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Unformatted text preview: Thevenin and Norton Eguivalents Finding the Thevenin or Norton Resistance: 1. "Deactivate" the indeEendent sources voltage source : 0 volts 2 "short" current source 0 amps = "open" 2. With degendent sources, after you deactivate
the independent sources, apply a test voltage
or test current at the terminals: Rth 3 Ru 2 (ﬁtest) / (itest) In most cases, you can set one voltage or
current in the "test" circuit to a "convenient" value and then find vtestandit39¢ c)me amt vow» .. '5 '
Lb: b1='Lfb(40'bbh
103:2: 02. = *QSQZU 1 La: —40(:b Lb: 58.7114 in (32— :G [cm
is; : I14 M A Him n G
hq l ‘qlu
12m = 3 0‘1 : Lia'sz 3 ~11qu lea; Lyﬁmew mV ~ —'5
1983 bB‘x‘: ‘ +1.93: A
’40“: [mg 25mg“ 'R :b‘we’r‘
TH /Lk9+= 199—;—
zamauﬁ = L13IZ.C:G 1.2 Mallard EE 2i]! Spring 21303 MW]: mm — Homework: MARI Page I of: E Homework: M.4.71 You may submit this Homework :1 tom] of [U Limes and receive full orediL.
1«Fistr eumuIaLix‘e enm ietinn status This is similar to problem 3153 in OUT Lest This question LT adopted. with penniﬁiwiﬁ'om the publisher. ﬁres: problem 4!. FE nfaur ieuboek, Eieen'ic Circuits. Misses out! Ri'edei. fill
edition. AddisonWesley, I996: ft is intendedfor use by registered “infants or fawn State University and may not its meti'for other
pitaposes without the CAPTEJFIL'I! permission of [he pubi'fs'hcr. Given that:
v3 = see 1colts,
R2 = lﬂﬂ ohms.
R3 =1ﬂ ohms. and
i11 =ss *1:
in the circuit below. R1 = 24 ohms.
114 e If: ohms. The variable resistor. in the circuit above is adjusted until power dissipated in the resistor is 250 watts. Two values of RC
are found. The solution Stratumfor this problem Ltﬁ'rsr mﬁnd the Thevenfn equix'ai'enr circuit Hear the variable food resistor sees
making bad: inn: rm: circuit. Ham to write the mothenmtierxi expressinnfor the power dissipated in the varmwe Road resistor
m aﬁmcrimr rafﬂe set this equal to 2‘50 Watts and solve:on R0. Note that the mathematical solution wi'ii' involve monotone numbers that are ofﬂie same order ofrrmnnimde. Therefore to
get your mumer m be within the 2% limit alianed deviation. you may need to carry settemi' s'fgny'imntﬁgm'es aiming in you
compurotfons. 1. Find the Thevenin. open circuit voltage as seen by the variable iond resistor, RCl {with peisrity e on top GER“ and  on the
hetiem of Ru} . I _ _ _ _ volLs. 1. Find the larger value of Ru . htips:Heee—oniine.ee.isstate.edw'EE2ﬂ1spﬂ3mwf]{Illi'webquizegi?SessionLD=It_ruempe]_142:3... 2f 1 "H2 [303 Mallard BE 20] Spring 2:333 MWF lﬂﬂl]  Homework: MAJ] Page 2 of 2 l _ _ __ uhms. 3. Flncl the value e1" veltuge across the larger value ef Ru . l _ _ _ _ _vells. 4. Find the smaller value uf RU . _ ___ __ _ ___ uhms. 5. Find the value ui' Unltage HEIDI15 the smaller value of RB . _ l wills. e'uemitjfdrﬁfeﬁieﬁ" ' % .uauumcegmzlelreeeeeeeee
Chums Material Cupyﬁgh! :3] I993  29133 Iowa Siam Um'vgrﬂlj.‘ httpszﬂe ce—enlt11e.ee.iesrate.edwEE2ﬂ 1 513133 mwf l ﬂfwebquiz.egi?5eesienlD=kruempel_ l 133. .. 2.} 1 W2 CID} MAN MUM ﬁpcmea T‘EAN%F6E . “0+ 6% (cm 0&1 mermsgzn
: S‘hﬂ‘ ¥mm¢ mmm ecbW MM.IMQWK
$9M}. ETEL. 3.24 (:1) Find the value of R that enabies the circuit
shown to deliver maximum power to the termi~
nals a,b.. (b) Find the maximum power delivered to R. USADigs 7/“: ..... ___#—mﬂ—W_m—— 3.le
0% Wm} \Joljggg Volstaqq $0M: 0?; =00 Nada—l1 U] 0; (5120 b1_(U;+_U'¢)
— 'i' ———'l" __________.._.__.._._::
H L! E: %.\}Q3\1h&b&21 bagS—Tmﬁzo .______m,.._......—._.__...__..——«—————— .W_—__mn—u_—___W—__—m_w.__.__.n—— T854 VGH‘CLQSL/ 445$ Qmw. 3.25 Assume that the circuit in Drill Exercise 3.24 is de—
Iivering maximum power to the load resistor R. (a) How much power is the 100 V source deliver
ing to the network? (b) Repeat (a) for the dependent voltage source. (c) What percentage of the total power generated
by these two sources is delivered to the load resistor R? ...
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This note was uploaded on 04/02/2008 for the course EE 201 taught by Professor Kruempel during the Fall '06 term at Iowa State.
 Fall '06
 Kruempel

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