Reading Assignment: Chapter 2 in Electric Circuits, 9 th Edition by Nilsson
ELECTRIC CIRCUITS II ASSIGNMENT I One of the questions ......ELECTRIC CIRCUITS II ASSIGNMENT I One...
Transcript of ELECTRIC CIRCUITS II ASSIGNMENT I One of the questions ......ELECTRIC CIRCUITS II ASSIGNMENT I One...
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ELECTRIC CIRCUITS II
ASSIGNMENT I
One of the questions included in this document shall be selected for in class assignment that will be conducted during the week of
Feb 29, 2020 - March 5, 2020
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Question (1) In the circuit shown in Figure 1 the current magnitudes are known to be: h = 29.9A, 11 = 22.3A, and
h = 8.0 A. If the supply frequency is 60Hz, obtain the circuit constants Rand L.
' \Tr I - \CS \ I 2. \
- \S 'I.. '8
\ \ \rT \ - j ?..o ~\
I .I \ = \~11\\41 \
\ '1 I -I
'i, -
I 't. \ 'L (_-) -t ( _ ) =
R ' w L
I 'IT I -
+ c
\r_ I
\ ) "' wl
\ j\.~L
.. Ir
► R • ►
•► -
( I \ )1.. ~ -t I~ -t"
.,_ . 2
(-' ) ; ('2.'i .°\) WL 12.~
l •· l l2
I .. ► .. ; i«>L :~ 1s n ... ~
1
2. 1'5 "4<- _ o- o'l3i"1
\-R = 5".T1 SL J
' vJL
L -
-- 0 .Ob ~ I}
\
, 2 0,;: ( 0 _06-=t-1,)
"
Question (2)
In the network shown in Figure 2, the two equal capacitances C and the parallel resistance R arc
adjusted until the detector current Io is zero. Assuming a source angular frequency co, determine the
values of Rx and Lx.
C C --t> o A I I I I
I I --. +2-!- ! lo Is r,: i rs -:cl -= 0
Zs I
>J i ) Zo .~
v,cp Rx -~
:-,L
Figure 2
. , ~s~ Cvc-t'en-1- <liv,'5 ion :
- j Xe. - JX<-T I~
:1 - R- j X c. - j X. c.
k.\JL
~x -
- -Z.x I s
Lx Is
I -
- l - ) Xe.. ) I I - D
- J Xe. _L ,
(
- j Xe.. -j Xe.. ---
i< - j 'l.X c.
X l. C.
K - j 2 Xe. R -- - J ~l.
2x X 2. Xi. c., c.. Xe..
~ x~ \ ~x
\ ~ x - ----t:,. - - -1> - --R
~)( x"\.
(.. 4
'2. - \ Lx Xe... \ L~ - -t, =- ---t> -Xe... Wlx 2w
I
-
=
\ \ - ·+- -i<x Jwlx
\ 1
t 'Z, 'l. R we
\ l 2-w1.. C..
Question (3)
The circuit in Figure 3 is operating at sinusoidal steady state. The capacitor is adjusted until the
current lg is in phase with V g. Specify the capacitor in Microfarads if the phase angle between lg and
V g is 40° leading.
\r<t - -z._, 'I~
~-\- j ~ ·,~ ~
z~
--
-- jWL +
Figure 3
R ~ Jv.JC~ 1..
j -t (wcf<. )z.
'T j I WL -\ -t- l vJ c.,P- ) 'L L
LZ~
(i)L -1 + ( w c..RJ i
=-t>
'- 1 wcR
\ t ( Wc.1<. )"L
- CJ
L -+ L( w CR )1.. -= CR
C -'K :t J ~ ~ - 4 w~ it c-
2 w 2. ff· L
Y<>-<+ '2 'el'""" -ttcJ- ii;._ rhaS.e.. U'...__ ~~ l.\ 0 ~ LG>-.d.t~
Lz'i = e.,. - G, - - 40°
l"" l ~\ tOJ\ ( -40) =
Re.. 1. Z~J y.,_ \"zi 1 -h,,, c -\.\ 0 ) = __1_"' \ Li J ~ "tevl\ ~40)
\ -+- (wcR) 1.
WC R--i. wl ----
\ + lW CR) ,_
K-\,~ (-4 o) = wL l \ + ( L-:>C: A.l·} - w c.. R "L,.
l "2. 'l. ' ( w L T wl we R ) - <...JC R - R. ~~ - 4 o) -=- <:)
[ c.,}--,;t L] C ,,_ + [ - G..)R 1 C. + ( wL - R -lov... l -40)] =- 0
wR = J ( -w~ ) "l.. - 4 ( w3 R"?. L) ( wL - R-b(-40))
1 u:l:, Q'- L
..
Question (4)
In the circuit shown in figure 4, i {t) = 10 sin t . Calculate the value of C such that v {t) = Vo sin t V. Find Vo.
1n r-----.------.....--",l\J\~
i(t) t +
1{1) l F C
w = 1 ,o.J. ls Figure 4
~~ \"
\.. l-t) -\iltl -
G
..l...
t i,.._.._ ~o~c.,.a~ -=-\> I" '? ho,.~0.- do(","\~..,..
Jo CS ,1' (~ ) T -
\jl) ~ \Y"I ( ~} V --
-- + . 2 J J
_ ~ \ I l • I + j Xe_
- j \ - j X~ \ ,,_ \ ~ i X c:.. l l
. I - j - -r
L.
j
\ + x~ c..
3o \re
\-t i ~ <:..
. t X, -r J
\ + X "\. c..
- G
-~io = 1c l- qo"
-~~ =- \ro L-9o~
U.si0 @ : 'fkse.. '«\ ".,J.;; 0 ~
.. L \/ = -la.V\-\ [I~\ ·Y-to\; \ l t--c~ot Re. 1 'f ~'c 1 ~
:. Trv-1 \ ~ "tot J =- c,
Xe_ l --\ + )<'l. 'l.
c..
'l Xe.. = \ + x~ x-z.. - 2. X '- + j - ~
C,
we_
\()
\To
o'.. \ . -s O _ 2 o ""J
Question (5) Figure 5 shows a Wien-bridge network. What is the frequency at which the phase shift between the
input and output signals is zero? What is the ratio Yo/ Yi at that freq uency.
7 - ~ T" I - ' J WC..
i<
.. \r . l-, L -
~, -tt<1. . \r· L'l. I -
zl -t -z,
Figure 5
I I
¥..\fl - \.r6 -I').7-, -t- "I,R1 -= Ci
-=-\f. 7, \f·
- \f, -r \ 0 I z , -t z.,.
\f"o R., "Z., - - R,-r (<..,,. \r- z, + z,.. \
R, 'R ,;-Ri.
for ze.ft) f'~C cs h\~-\- I(Y\ \ ¼ 1 = 0 \
K-1- \/jwc
R;- y -t R J'v)C. l T J t.vCR
~ J WC.
-
't. ( \ -t ~w c R)
-=
0
( j + )w e~) i. -r jwcR
. . \r-t
::i- w1.c'l R'.l ~ ) ':L.wC.~
1 - w '-c'l.. ~'" ;- j owc..R
"' ( 1 - w,.ct R1..) - S 3DJc.R
x ( (.1 - w"l.ci.R""t. ) - ~3wc.RJ
~ ( \ - 1.,:}-c 'L 'i<.' ) ~ ~ '.lwc.R 1 ~ (_ I - w..._ c"-fl_-... ) - j ~ w e,~
( \ - w"\..c.-i. ~ ) 1.. -t ( 3wCR ) 1...
.
( 1 - 1._,}-c_ 'L ~ ~ - ~?, w c_ ~ ( \ - w .._ C 1/ ) T j '.2.W cR ( I -w .,__ c' R ) + (j 2""<:.R) b;,, WQl.) "2.
:Irr> 1 ~i ~ = 0
7 ~ (
1 _ c,;-' c, .,_ f<. '- } -1--~ (1 - w'-C p_" ~ =- 0
"2.. t. 'l.. n, . ..._"2... c °t.. o"'L 0 - 3 -t 3 ~ c R ;- 'l. -, J...V'J " -
vf c., "Z. ~ "'L. - 1
1 w --
'KC
j j =
2-r::RC
\Jo
\f· I
'-.r· I
~, ~~1 z, ) 'i<' ~ R, z,-+ 2'L
R, ( "l.l-4. ) \-WC~ -+ b u)7.. c 1. R "l..
R, + R1.- ( \ -W1.. C 't. R."1.. ) ,_ -t ('~ v-JC R) "l..
Question (6) The ac bridge circuit of Figure 6 is called a Wien bridge. It is used for measuring the frequency of a
source. Find an expression for the supply frequency when the bridge is balanced .
.....------------ll \f"s
Figure 6
Ob L b,,d~e. ,~ boJO-K\C'.-'2- J 0
\;. C> \r°'- ·- \r b -:=. Qi =t> ~ - b o C
L.2. ~ L."' \Ts \r'S ::: Va. - -z..a + z\{ -z.., + -z'l.
L2 rz'1 --
72 -Z3 = Z ,L.4
-X ( n ) f __: j'X c.l.\ R 4 \ ( Ri. - J C.1. )( R3) ::::. f( I \ \"')~ ' X ' ) ~" J C'.'\
( ~19-~ - ~ Xc."l-~ 3) (R- j X e_~ ') =- - j Xc.y~, R"' "\
~l.~~ ~'1 - ~ X C.'2. K~ R" - ~ Xe.'-\~ '2. \{~ - Xc..4 Xc..1 ~ 3 = -j X~~ ~, R~
~· ~3 ~ l\ - Xc.~X. u ~31 -j l Xc.8 ,3 R'I .,,. X ... s ... ~31 = - ~ X '-'\~ I~ '\
Xc).~ ~R'i -t x~ Q.,~ = x'-'< '<, ~~
t<-i ~~ '<-z. '<3 - ~'~" G -t -wc"l... W C t.t we.\{
~1. ~"
\ \ = -we'\ wc'2.. R~~~ ~-z..~l ~,\<.~ 1" -
c'1.- ~ c \\
= "Q'" \<" C \\ e,1.. (0'2... - Co.M.-N:1 \ ~e.~c.(L w j,.C) ~
w -- j ~LR;C,.C,_
~ \ f't'\~eK\~J Co"" f ooe.--\-
l= \ \
LI\ ~,.R'1c"\c'2.
Question (7) The ac bridge circuit of Figure 7 is known as a Maxwell bridge and is used for accurate measurement
of inductance and resistance of a coil in terms of a standard capacitance Cs. Find an expression for Rx
and Lx when the bridge is balanced.
LJ-z .- _- j Xc.s R
I -
R- ''x J c~
7 - ~.., '-l... - ....
23 - ~~ Lx
~\',J~ \:}~~ ·= 0
"3"s Lz
Figure 7
J
z. / 3
z; 7_ x ;- Z2. 7_~ - L.7. L.3 t- L'l.L~ ./"'
Z2. -Z.3
\ -j )(C5-« ]
l R- J X c_~
R2R3 ( R -j X~ )
- j Xc.s R
-Zx ~\<L ~1
_ - - ~ ~; ~1.~~ - -t"
-~ A<-s~ _ - ~~~Q.
'Zx ~,. \< 1 . 1<z. R~
- --t- J R xcs
. ;- j
A.x -
G{(L>< = \<,_ ?.~ e,,6cs
\ Lx = ~LR3 Cs \