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Unformatted text preview: Problems drill P 4.2 18V [a] 11 branches, 7 branches with resistors, 2 branches with independent
sources, 2 branches with dependent sources [b] The current is unknown in every branch except the one containing the 5
mA current source, so the current is unknown in 10 branches. [c] 11 essential branches each containing a single element. [(1] The current is known only in the essential branch containing the current
source, and is unknown in the remaining 10 essential branches [e] From the ﬁgure there are 5 nodes — four identiﬁed by rectangular boxes
and one identiﬁed by a triangle. [f] There are 5 essential nodes, four identiﬁed with rectangular boxes and one
identiﬁed with a triangle [g] A mesh is like a window pane, and as can be seen from the figure there are
7 window panes or meshes. P 4.4 [a] From Problem 4.2(d) there are 10 essential branches were the current is
unknown, so we need 10 simultaneous equations to describe the circuit. [1)] From Problem 4.2(f), there are 5 essential nodes, so we can apply KCL at
(5 — 1) = 4 of these essential nodes. There would also be two dependent
source constraint equations. [c] The remaining 4 equations needed to describe the circuit will be derived
from KVL equations. [(1] We must avoid using the meshes containing current sources, as we have no
way of determining the voltage drop across a current source. P 4.5 [a] At node 1: —é§+é1+ig:0
At node 2: —ig+i3+i4:0
Atnode3: égil—ég—i4:0 [b] There are many possible solutions. For example, solve the equation at
node 1 for i9: is. = 3'1 + 1'; Substitute this expression for rig into the equation at node 3:
(i1+é2)—£1—é3—i4=0 so ég—i3—i4L—O Miﬂtiply this last equation by —1 to get the equation at node 2: SO ﬂi2+i3+i420 P 4.6 Use the lower terminal of the 5 Q resistor as the reference node. '00 — 60 + E33
10 5 +320 Solving1 no 2 10 V 'no — 60 Uo
10 + 5 [b] Let 11m = voltage drop across 3 A source “or = ’00 — = ——20 V P4.8 [a] +320; vo=1UV p3A (developed) = (3)(2D) : 60 W [c] Let 8'9 = be the Current into the positive terminal of the 60 V source 3'9 = (10 — 60)/10 = #5 A
Pew (developed) = (5)(60) = 300 w [d] 2% = (5)2(10) + (3)2(10) + (inf/5 = 360 W
21085 = 300 + 60 = 360 w [e] no is independent of any ﬁnite resistance connected in series with the 3 A
current source ’01 v1 " U;
P 4.9 2.4 ~— 2 U
+ 125 + 25
'02 — ‘Ul 92 ’02 25 + 250 + 375 3'2 :0 Solving, 111 = 25 V; ’02 = 90 V CHECK: £01259 : (3:: :— 5 W P259 : Eggs—25‘): z 169 W
P2509 = (:22: = 32.4 W
103759, I = 21.6 W 122.48 : (25){2.4) = 60 w 2pm = 5 + 169 + 32.4 + 21.6 + 60 = 288 W 21088, : (90)(3.2) = 288 W (CHECKS) P 4.10 [a] 1:1 — 128 111 1:1 — 112 8 + 48 18
U; — 1'31 “0—2 U2 — 18 +20+ 10 ’ 0
111 standard form? 1 1 1 1 _ 128
v‘(8+18+18)+”2(—1_8) — 8"
' 1 1 1 1 # 70
“1(*I§)+”2(‘18+§a+1ﬁ) — m Solving, 111296 V; 1J2=60V
128—96
=T:
.‘_96#
‘Bb—Eg—
._96—60_
“‘3’ 18 _ 4A 2'8 2A 2A .__60_
“‘ﬁ‘
, 60—70
3“: 10 pdev : + : W 3A =—1A P 4.11 {a} In standard form:
1 1 1 v( +—+—)+1 ( 1)+ ( 1) — 125
‘ 1 6 24 ’2 5 “3 24 “
(1% (1+3+1)+ (i) — 0 “1 6 v2 6 2 12 v3 12 _ 1 1 1 1 1
“I (‘11) + (“11) + '03 (1 + E + a) = ‘125
Solving, 1:1 2 10'124 V; '02 = 10.66 V; '03 = —106.57 V
Thus, 1, : £51131 = 23.76 A 2'4 = ""1 ' “2 t 15 A . JE: ‘ a, z'Uz—ﬂ3: :2 u 2 5.53 A 15 12 9.17 A 2 _ .
13 x M = 18.43 A 126 = 1'1 “5 : 8.66 A 24 P 4.15 Solving, 01 = 76 V; ’02 = 46 V; 03 = —2 V; ésuv = U A
111A = —4’ul 2 —4(76) 2 —304 W (del) pm = (1)(—2) = “2 W ((181) Paw = (30)({]) = [l W P159 : (0)2(15) = 0 W 2 2
I _'vl H76 " ‘_
P259——25—w25 —25L04W (’01 — 62):; 302 303125” 2 W _ 31.25 : 28's W ('02 — v3)? 482
1 we 2 ——m— —_ _ _,— ,
Fwd 0 1") 50 50 46 08 W Lg . 4
Psmmght) — E — E _ 0.08 w
Xmas = 0 + 23104 + 28.8 + 46.08 + 0.08 = 300 w Em” = 304 + 2 = 306 W (CHECKS) P 4.17 —3 + 11,0 IUD + v0 _ _ . , _ {no _ 200+ 10 + 20 "‘0‘ “‘— 20 [3] Solving? u, = 50 V . __ ’UO 4‘ 5%.}; 1,3 = (50  80)/20 = “1.5 A ids = 4.25 A; 533 = —7.5 V: pds = (—5éA)(éds) = 31.875 W P 4.18 [a] . 32—03
to: v1 — v2 2' + “1 +
‘3” 100 25 '02 — '01 '02 1’2 * ’03
25 + 200 + 50
v3 —— '02 '03 # 5%", + ’03 — 38.5 _ 50 + 5 20 = 0 so 5111 — 8'02 + 41:3 2 0 so —8‘U1 + 13% — 41,13 2: 0 0 so 001 ~— 4192 + 2903 = 192.5 Solving, 1.11 : —50 V; ’02 = 430 V; :93 2 2.5 V P 4.22 $6 + 10 ‘ + 5 = 0 so 221:1 — 7112 = 1120
U2 v2 — U1 '02 A 80 _ ‘. _
12 + 10 + 5 w 0 so —6?11 + 251,12 — 960 Solving, v1 = 70 V; :22 = 60 V Th115,z'0 2 v1 #U2 2 1 A
10 P 4.23 [a] [b] 50
5f}
V
20 4
500 3 m
V
20 5
30
m
— 500 — ~—
02 4 + U” 2 “4 “2 31’3 = 0 so 1302 — 403 a 004 + 005 = 1500
U3 — W E + [03 ” 1J5 : 0 so 2?Jg + 6013 + 0114 — 3'05 = 0
3 6 2
i — 500 —
“4 2 1’2 + “4 11 + v“ 4 “5 =0 so —2202+01:3+:md — 1105 = 2000
U5wv3+ﬁ+vﬁim=0 so UTJ2—203*114+4U5:0
2 4 4
Solving, 02 = 300 V; 03 = 180 V; 04 = 280 v; 05 = 100 V
. 500 — '04 500 i 280
150 — 11 — 11 — 20 A
p55; = (20)2(5) = 2000 W
_ _ 'Ul ' ’02 '01 — U4
150W — 4 + 11
500 ~— 300 500 w 280 P500v = 35,000W Check: SUD (50PM) + (4(})2(3) + (30)2[6) + (20)2(11) + (10)2(2)
+(30)2(4) + (10)2(2) + (40)2(4) : 35000 W ZPdis = P 4.27 Place 1),; / 5 inside a supernode and use the lower node as a reference. Then '01 — 50 v1 U1 — 03/5 01 — era/5
I ~— = U
10 30 + 39 + 78 134191 — 611,5 2 3900; ’03 = 50 — v1 Solving, 1:1 = 30 V; v3 = 20 V; rut, = 30 — vA/S = 30 — 4 = 26 V ...
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This note was uploaded on 09/23/2008 for the course ECE 202 taught by Professor Xou during the Fall '08 term at Clemson.
 Fall '08
 xou

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