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Unformatted text preview: Version Dﬂi — Homework 7 — Hoﬁnianr] — [55745) i This print—out should have 12 questions.
Multiple—choice questions Ina},r continue on
the next column or page — ﬁnd all choices
before answering. Internal Resistance [I4
ﬂ'ﬂ'l 19.0 points
The emf of a batter}.r is E = 11 "v' . When the
batteryr delivers a current of [1.9 A to a load.
the potential difference between the terminals
of the battery is 9 "'v' volts.
Find the internal resistance of the battery. Correct answer: 2.22222 ﬂ. Explanation:
Given: 5 = 11V.
WW 2 9 V. and
I = [LB :1. The potential diﬁerence across the internal
resistance is E — Wand. so the internal resis—
tance is given bv 43—Hour! 3" = ——— I
llVQV {3.9 A 2.22222 ﬂ . Four Series Copper Wires
["32 10.0 points
Four copper wires of equal length are con—
nected in series. Their crosssectional areas
are 1.5 crng. 2.2 cmE. 2.? crng. and E: cmg.
If a voltage of 7'3 V is applied to the ar—
rangement, determine the voltage across the 2.2 cm2 wire. Correct answer: 19.15158 V. Explanation: Let : :11 = 1.5 c1112.
A2 = 2.2 c1112,
A3 = 2.? cm‘v’.
A4 : 5 ctnﬂ. and trim: 2 T3 V. The resistance of the copper wire is propor—
tional to the length of the wire and inversely
proportional to the ertrss sectional area of the wire. Since the lengths are the san’le1 and the
current i is the same for all resistors connected in series1 taking 1" out —._ etc, we have A i_ We! _E
rl+r2+r3+n rg'i where Via: is the diﬁerence in voltage across
all for wires and ler is the difference in voltage
across the second wire only. Solving for V2._
we have 7'2 L2 _ r1+ T2 + 1'3 + “Ir1.er
1
Ar
=a+ifi+im'
A1 :12 A3 A4
1
= 2.2 cm2
1 l. l l.
lbemz + 22cm2 2'r'crn2 +5Cm2
x [73 it] = Four Resistors JMS
nos 10.!) points
Four resistors are connected as shown in the
ﬁgure. at“ Find the resistance between points a and t. CoIIEct answer: 35.9911 fl. Version [1131  Homework T — Hoﬁmann — {EEHEJ 2
Explanation: = 14.11194 Q + 52 ii
= [56.1094 ii .
3:223
3 E. The parallel connection of R123 and R4 LEE : R1 = 21 [1‘ gives the equivalent resistance
Hg = is it,
{11325211. #2 1 +L:R4+R123
H4 2 To ft, and Rob R123 R4 Rl'23 R4
ii 1'
is = 97 V. Rab = L“
. _ . H123 + H4
DhﬂlhlﬂWlSVZIR. [ﬁﬁmmrﬂnmu] A. good rule of thumb is to eliminate jtlnc— = tions connected by zero resistanoe. ﬁﬁ‘mm H + ‘9 ﬂ = 35.9911 11. or combining the above steps, the equivalent m e R3 resistance is
n. b
( R] R2 + Rn) “ R.
d H1 + H: t
Hob = R1 R2
— R R.
£1 + R2 + :i + i
The parallel connection of R1 and R2 gives (21 Hi {43 it) + 32 H {W m
the equivalent resistance _h 2] 11 + 43 ft '
1 1 1 R2 + R. _ [:21 it) [43 o) n
_=_ _= —— rest so
H12 R+Hg HIE: 2152+43ﬂ+1 +r
RI RE — as ILiriii o
R . = _....__—._— — II I: n 1 .
12 RI + RE :1
_ {21 ii} {43 ii]
_ 21 it + 43 ﬂ Holt SF sortev 24
: 14.109411. [1114 111.11 points
H12 Rs Find the equivalent resistance of the circuit.
shown in the ﬁgure.
ﬂ 1: 11.5 i'i 3.511 The series connection of R12 and Hp, gives
the equivalent resistance Risa = R12 + Rn 1Iriersion {ltll — Homework T" — Hoffman“ — (SET'15) 3 Correct answer: 31.54154 ll. Explanation: Let: R1 = 1351, 32 = l3 ﬂ .
R15 = 14.5 H.
R4 = 3.5 ﬂ. and
Hg 2 14 ll.
R34 = R3 + H4 2 {11.5 Q] + (3.551] = 18 ﬂ
1 1 '1
3.. = __ __
m (Re i 334) 1 1 1
= [111111) + [11111)]
=?.51839 11. so
Rm = 31 + H.234 + Rs
= {13 n) + [rsesssnj + (14.11] = 34.554114 Resistor Combination Ill
BEE lilﬂ points
The following diagram shone part of an elec—
trical eirenit. '26 ll .‘l'i' El
A 71111 B . Find the equivalent resistance R“, between
points A and B of the resistor network. 1. RAB = 13 11
2. ft“; = 42 it correct
3. R“, = 11' n 4. R” s s? n s. RAE, = 74 n s. RAE = .13 n ’3'. RAB = 45 ﬂ 3. RMg = 53 fl 9. RH = at} it
113. R” : €139
Explanation: R5 R:
A
é.
B Let: R1=35H1 H2=3Tﬂ1
33:12.”.
R1=ti35L
R5=26§L
Rﬁ=29ﬂ. and
37:?851. Start from the rightrhanrl sirle in determining
the equivalent resistances.
Step 1: H1. [132. and Rs are in series. so
3123 = 31+ Rs + Rs
—_ :1511+3Tﬂ+ 1:211
= 84 ﬂ . Step 2: Hg; is now parallel with R4. so
1 __ 1 _i_fi’1+R123
Hl'ﬂl 3123 I R4 lites H4 Version ["31 — Homework 7  Hofﬁnann — [fiﬁTalﬁ] 1 Rurqu
R123 + R4
[at n} [as n)
2 at n + as it
= ﬁll: 51. R1234 = Step 3: R1234 is in series with R5 and Hg, so R3=R12M+R5+Rﬁ
=3ﬁﬂ+2ﬁﬂ+2liﬂ
=91ﬂ. Step 4: Finally R2i is parallel with Ry, so 1 _ l +1_R1+R5
Esq—R5 R»?— Rer _ are?
R“? * at + H7 m {91 nurse]
_ tnti+rsri = 429. AP EM 1993 MC 7!]
ﬂﬂﬁ 1].l] points
The following diagram shows at closed elec—
trical circuit. The nmmeter in the center of
the resistive network reads zero amperee. Find the electric resistance Rx. 1.Rx=14n.
2.Rx=iﬂﬂ.
s.Rx=sn.
4.Rx=lﬂﬂ.
5.1L =sn. {LRx '—ti£l.
T.Rx =11 ﬂ.
S.Rx=22ﬂ. 9. RT = 24 ﬂ . correct 10. HI = if: Q.
Explanation: Let: 31:3!!!
32:31}! and
Hg =95}. If the anirneter reads zero I_.. = G A, the two
ends of the amineter should he an equipoten
tial: Vuﬂ = $4. This means that the potential drop from
.J’: or y to each side of the amineter through either the upper it or lower t5 part of the circuit
must he the same. II"(Jr—M:=InIr1‘r1 {1}
i;a,—ts=neiLi {2}
Vii — Vy = It! R2 {3}
VeVy= “R: {4} Setting the potential across the ammeter
equal l—‘L = 1”} (Le, Eq. 1 = 2 and Eq. 3 =. 4],
we have In £1 = If R3 and In Hg = £31133, Luzﬂ'izﬂ
n at n2 Version [FIJI — Homework 7 — Hofﬁnann — {5131745} 5 Ra 33
RI _ R1 _WQHMH
_ so
=2ML Current in Circuit
ﬂﬂT (part 1 of 2] HM] points
Consider the circuit shown below. ‘What equation does the loop DCFED
yield"? 1. £2 ! igrg r it? = L']
2.51—tgr2miR=D
3.52—t2r2+iﬂ=ﬂ
a—o+am—ne=n 5. —£2 —t?1"g  2iR=ﬂ 6. —£'; t21"2 ~—tR= I] 7. £2 — tg’r'g — if! = I] correct
8. —E.'g — igrg + 2151?. = I]
Q.£g+t3rg+iR=ﬂ 1"]. —51 + tzfr‘z — Hi = ﬂ Explanation: Recall that Kirchhol'f's loop rule Slill'lLii that
the sum of the potential differences across all
the eiernents around a closed circuit loop is
zero. If a resistor is traversed in the direction
of the current. the change in potential is — IR. if an emf source is traversed from the — to
+ terminals1 the change in potential is +5.
Apply the opposite sign for traversing the
elements in the opposite direction. Thus by inspection, : £2 — “IiR —t21"2 = [1 {ME {part 2 of 2} 113.0 points
Let 51: £2 = E. and r1: r3 = r. where
5 = II Vandr=5ﬂ AlsoletR= 1.851.
Find the current i. Hint: From symmetry.
one expects i] = 112. Correct answer: 2.55314 A. Explanation: Let: £1:£2=£=11V,
rl=r2=r=ﬁﬂ1 and
R=1.8ﬂ. E] = £2 and r1 = r2. This implies that t; =
i2. (Why? Look at the loops DCFED and
aeeea. Notice the identit}r of the two loop
equations.) Hence the junction rule yields t1+t2=2i2=t
_ i
32:— 2 Substituting this into the loop equation
DCFED. . r
Eg—tR—Erg=[l Solving for i yields Charging RC Circuit I11
DDS 10.0 points Version ﬂﬂl — Homework ? — Hoﬁmsnn  {56?45] I Thus the corresponding potential across R; i E .r Discharging RC Circuit {11
DIG 19.9 points Let Rl=R2=Rs=R Initiallyr the switch is in position "b" and the
capacitor is uncharged. At time t = U the
switch is moved to position “:1”. Determine the potential difference across
the resistor R1 at r = R1 (3. Find the characteristic time of the circuit
when the switch is in position ‘13". 1J54
e 1 1
. T I —
2.25sJ os,ego
i
s. or. 8.3—9 2‘ T _ REG
1
_ 3_ = —
4.osse 1 T (R14joﬂ
5' 153—05: 4. r = {R1 + R210 correct
5 2 5 T — _.w...%.__.m
6'1 E ‘ ‘(s1+s2)o
1
rec” . =
F 5 T RIC
3.28e'ﬂﬁ 7.T==ngc
9.1.EEE_L5 3.T=i..r‘R] It}. 115 E's—“'5 correct 9_ T = RI R? C
Explanation: '
'When 3 is at a. the loop equation is given 10 T = R1 C
by Expianation: 5—i—4Uh+ﬂﬁ=ﬂ C In charging E111 R C circuit, the characteris— . . tic time constant is iven be
This gwcs g ” s c s o T=RC1 2 —R R 9W : E'JHI‘.‘
'3‘ . . . .
1+ '2 hR where In thls problern R Is the equivalent
At t = RC“. resistance. or i i=ﬁf'! R=m+nt Version {lﬂl » Homework 'r'  Hoﬁ'inann  [55745] T Four Resistors and Capacitor
011 [part 1 of 2) 10.0 points
The circuit has been connected as shown in
the ﬁgure for a "long" time. 'What is the magnitude of the electric po—
tential across the capacitor? Correct answer: 10 ‘2". Explanation: Let: 31:45“.
H3=52£L "After a long time” implies that the capac
itor C is full}r charged. so it acts as an open
circuit with no current flowing to it. The
equivalent circuit is ___.t it ag=e1+e2=45n+52o=1mio and
Ra=R3+Ri=2ﬂ+23ﬂ=25ﬂt 55 s 25 v
I = — = = [1.2 A. d
* 221 men 5 an
s 25 ‘v'
r = —=—.——=1.5.
'4 ab 25 52
Heroes R1. and across R3
53: IbR3=lilliliII2ﬂ1= 2V. Since 51 and 53 are "measured" from the same
point ""rt"._ the potential across C must he 5o=53—51
=2V—l?V=—lﬂv licl =. 012 [part 2 of2) 10.1] points
If the hatter}.r is disconnected. how long does it take for the 1.roltage across the capacitor to drop to a value of V[t] = where Eu is the
e initial voltage across the capacitor?
Correct answer: 3'3"] as. Explanation:
1ﬁﬁth the battery removed. the circuit is where R§=R1+Rg=rlﬁil+2ﬂ=ﬁﬂil
at=eg+ad=...2o+25o=75n Versiﬂn EH11 — Humewnrk T — anfma and an the time constant is T '=" RWC = {3UﬂH13pF}
=39ﬂ ,ns. The capacitor discharges according to g: —m Qt:
ﬂ ﬁrm =l
E}. c .t 1
—— = In (—j = —lne
T I? t= THHE} = I[3‘Elﬂ ,ns] (—1] =Eﬂﬂﬂn5. —l __ r1, +IZERH in Mamie! ,. 59 IL]: 1 2. £1... LIE{1.1L QIHZL _
11.1,” it")
a. a": 1; 47—” +5123. Mt in..$t.riﬁr!
1 at”. = _..E_.+ 9.} any;
{—ﬂ'ﬂ
.. WEN1 arm E" {1+ “1!. J“: FH‘U'LLI
. {1:— _
Qum ‘ ﬂrup‘f‘; 1 LL’ “Ah£33..) “ma —
11'” r 3* EL“ “571....
/ [1ij : WC‘M
____________H________ . ﬂr r;
‘13. r L eq— 29:2,... _ _
_ , I11 4 11M .:._ Attila? _ _
n? f Fich— . EH1. _
Inn. REL—5
., _ .. 2'1;+{' : ‘2! w . [a $..¢.~a..i'_._ﬂ,__
— = (might: m + ea —— 1& 'r
_ 9, +..’21,+!3r._'5_ E h Jam“, ' .Hni'.HQJS. 1111 H .41. ﬂu = “>41 ._ Nut+1.? M's .,=::~:F é} ) 7 This _£K_;Fa_ wﬁ_5__JuM [*5 EA”.
L111? _.= LL'rLﬂL t 3 r1. .  @HCC‘; ( 1: nag/ﬂ." )__I_%. ‘ﬂg_'..'1:'T.ﬁ_‘“’>._ ... «ﬁx—sf. .. .__.... 39..? T “49.! a” r ..i.r‘~ £361.55 at 2.11 ‘1 El Eu» :_._(_HSJ)(‘?_{) 7—: 3D  . mm _. MW 3% rid LC‘SGJUJMQ‘T) .s._
f f: =.. Aﬁwﬂﬁ) . ...
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This note was uploaded on 11/10/2011 for the course PHYSICS 303L taught by Professor Hoffman during the Fall '11 term at University of Texas.
 Fall '11
 HOFFMAN

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