Energy Storage Elements - Circuit Analysis, Notes for Exam | ECE 2040, Exams of Electrical Circuit Analysis

Material Type: Exam; Class: Circuit Analysis; Subject: Electrical & Computer Engr; University: Georgia Institute of Technology-Main Campus; Term: Fall 2011;

Typology: Exams

2010/2011

Uploaded on 12/11/2011

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Problem 1:
Energy storage elements
Show that the parallel combination of two inductors with inductance
Ll
and
L2
is again an
inductor and determine its inductance.
Se.e.
CI().$.S
~~..l
,
2
pf3
pf4
pf5

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Problem 1: Energy storage elements

Show that the parallel combination of two inductors with inductance Ll and L2 is again an

inductor and determine its inductance.

Se.e. CI().$.S ~~..l ,

Problem 2: AC Circuits

lOkD ~^ jX

+E, WA~ _j8-+

In the circuit above, it is known that the magnitude of E; and I are respectively 100 Volts (effective) and 6 Amp (effective). Determine X.

~S s~W

~ =:- 104 + J ~ -s)

1=- Cs

e,

10"" + j (z:..g)

t:) \ I :::l Es I . l~ 1

100 6 ~ ~ -s) 'L

..,...1.'£lA' (-v 8)'L ¥ _,J g J V..\ s.~ v~.s ~ ~... .::: _ D. '3~ '33:> 10

LwJ..:c' C& ColJrre..- (s l.D~ce- / ,,~ ~ s.,lvoJ wo..JJ '4"'''''-)

X -::- S -+ 9 ~qlj • £j&§ ~

16.6-

Sow.-~ ~ 1k- yA.o~ :r {It~~, /0 k. .[L. ~ 10 n.

a~ :.(,o.e.. )Z: l z: J1<90 + Qi -3)t =" 16.6i ..0-

~V-'"f~~^ o^ •^ X^ z: 1~.1~~

Problem 4: AC Power

A generator delivers power to two parallel loads at 220 V with a total power factor of 0.75 lagging. One load is 4800 VA at a power factor of 0.85 lagging. The second load absorbs 4 kW. Determine the apparent power and power factor of the second load. :t

1..10 v ~. 1: Z: l 9~'9~Z:"Z...

Co\l~V~t,~ l~W!;: '. T~ == '::El +~

~ :=. ~\ -to ~L

S~ :::.'3.-v S2..

1:" \S.~ +h>tY~t +k ~Q'r-.:;r.." V=- =«, pf::.. t:>,t-~ b~~·""J -:::.Cclf(.

.; 1?--=-~~__ =- 0, 'is

_\S'~_

1.) a~lo~c:t1... ~ v- t.'U;)V oJ \St \ -:::. 4go~~ ?~:. 6A~ ~'}A ..••~ = ~~

=".;> l?l ~ 4.Roo K D ,g S .::!' 1-o.g0 kW __ k~_

~\ ~ 4800)( V;- 0.8 S'L :=. 4 goo x D. 5"'Z.66> = eses. S 6

$~~ J J at lo~az..: V::: 7...2.1Y. '::P.'t.,~ '4000 k W

~\I ~ ~W ~ 4~/ _~N,~ L=:..~_ -t- J?L =:. 4 ~g() +- 4-000 =- Jb€o ~-L-'ef-

::) \S)"Ir> ::0 8o,go =- (oH3. 3$ vA

(').7-5 - kVAR.

Q%,,-, =- \ ~ \ ~ <f. =- (0 1-1-3. 33 V, - o. H'l. =- 7- m;,8'j

12 kllA~

.; 02 z: Q~ - ~\ =- i-\2.S,~Cj-- 2 S2g :~;;-6= 4- S":l t- 33

I r;::c--;::. t6> kYAJIP

Sz t: y:t;. +6<t.:::: 6o~;.S<S'

'?~ z: Co.s { z: ~~::::: 4- 0

0 0_ -:::: 0.656't 2- z, S 609~.~ ~

C.

I

l s-~ .,,01 -e,^ "?1I6^ .•..-;)~]^ J

$) (^) oooJ" (^) - ~~ +^ JA ,

ct

*S",* oJ'5 (^) - _~I_^

01. (^) 01 '.51 4-^ _J/_^ - (^) .-

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