Ch 04 Solutions Practice Problems m4j24

Enginssrilg Circuit Analysis, Eighth

4.r

Edition

Practice Problem Solutions

(p83)

o

The bottom node is the obvious choice for a reference node.

At the top left node

vt lll -5= 2+3+vr-vz 15 and at the top right node,

2

-vt =L+rz 415

21

Simpliffing, we obtain -75

and

4.2

: 4\ - vz

120=-4v1tI9v2

I1l

t2l

(p86)

o

we number the top nodes 1,2 and 3 moving left to right.

Atnode

l,

Atnode2,

-3-

ry*" 21 -"

tl]

ry*)+vr-vt 134 -," +vz-v' Atnode 3, 7 -+*" 542 Q=

l2l t3]

Simpliffing, wo obtain

- 3r, -2r, -v3 0 - -12v, +19v, -3r,

-6

140=-10v,

-5rr+19v,

tl]

I2l t3]

Solving,

and

Engineering Circuit Analysis, 8ft Edition

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Engineering Circuit Analysis, Eighth

4.3

Edition

Practice Problem Solutions

(p88) There are two basic stategies here: (l) insert the quantity for A and solve; or (2) write two general equations in of A, then substitute and solve. Here, we'll go with the second option.

Thus, at node

I

we can

write

-.u' or 3v, -2v, =1g 1* 21 "

S=

tU.

At node 2, we write

l=2+ry 21

or -2vr+3v2=2A l2l.

(a) Substitute A = 2h nto Eqn [2]. We then note that Eqn [2] becomes

h: rr - vz. So Eqn []

(b) SubstitutpA=2vlintoEqn [2],whichthenbecomes 6v, and Eqn

4.4

[2"]

is unchanged but

6vr-7v, =0 l2'1. Solving Eqns [1] andl2'1, we find that

-7vr=0

12"1. SolvingEqn

[l]

we thus obtain

(peO)

Define the top left node

l,

and the top right node 2.

Treating these two nodes as a single supernode,

or

4+9 - 3(v, -v2)+3(vr-vt)+2vr+6v, 13 = 2v, + 6r, tl]

The remaining equation is simply

4.5

,t -

v2

-5

tU

I2l

(p9l) By inspection,

vr:3

For the 2,3 supernode For node

4: 0 -

And, finally

vq

V.

: 4 -+*" 241

,'o +v'-v'

-rt *'o +vo -vt

234 i vz - v2: 0.15v* :

0.15(vl

- v+)

Solving,

Engineering Circuit Analysis, 8ft Edition

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Engineering Circuit Analysis, Eighth Edition 4.6

Practice Problem Solutions

(pe5) mesh I mesh

:

2:

-6 +14\+ (5 + 5)r, - (5 + 5)i, - 0 (5 + 5)i, - (5 + 5)4 +10i, + 5 - 0

Simpliffing,

24\ -10i2 - 6 -10r, +20i, - -5

t1l

l2l

l1l

I2l

Solving,

4.7

(pe6)

o

define clockwise

:

il

in third mesh.

-10 + 4ir- 4ir+ 3 - 0

tu

2: (4+5+9+10)i2-4ir-10r, -0 mesh 3: -3 + (10 + I + 7)i, -10i2 - 0

l2l

mesh I mesh

Simplifying,

4\-4i2-7 -4ir+28i, -10r, = -lDiz+186 -3

Solving, 4.8

t3l

tll I2l

Q

t3l

and

(pe7)

- 2iz, we can write two mesh equations: or 7h-7iz:2 tll 2- (2+ 5)ir - 5iz-2iz or-5h*l2iz-6 I2l 6-(4+5+3)iz-5ir

(a) With A

Solving, we obtain

(b)

A- 2v*:2(h- i)(5): 10 ir - l0 iz, we can write two mesh equations: or -3h+5iz:2 tll 2-(2+ 5)ir-5iz- 1Oir +lOiz With

or-5h*l2iz-6 I2l

6-(4+S+3)iz-5ir

Solving, we obtain

Engineering Circuit Analysis, 8tr Edition

.

Copyright @2012 McGraw-Hill, Inc. All Rights Reserved

Enqineering Circuit Analysis, Eighth

4.e

Edition

Practice Problem Solutions

(pee)

o o

define clockwise current lz in top right mesh and clockwise current form supermesh with meshes I and 3.

il

in bottom right mesh.

Supermesh: -10+ 4\-4k+Q0+1+7)L-10i2=0 tU

(4+5+9+10)rr-4\-l0L=0 remaining equation: h-\= 3 mesh2: Simplifying

,

t2l t3]

4\ -14i2 + 18r, = 10 -4\+28ir-10i3 - 0

tll l2l

aiz-3

-ir

t3l

Solving,

4.10

(pl00)

e

define two clockwise currents: iz

ntop right

mesh,

j:

in bottom right mesh.

-80+ l0ir +20(h-i)+ mesh 2,3 supermesh: -30 + 40a + 30Gr - ir) + 20(iz- ir) =

1:

Supermesh equation:

Solving, Thus, v, =

4Qiu

h-

iz=

l5h

30(ir-,r)=0 0

tU t21

t3l

is=2.604 A =

Engineering Circuit Analysis, 8ft Edition

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