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Unformatted text preview: thR 2430 Name 1 [Mt ALMSJ/t
’l‘liixd Exam
Nouemher 30, 2007 Must Sectiun: Problems 171‘) ﬂ SecohdSecu‘on:1l .39:
12 39' Tomlﬂ/ Closed hook. closed class notes.
three 35” x 11.0" sheets of notes permitted.
Calculators permitted. Problems 140 nre 5 points ench. 0n multiple choice questions, circle the letter for the
auswcr ofymlr choice 1. The current through a 100 tnll inductor is t'(t) = 505”. How much energy is stored in
the inductor at t : 0.2 s 7 w: EI'V w , ﬁ'lavwhl'iloe'zm l
/t,: iotmj , 2.
Z :
Z5
(1,.”ij "ﬁfe in,
3. In a linear . with n sinnsoidnl voltage input operating at 1000 ind/s, what is the
mathematical fomn ofthe current in the component farthest from the source? Vénj‘n sinusoid operating a! 1000 rad/s
b. A sinusoid opernting at 2000n had/S
c A decaying expunenlial with a lime constant uf 1000 s
d. A decaying exponential with n time conslanl of 20001: s 8, The impedance seen at terminals aeb in the circuit below is about 100 n
u . “My 1
=<
j75 Q a“
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r .—L_ e. mousse
b, 10.9:1'3129 ) l
lace/Sim W l 30' J37
d, 0.1+j35S2
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// J / .0117, 9. The KVL equation for a singleloop circuit is d1i(t)+£dii(t)+it(l)=ll
all2 L dz LC Find the values of: inthe solution i(t) = Ale“ + A,e"' if R = 800, L = 2.5 mi , entl C=100p1ﬁ
CAL; , Li Z .
glhpr t {000mm .J
it ‘ et :1,» gloat); Ari(Udﬁdnﬁr’ r; 5, —3lg14750% om» ’lZﬁq/‘ll
/ 10. RepealProblem9ifK=8Q, L=25mH.nl:td C=1011F. .z, s 4:“ + '510% n Llanodwﬂ
_. h,
.o
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31* $105 >L/vta J
g; ~llo 110455”;
3 4. Which stalemcm is true? /éQl‘he voltage across a capacitor cannot change instantaneously.
Jet The \ultage across a resistor cannot change instantaneously
z The voltage across an inductor cnnnot Chang: insmnmneously.
a. The Voltage ncross sin ideal voltage source cannot change instantaneously. s. in u phasor diagram 1h. impedance phasors always form angles less than 90“. lhe lengths ofthe phasors ore proponionel to the angle orthe phenns.
79‘) The voltage and current phasors nre ﬁxed in position and do not rotate.
X All 0f the phnsnrs rotate clockwise at the excitnhon ﬁequenm/ 5. In salving electric circuit pmhlems that me described by differential equations, the initial
conditions found by an analysis oflh: circuit print to l: 0 are 4captwilol vullxge mid inductor current.
V cnpscitor voltage and induculr voltage. c. capacitor current and inductor currenlr /d cup ,iloreurretltnnd inductol voltage. 7. What is the voltage at tenninnls aeb in the circuit below't7 V, mum ll. (25 points)
before aperating at t = o. lihd i0) 2 2 o. loo 1) In the circuit helow, the switch has hccn in its position for along time «+61 3 : :1 J l o
12_ (25 points) In the circuit below, 7, :12040° V(rms) and I72 = 130120" V(rms) ‘
Find ix. ‘IWLM K _ I
'LQlo E‘Jio JSO’ O
a / : jsﬂ/jgwjy ,36 M : *LWH’ 3.9”} : 71(in bmun
4,01/7L"37r1‘4b‘ : IZJZL LImmqq" Tm [HHLW ...
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 Spring '08
 Farris

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