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Unformatted text preview: P1 .26* Elements A and Bare in series. Also elements E and F are in series. P1 .27 (a) Elements 6' and Dare in series.
(b) Because elements 6' and D are in series the currents are equal in
magnitude. However, because the reference directions are opposite; the algebraic signs of the current values are opposite. Thus, we have I; —/d. (c) At the node joining elements A, B, and 6: we can write the KCL
equation I}, = I; + I; = 3 +1: 4 A. Also we found earlier that E2.8 First write KCL equations at nodes land 2: . .1.
Node 1. 2 + 5 + 10 . V2 — 10 V2 V2 VI = 0
Node 2. 10 —5 + 10 Then simplify the equations to obtain:
8V1V2 =50 and —V1+4V2 =10 Solving, we find v1: 6.77 V and V2 = 4.19 V. 52.12 (a) Select the
reference node at the
lefthand end of the
voltage source as shown
at right. Then write a KCL
equation at node 1. Vl—IO
—+ +1=0
R R2 Substituting values for the resistances and solving, we find V1 = 3.33 V. IO—V1
= 1.333 A.
R: Then we have I; = (b) Select the
reference node and
assign node voltages as
shown. Then write KCL
equations at nodes 1
and 2. 3 Ra 95 Substituting values for the resistances and solving, we find v1 = 13.79 V and V2 = 18.97 V. Then we have I}, = V’évz = —0.259 A. £2.13 (a) Select the 5'
reference node and _n. '2);
node voltage as 7
shown. Then write a x 2A
KCL equation at node /0 V 9 5:“ o x
1, resulting in
_Vl v1 10 _2. ___ 0
5 + 5 ’* ___ Then use ix = (10 — v1)/ 5 to substitute and solve. We find 14 = 7.5 V. 10 —v1 Then we have ix = = 0.5 A. (b) Choose the reference node and node voltages shown: Then write KCL equations at nodes land 2: Vl—Ziy _ y_2 Vz—Ziy
2 +3—0 5+ 10 V1
— +
5 =3 P220" Combining resistors in series and parallel, we find that the equivalent
resistance seen by the current source is [2,, = 17.5 (2. Thus, v=8x17.5= 140V. Also, i=1 A.
{> 20 ID 73
3A0» 25 ,6
P‘s
2””
$20 9 P2.28* 1 . 20v
= —— = 4 :2 = —— =2 A
R“ 1/6+1/12 ’1 2/2,, 5
v, =v, =R,,/; =1ov f3 =10/6=1.667A
I; = 10/12 = 0.8333 A i, =1; — 1;: o 8333 A P2.39* Writing a KVL equation, we have V1 — v2 =10. V1
At the reference node, we write a KCL equation: —5— +—— 10: Solving, we find v1: 6. 667 and v2: —3.333. P2.48* VX = V2 — I/1
Writing KCL at nodes land 2: ﬁ Vl—va Vlszl
5+ 15 + 10
£2. vz—va vz—v1=2
5+ 10 + 10 Substituting and simplifying, we have
15vl — 71/2 = 30 and 1/1 + 2v2 = 20. Solving, we find v1 = 5.405 and v2 = 7.297. ...
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 Spring '06
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