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Unformatted text preview: 13.5 (a) The equivalent resistance of the two parallel 7.00 E)
resistors is Thus. Rub = R4 +Rp +129 =(4.00+4.12+9100) Q: 111;: (m 3.
(13) Int: "5: 40V=].99A,s0 14:19:1qu
R 17.19 ab Also, (AV)? = labRF = {1.99 A](4.i2 :2): 3.13 v (Ml 313v
Then. L=' “"2 ' = LUA
’ R 10.09 :i 0.818A 18.9 (3} Turn the circuit given in Figure P189 90° counterclockwise to observe that it is equivalent
to that shown in Figure 1 below. This reduces. in stages. as shown in the following ﬁgures. b i b )
120 _ 00 I20
2%0: 0Q Ill"GillT 500
10.0 5.00 C 10.0 Q 25.0
(2 Q Q S).
10.0 g 20.0 10.0 g
o o o
E I:
Figure 1 Figure 2
i1
1’ 1"
25.U__ 25.0__
V —— V ‘—
2.94 12.9
Q Q 10.0 g
o .0
Figure 3 Figure 4
From Figure 4.
i=ﬂ— 25DV=L93A R _ 12.9 o
(b) From Figure 3. (swim = mm = (1.93 A)(2.94 o}: 5.53 V (a) From Figures 1 and 2, the current through the 20.0 9 resistor is (2.1!) 5.68 V
l. = b” = = 0.22TA
‘0 Rm 253 Q 18.13 The resistors in the circuit can be combined in the stages shown below to yield an equivalent
resistance of Rmi = [63]] I) Q. 3.0 n
L? ﬂ IIIII II'IIIVIIII
_nJI".¥.'I.II,—C an— 309 3.051
W I.
t J
3.0 o *I 60 Q
a ﬁrm—b, 1 4.0 £2
vf 'l'v'l‘l "RN
2 ‘ 2.0 ﬂ
 Wt
12 £2 " 18V
Figure1 Figurez
6.051
30 as
—Q —Q
a 3!! b H d a 1 a,
I 1'
18V 18V
18V
Figure 3 Figure 4 Figure 5 {Av} 13 v
rom iguie _ Rad (63:51]) 9 Then. from Figure 4. {AVde = i'RM = {3.14 A][30/1] Q] = 8.5? V Now. look at Figure 2 and observe that (AWN _ 3.5? V =—— =l.?lA
3.0 Q+2.U Q 5.09 1 so {Al/lb: = QR,” = {131 A]{3.{] Q): 5.14 V 


.0
.p.
Lu
3" Finally, from Figure I. In RIB 12 ﬂ 13.13 Observe that the center branch ofthis circuit. that is the branch containing points a and b. is not a
continuous conducting path, so no current can ﬂow in this branch. The only current in the circuit
ﬂows counterclockwise around the perimeter of this circuit. Going counterclockwise around the
this outer loop and applying Kirchhoff’s loop rule gives —s.ov—{2.o (in—{3.0 n)r+12V—[10 cur—{5.0 oji=o _ 12 v—sov
209 Now, we start at point D and go around the upper panel of the circuit to point a. keeping track of
changes in potential as they occur. This gives or I = 0.20 A avg = v; — v; = —4.0 V+{6.oo)(o]—(3.0 o)(o.2o A)+12 V— (10 n)[0.2o A): +5.4 v Since AVab >~ 0, point a is 5.4 V higher in potential than point if). 18.23 (a) We name the currents I, . 13, and I3 as shown. 4 00 kg
5 C W d
Applying Kji'chhoff’s loop rule to loop abcfa. T
gives +£,—52— Rgrg— R11] = o It 91 El 82 R3 93 it;
0i“ 31’: + 21'] =100 m 70.0 v “60.0 v son v__
and I.=5.00ntA—1.5012 [1]
R2 g 3.00 m
Applying the loop rule to loop edcfe yields
+83—RJIS—82—R313 =0 2.00 kﬂ
or 3L+4l =20.0mA will :
_ 3 R R1 f L
and I3 = 5.00 mA— 0.7"50l’2 [2]
Finally, applying Kirchhoff ‘s junction rule at junc—
tion 6 gives
12 = llr1 + llr3 [3] Substituting Equations [1] and [2] into [3] yields
12 = 5.00 mA— 1.501"2 + 5.00 mA— 0.7"50l'2 or 3.25!2 =10.0 mA and I3 = 3.08 mA .Then [1] gives 1'] = 0.380 mA , and from [2] I; = 2.69 mA . (b) Start at point c and go to pointf, recording changes in potential to obtain VJ. VC— 8; R213— 60.0v (3.00X103Q][3.08><10‘3A}=—69.2V (I = 69.2 V and point c is at the higher potential or AV 13.39 From {P = { civil/R. the resistance of the element is (at!)2 _ (240 it)2
re _ 3000w R: =19.2 Q When the element is connected to a 120—3! source. we ﬁnd that av 120v
r=—= = sass . d (a) R .929 a" (s) e=tavjr=uzo Wises a): 750w 18.60 The total resistance in the circuit is —l —
R=[l+] = 1 + 1 =1.2ko
R. a. 2.0m 3.0 to and the total capacitance is C = C, + C1 = 2.0 ttF + 3.011]: = 5.0 MP. Thus. Q... = CE: (5.0 ammo V] = 600 pc 6.0
and I: RC: (1.2x103 o](5.0x10—5 F]: 6.0x10‘3 5: 5
 1000
The total stored charge at any time I is then
Q = Q. + Q. = Q... (1— at”) or Qt+Qz=(600t1C]i1 it [1] Since the capacitors are in parallel with each other, the same potential difference exists across
both at any time. Therefore. (Al/if = % = 01' Q; = [%]Q. = 1.5 Q. [2]
l  C1
Substituting Equation [2] into [1] gives ZSQI = (600 HC)(1_ e—Inoou'sns] and _ (240HC)(1_ e_I000=,'s.ot] Then. Equation [2] yields Q2 = l.5(240 yCl(l— 9—1000""6'°‘)= (360 Inc)“ _ e—l Wort6.05) 18.58 Consider a battery of emf 8 connected between
points a and b as shown. Applying Kirchhoff '3
loop rule to loop acben gives —(l.0)[1—(1.0][1—I3]+8= 0
or 13=2l,—8 [1] Applying the loop rule to loop adbea gives —[3.0]I3—(5.D)[l2+l3]+8=0
or 812+513=8 [2] For loop adcn, the loop rule yields 11H“; 43.0)12 +0.0):3 +(].0}l, = 0 or 13 = [3]
Substituting Equation [1] into [3] gives [3 = I. — 8/3. [4]
. . . . 26 , 13
Now, substitute Equations [1] and [4] into [2] to obtain 181’] = g8, which reduces to II = ES.
Then E uation [4] ives I — E— i 8— i8 and [l] ields l ——i&‘
' q g 2 2? 27 27 ' y 3 27 ' Then. applying Kirchhoff’sjunction rule atjunction (I gives I = {I ‘3": =£5+i5 = £8.Therefore._ Rab =§=LZ E Q _
27 27 27 l' (178.327) 1 ...
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This note was uploaded on 01/06/2012 for the course PHYS 205 taught by Professor Antar during the Spring '09 term at American University of Beirut.
 Spring '09
 ANTAR
 Physics

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