for an inductor.
When di/dt=0, no change in current, then the volt drop across the inductor is zero ... V=L(di/dt), so the current is dependant only on the resistor, as being i=V/R
A larger inductance gives a larger transient time.
A larger Resistor gives a shorter transient time.
Note:
The way in which the switch is turned off will make a major difference to the entire action... this needs further explanation.... eg switched by spst, pulse, logic train, capacitor etc.
Using standard formulae may be fine for many cases, but for ventures in to design, it is possibly better to have an insight in to why and how the current is acting in a particular way at any chosen point in time... and so make adjustments accordingly.
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