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Catégorie :Category: mViewer GX Creator Lua TI-Nspire
Auteur Author: vxddvdvv
Type : Classeur 3.6
Page(s) : 97
Taille Size: 3.52 Mo MB
Mis en ligne Uploaded: 16/03/2019
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Engineering Circuit Analysis, 7th Edition Chapter Eight Solutions 10 March 2006



8.
103 106
(a) iL (0) = 4.5mA, R/L = =
4 × 10−3 4
6
∴ iL = 4.5e −10 t/4
mA ∴ iL (5μ s ) = 4.5e −1.25

= 1.289 mA.

(b) iSW( 5 μs) = 9 – 1.289 = 7.711 mA.




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PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
www.elsolucionario.net
Engineering Circuit Analysis, 7th Edition Chapter Eight Solutions 10 March 2006



9.
100
iL (0) = = 2A ∴ iL (t ) = 2e−80t / 0.2
(a) 50
−400 t
= 2e A, t > 0


(b) iL (0.01) = 2e −4 = 36.63mA

(c) 2e −400t1 = 1, e 400t1 = 2, t1 = 1.7329ms




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PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
www.elsolucionario.net
Engineering Circuit Analysis, 7th Edition Chapter Eight Solutions 10 March 2006



10. (a)
di
L + 5i = 0 [1]
dt
vR = −2i so Eq. [1] can be written as
⎛ −v ⎞
d⎜ R ⎟
2 ⎠ ⎛ −v ⎞
L ⎝ − 5⎜ R ⎟ = 0 or
dt ⎝ 2 ⎠
dv
2.5 R + 2.5vR = 0
dt

(b) Characteristic equation is 2.5s + 2.5 = 0, or s + 1 = 0
Solving, s = –1, vR (t ) = Ae −t




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(c) At t = 0–, i(0–) = 5 A = i(0+). Thus, vR (t ) = −10e − t , t > 0
2
vR (0− ) = (10 ) = 6.667 V
3
+
vR (0 ) = −10 V
vR (1) = −10e −1 = − 3.679 V




PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
www.elsolucionario.net
Engineering Circuit Analysis, 7th Edition Chapter Eight Solutions 10 March 2006



11.

(a) i t I I t
= e − t /τ , = ln o , o = 10 ∴ = ln10 = 2.303;
Io τ i i τ
Io t I t
= 100, = 4.605; o = 1000, = 6.908
i τ i τ


(b) i d (i / Io ) d ()
= e− t /τ , = −et /τ ; t t / τ = 1, =a−e −1
Io d (t / τ ) d ()
⎛t i ⎞
Now, y = m( x − 1) + b = −e −1 ( x − 1) + e−1 ⎜ = x, = y ⎟
⎝τ Io ⎠




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At y = 0, e ( x − 1) = e ∴ x = 2 ∴ t / τ = 2
−1 −1




PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
www.elsolucionario.net
Engineering Circuit Analysis, 7th Edition Chapter Eight Solutions 10 March 2006



12. Reading from the graph current is at 0.37 at 2 ms

∴ τ = 2 ms


I 0 = 10 A




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PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
www.elsolucionario.net
Engineering Circuit Analysis, 7th Edition Chapter Eight Solutions 10 March 2006



13. w = ½ Li2, so an initial energy of 15 mJ in a 10-mH inductor corresponds to an initial
inductor current of 1.732 A. For R = 1 kΩ, τ = L/R = 10 μs, so iL(t) = 1.732 e–0.1t A.
For R = 10 kΩ, τ = 1 μs so iL(t) = 1.732 e-t. For R = 100 kΩ, τ = 100 ns or 0.1 μs so
iL(t) = 1.732 e-10t A. For each current expression above, it is assumed that time is
expressed in microseconds.
To create a sketch, we firs t realise that the m aximum current for any of the th ree cases
will be 1.732 A, and af ter one time constant (10, 1, or 0.1 μs), the current will d rop to
36.79% of this value (637.2 m A); after approximately 5 tim e constants, the curren t will
be close to zero.




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Sketch based on hand analysis Circuit used for PSpice
verification


As can be seen by
comparing the two plots,
which probably should
have the same x-axis scale
labels for easier
comparison, the PSpice
simulation results obtained
using a parametric sweep
do in fact agree with our
hand calculations.




PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
www.elsolucionario.net
Engineering Circuit Analysis, 7th Edition Chapter Eight Solutions 10 March 2006



14.

3.3 × 10 −6
(a) τ= 6
= 3.3 × 10 −12
1 × 10

(b)
1
ω = .L.I 0 2
2
2 × 43 × 10− 6
I0 = = 5.1 A
3.3 × 10− 6



i (5 ps ) = 5.1e −1×10
6
× 5×10 −12 / 3.3×10 − 6




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= 1.12 A

(c)




From the PSpice
simulation, we see that
the inductor current is
1.121 A at t = 5 ps, in
agreement with the hand
calculation.




PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
www.elsolucionario.net
Engineering Circuit Analysis, 7th Edition Chapter Eight Solutions 10 March 2006



15. Assume the source Thévenin resistance is zero, and assume the transient is measured to
5τ. Then,

L 5L
τ= ∴ 5τ = = 100 × 10−9 secs
R R
so R must be greater than 6.285 kΩ.
(5)(125.7)10−6
∴R >
10−7

6.285
(If 1τ assumed then R > = 125.7Ω )
5




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PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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