7 6.10 A series RL circuit with a nonlinear inductor is shown in Fig. P6.10. Recall that the following nonlinear function for inductor current was used in Problem 3.11 in Chapter 3 97.34+4.2A (amps,A) where is the flux linkage. Use Simulihk to obtain the dynamic response for current IL(t) if the source voltage is a 4-V Step function, that is, ein(t) = 4U(t) V. The RL circuit has zero energy stored at time t = 0 and the resistance is R = 1.2 Q. Plot current IL vs. time. Nonlinear inductor L mo in www

Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
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6-10) IL()) = (17.31² +4-2X)
Applying KvL for t70, get di(t) + RIL (X)=em(t)
d+
di(t)
d+
+ RIL()) = Rm(t)
dX(t)
(R=1·²5)
+1-2 [97.3 +4-2X]=4
dt
(t) =(4-16-76X²-5-04 X)
dt
4
S
4.2
5.04
97-3
116-76
7 6.10 A series RL circuit with a nonlinear inductor is shown in Fig. P6.10. Recall that the following nonlinear
function for inductor current was used in Problem 3.11 in Chapter 3
97.34+4.2A (amps,A)
tion
where is the flux linkage. Use Simulihk to obtain the dynamic response for current IL(t) if the source
voltage is a 4-V Step function, that is, ein(t) = 4U(t) V. The RL circuit has zero energy stored at time t
= 0 and the resistance is R = 1.2 Q. Plot current IL vs. time.
Nonlinear
inductor
L
000
ein/th
www
R
eq
Transcribed Image Text:6-10) IL()) = (17.31² +4-2X) Applying KvL for t70, get di(t) + RIL (X)=em(t) d+ di(t) d+ + RIL()) = Rm(t) dX(t) (R=1·²5) +1-2 [97.3 +4-2X]=4 dt (t) =(4-16-76X²-5-04 X) dt 4 S 4.2 5.04 97-3 116-76 7 6.10 A series RL circuit with a nonlinear inductor is shown in Fig. P6.10. Recall that the following nonlinear function for inductor current was used in Problem 3.11 in Chapter 3 97.34+4.2A (amps,A) tion where is the flux linkage. Use Simulihk to obtain the dynamic response for current IL(t) if the source voltage is a 4-V Step function, that is, ein(t) = 4U(t) V. The RL circuit has zero energy stored at time t = 0 and the resistance is R = 1.2 Q. Plot current IL vs. time. Nonlinear inductor L 000 ein/th www R eq
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