SOLUTION: the voltage VL across an inductor L in an electrical circuit falls exponentially over time t. The relationship is VL = Vmax e^-t/T, Where T= L/R. If L= 4.5H and R = 270 calculate

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Question 261326: the voltage VL across an inductor L in an electrical circuit falls exponentially over time t. The relationship is VL = Vmax e^-t/T, Where T= L/R.
If L= 4.5H and R = 270 calculate T?
Vmax=10.4 volts
Calculate V for values of t from o to 0.5 seconds at intervals of 0.01?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
formula is:

VL+=+V%5Bmax%5D+%2A+e%5E%28-t%2FT%29

T = L/R
L = inductance = 4.5
R = resistance 270
T = L/R = (4.5/270) = .016666666...
t = time
V[max] = 10.4 volts.

Calculate VL for values of t from 0 to .5 seconds at intervals of 0.01

When t = 0, the formula of VL+=+V%5Bmax%5D+%2A+e%5E%28-t%2FT%29 becomes:

VL+=+10.4+%2A+e%5E%28-0%2F.0166666666%29 = 10.4+%2A+e%5E%280%29 = 10.4

When t = .1, the formula of VL+=+V%5Bmax%5D+%2A+e%5E%28-t%2FT%29 becomes:

VL+=+10.4+%2A+e%5E%28-.1%2F.0166666666%29 = 10.4+%2A+e%5E%28-6%29 = .025779023

When t = .4, the formula of VL+=+V%5Bmax%5D+%2A+e%5E%28-t%2FT%29 becomes:

VL+=+10.4+%2A+e%5E%28-.4%2F.0166666666%29 = 10.4+%2A+e%5E%28-24%29 = 3.926139926%5E-10 

t	v[max]	T	        VL = V[max] * e^(-(t/T))
0	10.4	0.016666667	10.4
0.01	10.4	0.016666667	5.707641015
0.02	10.4	0.016666667	3.132419804
0.03	10.4	0.016666667	1.719108438
0.04	10.4	0.016666667	0.943466714
0.05	10.4	0.016666667	0.517785511
0.06	10.4	0.016666667	0.284166713
0.07	10.4	0.016666667	0.155953999
0.08	10.4	0.016666667	0.085589369
0.09	10.4	0.016666667	0.046972442
0.1	10.4	0.016666667	0.025779023
0.11	10.4	0.016666667	0.014147828
0.12	10.4	0.016666667	0.007764492
0.13	10.4	0.016666667	0.004261244
0.14	10.4	0.016666667	0.00233862
0.15	10.4	0.016666667	0.001283462
0.16	10.4	0.016666667	0.000704379
0.17	10.4	0.016666667	0.000386571
0.18	10.4	0.016666667	0.000212155
0.19	10.4	0.016666667	0.000116433
0.2	10.4	0.016666667	6.38998E-05
0.21	10.4	0.016666667	3.5069E-05
0.22	10.4	0.016666667	1.92463E-05
0.23	10.4	0.016666667	1.05626E-05
0.24	10.4	0.016666667	5.79686E-06
0.25	10.4	0.016666667	3.18138E-06
0.26	10.4	0.016666667	1.74598E-06
0.27	10.4	0.016666667	9.58214E-07
0.28	10.4	0.016666667	5.25879E-07
0.29	10.4	0.016666667	2.88609E-07
0.3	10.4	0.016666667	1.58392E-07
0.31	10.4	0.016666667	8.69273E-08
0.32	10.4	0.016666667	4.77067E-08
0.33	10.4	0.016666667	2.6182E-08
0.34	10.4	0.016666667	1.4369E-08
0.35	10.4	0.016666667	7.88586E-09
0.36	10.4	0.016666667	4.32785E-09
0.37	10.4	0.016666667	2.37518E-09
0.38	10.4	0.016666667	1.30352E-09
0.39	10.4	0.016666667	7.15389E-10
0.4	10.4	0.016666667	3.92614E-10
0.41	10.4	0.016666667	2.15471E-10
0.42	10.4	0.016666667	1.18253E-10
0.43	10.4	0.016666667	6.48987E-11
0.44	10.4	0.016666667	3.56171E-11
0.45	10.4	0.016666667	1.95471E-11
0.46	10.4	0.016666667	1.07277E-11
0.47	10.4	0.016666667	5.88747E-12
0.48	10.4	0.016666667	3.23111E-12
0.49	10.4	0.016666667	1.77327E-12
0.5	10.4	0.016666667	9.73193E-13

the graph of your equation looks like this:

graph%28600%2C600%2C-1%2C1%2C-1%2C12%2C10.4%2Ae%5E%28-%28x%2F%284.5%2F270%29%29%29%29

the following graph shows the smaller values a little better since the y values are taken up to 1 max only.


graph%28600%2C600%2C-1%2C1%2C-1%2C1%2C10.4%2Ae%5E%28-%28x%2F%284.5%2F270%29%29%29%29

didn't matter too much as the drop off is very steep and the numbers after t = .2 are extremely small.