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Unformatted text preview: University of California San Diego
Department of Beatrice! and Computer Engineering ECE 35 Midterm Exam
Winter 2009 Name (Last. First) LE1;— Student ID Number Signature This is a closed book exam, but you are allowed a single 8.5" x 11" page sheet of notes. You can use a calculator, but cellphones and computers are absolutely forbidden. You must show your work to receive credit on any problem. Write clearly: work which is not legible
 will not count towards your grade. Numeric answers MUST HAVE UNITS to be correct (Le. 35), 2A. 4mV) Good exam policy: (‘1) READ THROUGH THE ENTIRE EXAM BEFORE BEGINNING
(2) WORK ON THE SIMPLER PROBLEMS FIRST (3) TRY TO ANSWER ALL PROBLEMS L Car/Y. cinep.15 metric 'u. chic”; +0 fimtweeu— 5?
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2,5) miqiun‘LL ital«1‘11" A155'.'ﬂq‘I¢.J (“of awry“1”?» Problem ‘I (20 pts, 5 points each): A) Calculate the power absorbed by each element in the drive circuit below when terminals A 8: B are unconnected 2 2.
Pas)= " is?“ P(R1)=__._L... 2~ P(R2)= 0 Curves—l $44“ 4lion—as aground an? 4'. anal Vm': 1'.,Ri POLO: CU‘ ESLE\
P(IS)C .. P(a) B) Express the Thevenin equivalent voltage and resistance. and Norton equivalent current, to the drive circuit at left ML
Wei Rm=ﬂl_ In=.._a¢2..__ 1 2—. 9a 9;er
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source. ‘5‘“:ng = Li. UH... liev.
C) Express the Thevenin equivalent vottage and resistance, and Norton equivalent cunent. to the load cirCuit at right
2..
V=s..,_0__ mi 1.5—0“
2..
 : EEL  z.
a“, 2mm, 91? a  3—K“ D) What value of resistance R, will maximize the total ‘power transferred from the drive circuit to the load resistors?
R 2 %( 2 * i" 91) k 3%“: Kit. 2 KWz “" 91:15(E'+EQ Problem 2 (20 pts): An ideal current source is connected to an ideal voltage meter as shown below, and
used to measure the voltage across the resistor R1. A) if the 1 M0 resistor R1 is accidentally disconnected from the “A“ terminal. as shown, the cirmit as drawn becomes
physically impossible (2 points) Why? Curmud— musi' Plat4 “trotJ L in purlLe NJisluucc' “seals Luci«Ht vquj; . B) Modify the circuit above by adding internal resistances to the current source and to the voltage meter. showing how
they are wired, so that the dashed boxes contain more realistic components. (8 paints) C) it the voltage meter is 99% accurate (the value read is eff by at most 1%) when meaSuring the voltage across a
1 Mn resistor, what is the numeric value of the intemal resistance? (5 points) Maia rm“! KNEW, =3 V.M == 3: K,//IZh = 0.14 IR, =9 Era,“
1&2” = can a, 2M: o.s‘i(K,+a.J
_ 95L“! _
o.or RM= ofl‘l a, 2,; 0.0, R, = ‘l‘iR,11_N D) lf'the current supply is 99% efﬁcient (at most 1% of the cunent provided by the ideal source is flowing'through the
internal resistance) when driving a 1M0 resistor, what is the numeric value of the internal resistance? (5 points) comel at:c.44r: cuntair Hanaﬁt but is 1W.
1 HR— 'l (rE)=i
I... :> 034: ..———§: =9 on“: i R, .l. .1. 1+
age, I; Dﬂ‘l 11.1 I, 19 O.°l‘i % ’0‘9‘
S Problem 3 (20 pts, 10 points each): Express the answer to the two questions below in terms of the
variables deﬁned in the circuit diagram. A) How much power is deiivered by the voltage suppty V2 in the circuit above? 1301:): 12% ““A Wu“); ”‘1‘ V z “I.
1" ILU'I.
1705th 01'" “11“.“ ? EKG—HASH») roku‘ﬁa—Q =9 meals {”0 varoLq *‘Fouu B) What is the voltage across the current supply. V‘. in the circuit above? v,= "(31JLN.) Kw. mm! in? an.“ VJ" + 11(Q)+Vl+ 11(2) 4 11(2‘) :0 V: + 3 1‘.\R‘ 1U so 59 Ux "'  51.31.41. Problem 4 {20 pts, 10 points ead‘r): A) Draw the Norton equivalent to the drive circuit left, expressing the current and resistance in terms of Iabeted
values. $aurcd. Muff—urn... Contra Mala,“ LEN" 2.. +1..
1.. =9 232.;
'1 u
I H H.
Power
B) Calculate the value of current supply 1 which MiNiMIZES the Wtdissipated in the ioad resistance.
V i1
[1 = _ 3.— M‘niuuu fomr {S 1<ro 'LJL‘H Lynm“ :5 intro. Curdsi' Rib Wu. 1:” =0 Problem 5 (20 'pts, 5 points each) A) Wriie an equatiOn for KCL for a supernode containing
the two voltage sources and the dependent current source. \l
v, r + + 1 =3
R. V; ' ‘ Ur”.
(i. ran} =0
(R I
f
B) FindV 0) Calculate the power P supplied by the dependent current source.
szolv [(U V) “61“!) OH) (“V—Vt) ViaUL
[O V “’1. gar Ur
i9 = ____w—' —— E." ‘—
i L to ) Fﬂa.( R.) D) Using the following numeric values calculate the numeric answers to parts B and C above.
R1= 20 R2— = 10D V1: 20V Vz— = 10V and l1= 4A. V I! K _ I01. .312 > if. z 10v
Vx’ “3+; (7. ‘1 I2. (6) #
\J‘t‘dl _ IOU
Fr: 2 _ .c} ‘Ug Vx _..__...._
2.0 +LC IO PX: FBE—LJ
: "I If.)
, 420W ...
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 Spring '07
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