example problern to cal c ate he
he
the seri ex connection of
in
votage drp
7esis tDTS. The ircuit u detgnedioith
sO uce two gkors R-2as
a l v povwe
suies,hen what the
in
aCrosS TesiStoi
every Tesi sto c
vo It age drop
2
|2V
Vei
V = 2V
tolotoing seieieonne otion V
tor the the folleusing 9anttie
CíY Cuit, calculate
&hown below..
Resutance
) Equivalen t
current
(b) Flow of ac v0sS evey.
resttr
otagedrop V20
c) The in eve
and the power dicsipa tion
Yesitoove2 22983
too
V
30
(a) Equivalent Rerietance
hoith
Res = to +20 +3o
the
60 -0
(a)
() (6)
Req
Cc)
() Vo7
QO+3o
(a
2
V
10tJ0+30
3
30
Vson eta0t30
ac'it
Vgon=
hows
belovo Testtora 5o
dagTam
The o
battey and (0 9ht detemine,
bth
oh
one and on the
the l e c
though
Cen t
a) he drop ac oach reje
hbr
voltage by
(6) fhe dirsipated
power
c) the
O.64
Layv
3
2
2
(6)(4) 5 5
6t4
ttuoughi
tte tinF
(
0 T tlo
T 0.6 t -3T2 I| 3 - Qt
vottagotop (G
acDs
tor ress each
3
3n
O.6
fmp a 3
Co.c) (e) Vo.6a-
)() =(. VIa
M.8ttmp Tas=8--
(
5 (
al6)
I
=l20
5
T 6 4.
I
120
Ty
ay= 5.5 I - 5 T
0
5 T + 0T,
2
5.5T -
-SI|
To-saz
Vea 4) (u)
16V
Vn(u) C) 4
Vsn - (u) (3) =|2 V
amp
Po:sX 4:4 = 69.12, VAmp
O30y VAm
e6a= (8) (4-5) 38 y VAny
Po.6
(3.2)(9) - 30.2VEny
PaR
(39(3 2) 024VAmnp
1o24Vprnp
Posa (:)(
(8) (4)) 3 2 VAmy
( )(1)= 6y VAnp
fya léVAy
(4) (4)
(4) 20
Pan= (2)(y)
cuit 6 elow
Gíven the
a'r
cqivalent re ictance
the
(a) Cal culate
of the circit
Cunent thuough the battey
lb) Calculate the
fun ction bf location
Cc) voltage akn ng that Va =0 V
on the ciIcit
at the
Gaph Cuent ad a fun
lo cation ction of
on the circuit
Vao
Ve- Vf = (2 -
27Vf 2V
Va= V
VL = laV
Ve= |a V
Vde = aV
Vo
=Vfac
10V
Ve=
10-V; 6
tov
10
=l0V /
lf-V! F 4
b
VK= DV
d e
Combined
7estsor! ae
identtcol as AhoLon in
5 ou
four diffeunt Iay ot
in the or
sake
below. TesistanQ
the hove a
ce, let cach rins how the
nien
Con ve -
0f The t end
Combin ation.
of a ch
atictly paalu
tuminals
combination
(0 Wich total
rettante
Detimine
1t seiu!
stictty
combinaton a
lich esstance -o
() eitance
Deleenne
tth total you
combinatior
two
Wich of the b ha the (b
Cc) a
and
analyzd fn pat, does it have
lower Tetstance
-
lowea nesistance ?
tte valen t
etono
Detesn ne the egui
(d)
com binoations
emainng
of t wo
too cornbi nations you
(e) whícl of
in pact d. ha the loa (c)
Kesistane
Teaistana' th
docs t has ta lowe
flo
the
(a) )
=0.25
sticty seaies
e
stonc
Rer 4R
ley
(o) for fig (b) has lowe rettance. Bec aue
n! they provide malttple paths tor cueRent
the oveall
fow, effectively increasing
Conductance of the cir cuit , thus Te duing
the reutan ce.
Ce
th
p a r a l l e e
(R) (e) 1
R+ R
(o
Roy
SR
Paallel paralll
|Rer 2
R
2
R
) for tAg (a) has lowe reik tance. ince
pauallel combination but
ase
theu
o) ondy paalt om binatin
teee.
Gr cuit below is made p of elever
tdnicoat Telstois Connected to a
15 V poude Determine
aTotal resetan ce
() the cwnt trom powet suppy
nestoy R,
p ) the Cuuent thuough
the cuent
theouab resy to Ri
a) the cuenbuthough esis tor R3
R
Series
(3) (R)
(3R+)
ISV
+R=p+ &R -15R ( R=)
Rer
(5) Viowce =|5V
Ren =L5
p-v= (5)(I5) 120 Wate
P= VT
190- (15)I
15V
mm
15 = 2i+ Ri
8
15
&
BR
(e)
3 R 0 c
ate
at
fi
t,3 when
R the
when
the
R+ 3R
cooatY in senles witha
e ou have a capacitor
a
poer Sppy
Swit ch , a esis tor and witch a n
he
thw So
At to, you
Cuent begins to fow.
bad been hafa So
Capacitor
(o) If the capac Cuvent have
the
how ODulol,
the coP have
flowed ?
That B,
Alowe,
Tutity
fastes 0r
charged
7ou esponse
loo volta
+=0
w h e n
acos cap
he
volbage
to O atonda atate
upto to ovotto
reaChes
t
when
complety
changes cor
e copac To
t0
n t ik changed to
shen
itha deCreased
se Vaue
So, ecreaied
?t neaches
to steady state fastenà
Sor;ola chaqe
chage faster o
so, cap
for c=
we
ciTCit shouon,in
Copadtoi4 in the
9. The
triti allt cunch arg ed; the
switch. k closa. kno oin
to) the capa citvr values!
yesistoy and
Ca) Detemine all hee initial cusents
m he cir Cit cuut jut afte
the
the switch u clozed) in the
threp'cunt
Detemine a time
C
a long peo
cir cuit, aftertheontiCal point t at
(ie-y at the
n,
them wn te out
(o) klithoutsoluingwoulaned to solve
the eu ation you the cuut
to oletemin e
if you waoted
arbitary point
In the circit Comple te.
ae
Be dwe you the (b
1n tirn. e
charg
the totae
(d) Deteenine accumulate S
6mfcapacito, ill on its plate
of charge
amount
Cie, the Rwitc clos es
tt ) at to
a 30
12
raoV
1246
at t=o
capa cttoTs ae s.c
i3-0
( copacitor3 aqe o.c
120 501 al
Voltage acmAs 30.2 resistr =
voltaqe
(G f) (48) a88xI0 Colnbs
20xI|
30Ta =0.
You
7. Ue a battey
ba ttey kohoge voltage Vo
to change upp aa Capacttor C when
tully chaged, thereYoui then
ga diCanneck
Chaige orn the cap. You then and
the battey
the capactor orn nchage ol
<ecod
Deconnect 1t to a
ce i
Capa citan
Capa citor ohose
Awit c
voHage
acDk cap actby
4+ st at t(o) ttu the
voltage
t(oo) -the
at
ateady atate) tt
s v[at
copacitor of sc paalet
connect a cagacitor
we capacttos diu chauges
then -the
to c
at
&0me te rme (aftet te ady
in to
acOs and sé beco me
state) voltoqes
and Vo repe ctivey
Vottage arA
Vo
Vo
stored
1o. Obtain the enegy ene
Cundee
a ch Capa citormF:
3k
CapacitDI
ot
conition
nole DC afte
Ans: circuit. Because
fer the
A(oNYTSANAMCAAM epen
chasged, i preverst
dup of
buil
GATE P¦ to th
due tively
CuurRent flovo plates,
effec
on is
charge cuerent.
the DC
3kn
a e deie
Tka 3 X 6 mA = 3x GmA2nA
3+6
Va=
3
enegy stoed
ehg ato ied In 2m F cap = x Xro»X
s tored
eneg7
3
Én
