ELECTRONIC PRINCIPLES - OD1647 - LESSON 1/TASK 1
(c) During the third constant, current again increases:
10 amperes 8.64 amperes = 1.36 amperes
1.36 amperes x .632 = 0.860 ampere
8.64 amperes + 0.860 ampere = 9.50 amperes
(d) During the fourth time constant, current again
increases:
10 amperes 9.50 amperes = 0.5 ampere
0.5 ampere x .632 = 0.316 ampere
9.50 amperes + 0.316 ampere = 9.82 amperes
(e) During the fifth time constant, current increases as
before:
10 amperes 9.82 amperes = 0.18 ampere
0.18 ampere x .632 = 0.114 ampere
9.82 amperes + .114 ampere = 9.93 amperes
Thus, the current at the end of the fifth time constant is
almost equal to 10.0 amperes, the maximum current. For all
practical purposes, the slight difference in value can be
ignored.
(2) Deenergization of an LR Circuit. When an LR circuit is
deenergized, the circuit decreases (decays) to zero in five time
constants at the same rate that it previously increased. If the
growth and decay of current in an LR circuit are plotted on a
graph, the curve appears as shown in figure 21 on the previous
page. Notice that the current increases and decays at the same
rate in five time constants.
The value of the time constant in seconds is equal to the
inductance in Henrys divided by the circuit resistance in Ohms.
The formula used to calculate one L time constant is:
R
L (in Henrys)
t (in seconds) =
R (in Ohms)
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