circuits sol ch 01 circuit variables oct 2010
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Mechanical engineering majors, electric circuitsTRANSCRIPT
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
هيادين
كـل
عى
بنت
ا
ـنع
ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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Chapter 1
Circuit Variables
ي + _ ة (N &W % ا& ك'
f د4 ا +(اد يV ا ت _ g! , % ة
f د4 ي dعى ا V*+(ا '(ك + ت Yة
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ا
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7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
1Z1 :=ectr6cB= :;<6;eer6;<P I; OHerH6e• Electrical engineering is the profession concerned with sste!s that
produce" trans!it" and !easure electric signals#
J>e p6He ABqor c=B886p6cBt6o;8 op e=ectr6cB= 8E8teA8P1# 5oAA7;6cBt6o; 8E8teA8P $uch as ca!eras" trans!itters" recei%ers#&# 5oAM7ter 8E8teA8P $uch as personal co!puters#'# 5o;tro= 8E8teA8P $uch as control of te!peratures" pressures" and flow
rates in an oil refiner#(# oer 8E8teA8P $uch as generators#)# -6<;B=sMroce886;< 8E8teA8P $uch as i!age*processing sste!s#
I; e=ectr6c c6rc76tP+s a !athe!atical !odel that appro,i!ates the beha%ior of an actual electrical
sste!#
@ote t>BtP the ter! electric circuit is co!!onl used to refer to an actual
electrical sste! as well as to the !odel that represents it#
+n this te,t" when we tal- about an electric circuit" we alwas !ean a !odelZ
0234ا /.ة# 78ا 6 5 9;<= >?;< @?4 ة ABD م 9 ماF?< @?4 ة ;2A4 GH< @I 0234ا /.ة JK;24ة
هيادين
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'( )*+( ةا
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3ا
ي
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! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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ل
*"! &N 3u
( ت
ا
Zv ـ
#( ا
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
1Z2 J>e Q;ter;Bt6o;B= -E8teA op w;6t8• Lhe $+ units are based on se%en basic units:
JILK: 1Z1 J>e 6;ter;Bt6o;B= -E8teA op w;6t8 x-Qy
z7B;t6tELB86c w;6t-EAbo=
Ke;<t>.eterA
.B88{6=o<rBA{<
J6Ae8eco;C8
:=ectr6c c7rre;tBAMereI
J>erAoCE;BA6c teAMerBt7reCe<ree {e=H6;|
IAo7;t op 87b8tB;ceAo=eAo=
K7A6;o78 6;te;86tEcB;Ce=BcC
+n addition" defined Muantities are co!bined to for! deri%ed units#
JILK: 1Z2 Der6HeC w;6t8 6; -Q
z7B;t6tEw;6t @BAe x-EAbo=ySorA7=B
Sre}7e;cE~ert• x~•y8s1
Sorce@eto; x@y|< Z A R 82
:;er<E or or{ Go7=e xGy@ ZA
oer€Btt x€yG R 8
:=ectr6c c>Br<e5o7=oAb x5yI Z 8
:=ectr6c Mote;t6B= xHo=tB<eyo=t xyG R 5
:=ectr6c re868tB;ceO>A x‚y R I
:=ectr6c co;C7ctB;ce-6eAe;8 x-yI R
:=ectr6c cBMBc6tB;ceSBrBC xSy5 R
.B<;et6c p=7ƒ€eber x€by Z 8
Q;C7ctB;ce~e;rE x~y€b R I
• +n !an cases" the $+ unit is either too s!all or too large to use con%enientl#
$tandard prefi,es corresponding to powers of 1N are applied to the basic unit#
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5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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aNــت # ) ــل ــ ا !'„ كلZنع „+…a †4 ‡/W
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
• Oll of these prefi,es are correct" but engineers often use onl the ones for
powers di%isible b 'P thus centi" deci" de-a" and hecto are used rarel#
JILK: 1Z? -tB;CBrC6•eC rep6ƒe8 to -6<;6pE oer8 op 10rep6ƒ-EAbo=oer
BttoB10s1ˆ
peAtop 10s1‰
M6coM10s12
;B;o;10s\
A6croŠ10s^
A6==6A10s?
centic1N*&
decid1N*1
de-ada1N>ecto>102
{6=o{ 10?
Ae<B.10^
<6<B‹10\
terBJ1012
هيادين
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& ك
'( )*+( ةا
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3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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Œ ( ع ىع X& $ا ةa( ا )Ž ZŒY
dل
ي
' ا
ت
ا
ة
!
‘(!
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
1Z^ o=tB<e B;C 57rre;t
8oAe 6AMortB;t c>BrBcter68t6c8 op e=ectr6c c>Br<eP
0234ا /.ة ;QRH4ا 5 .S./4ة T/ اU4
W Lhe charge is bipolar" !eaning that electrical effects are described in ter!s
of positi%e and negati%e charges#W Lhe electric charge e,ists in discrete Muantities" which are integral
!ultiples of the electronic charge" 1#XN&& Y 1N*1Z C#
• o=tB<e is the energ per unit charge created b the separation:
v=dw
dq
[here:
v=the voltage∈volts . w=the energy∈ joules. q=the charge∈coulombs.
• :=ectr6c c7rre;t is the rate of charge flow:i=
dq
dt
[here:i=the current ∈amperes .
q=the charge∈coulomb .
t =the time∈seconds.
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: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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! +N (` اد ةg g_ ا ةa’ ن ( اZ
V’عى
د
* †4
Vn! †(*
• \uestion 1#Z:
Lhe current entering the upper ter!inal of ]ig# 1#) is:i=24cos 4000 t A
Ossu!e the charge at the upper ter!inal is ^ero at the instant the current is
passing through its !a,i!u! %alue# ]ind the e,pression for M_t`#
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
1Z‰ J>e QCeB= LB86c 56rc76t :=eAe;t /02 ا /.ة U; ع وA;4 4 U4ا 5
• On ideal basic circuit ele!ent has three attributes:
1# it has onl two ter!inals" which are points of connection to other circuit
co!ponents#&# it is described !athe!aticall in ter!s of current andor %oltage#'# it cannot be subdi%ided into other ele!ents#
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مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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Solution:
i=dq
dt
∫q
o
q
dq=∫0
t
i dt
[ q ] qqo
=∫0
t
24cos 4000 t dt
¿ 24
4000[ sin4000 t ] t
0
q=6∗10−3
sin4000 t +qo
Current is !a,i!u! when cos 4000t is !a,i!u! @t =0
q=6∗10−3
sin 4000 (0 )+qo→ qo=0
q=6∗10−3
sin4000 t C
¿6sin 4000 t mC
_ـــ + ا ت )ـــ uا ـــ 'ي ي aـــ +Vدا `)ـــل ـ ـ ـ ـ ـ
m’ا
+_ ـــــب
U ـ ي ـ ـ ـ ـ
`
g
N “‘ ي
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
• passi%e sign con%ention:[hene%er the direction of current is sa!e as the direction of %oltage drop
across the ele!ent _as in ]ig#l#)`" use a positi%e sign in an e,pression
relateing %oltage with current# jtherwise" use a negati%e sign#
.?4ا #6 م k4 و4 0< ` @ _qع mة 4و K?S k4ع مة Tm ا;< ا< .?4ا I 6و ا< v ا
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مو
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5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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j بa! ) m ي d ا ل(V ا & ‘ اZ“ * &‘ ك4 + ه „'! ع udV ا
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October 2010
هيادين
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بنت
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ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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• Ossess!ent proble! 1#':
The current at the terminals of the
element in Fig. 1.5 is
i=0 t <0
i=20e−5000t
A t ≥0
Calculate the total charge (in
microcoulombs) entering the element
at its upper terminal.
-o=7t6o;P
i=dq
dt
∫0
q
dq=∫0
∞
i dt
q=∫0
∞
20e−5000 t
dt
¿ 20
−5,000[ e
−5000t ] ∞
0
¿−4∗10−3
(e−∞
−e
0
) ¿−4∗10−3 (0−1 )
¿4∗10−3C
¿4 mC
¿4000 μC
ـ ـ ـ ـ– ـ uع )“ !—ــــ ل ـ ـ ـd نN 4 % د 4 ̃ا ™†
ـ ـ كـ
ـــ
( f ـــ
ـــة
, ت ـ ـ ـ
a’
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
1Z^ oer B;C :;er<E p=
dw
dt
where:
p x the power in watts" w x the energ in youles" t x the ti!e in
seconds#
Lhus 1[ is eMui%alent to 1 zs#
• Lhe power can be related with %oltage and current as follows:
p=dw
dt =( dw
dq )( dq
dt )so
p=vi
where:
p x the power in watts" v x the %oltage in %olts" i x the current in
a!peres#
• Olgebraic sign of power:+f the power is positi%e _that is" if p { N`" power is being deli%ered to the
circuit inside the bo,# +f the power is negati%e _that is" if p | N`" power is
being e,tracted fro! the circuit inside the bo,#
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.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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ل
’ j ه
'N ي
4 T
VN T̃ا
™Z&œ ! /nN &# ا
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October 2010
هيادين
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3ا
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! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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• Ossess!ent proble! 1#):
Assume that a 20 V voltage rop occurs across an element
from terminal 2 to terminal 1 an that a current of ! A enters
terminal 2.a)"pecif# the values of v an i for the polarit# references
sho$n in Fig. 1.%(a)&().b)"tate $hether the circuit insie the bo' is absorbing or
elivering po$er.c` o$ much po$er is the circuit absorbing
+ ـــ ي 4 T ـــk( ا ـــ ــ
žg تــ
aNــ
Œ ــي
-o=7t6o;P
• $tart b drawing a picture of the circuit
described in the proble! state!ent:
a` }ow we ha%e to !atch the %oltage and current shown in the left figure with
the polarities shown in ]ig# 1#X#(a ) v=−20V i=−4 A (b ) v=−20V i=4 A
(c ) v=20V i=−4 A (d ) v=20V i=4 A
b"c` ~sing the reference sste! in ]ig# 1#X_d` and the passi%e sign
con%ention" p=vi=(20 ) (4 )=80! (absorbs)
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October 2010
هيادين
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& ك
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3ا
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! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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• Ossess!ent proble! 1#X:Ossu!e that the %oltage at the ter!inals of the ele!ent in ]ig# 1#) corresponding to the
current in Ossess!ent •roble! 1#' isv=0 t <0v=10e
−5000 t "V t ≥0
Calculate the total energ _in youles` deli%ered to the circuit ele!ent#
• Ossess!ent proble!s 1#':
The current at the terminals of the clement in Fig. 1.5 is
i=0 t <0
i=20e−5000t
A t ≥0
-o7=t6o;P p=vi=(10,000 e
−5000t ) (20e−5000 t )=200,000 e
−10,000t
¿2#105
e−10,000 t =
dw
dt
w=∫0
∞
p dt =∫0
∞
2∗105
e−10,000t
dt = 2∗10
5
−10,000[e−10,000 t ]∞
0
¿ 2∗10
5
−10,000 ( e−∞−e0 )= 2∗10
5
−10,000 (0−1 )=20$
4 ! 3 Tتd ة@o B6;8 6t>o7t B6;8Z
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October 2010
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! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
• \uestion 1#11:
jne Z V batter supplies 1NN !O to a ca!ping flashlight# €ow !uch energ
does the batter suppl in ) h
Solution:
p=vi
¿9∗(100∗10−3)
¿0.9!
p=dw
dt w=∫ p dt
¿∫0
5h
0.9dt
¿0.9 [t ]5h
0
¿0.9# (5#60#60 )
¿16,200 $
¿16.2"$
ـ ـ ـ ـ ـ ـ ـ ـ ـ ا ـــــــــ
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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بنت
ا
ـنع
ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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• \uestion 1#1&:Lwo electric circuits" represented b bo,es O and ‚" are connected as shown in ]ig#
•1#1&#Lhe reference direction for the current i in the interconnection and the
reference polarit for the %oltage % across the interconnection are as shown in the
figure# ]or each of the following sets of nu!erical %alues" calculate the power in theinterconnection and state whether the power is flowing fro! O to ‚ or %ice %ersa#
a` i=5 A v=120V
b` i=−8 A v=250V
c` i=16 A v=−150V
d` i=−10 A v=−480V
ا
` اš †4 † نك ا ىع … W uN Tد ™
ع
V’ ا
ا
د
Ÿى
n#U
• -o7=t6o;P
$ince the current i flows into +ve ter!inal of %oltage v " we can use the
eMuation p=vi without % ve sign# if the power is +ve " so it flows fro! O
to ‚# if the power is −ve " so it flows fro! ‚ to O#
a¿ p=(120) (5)=600! 600! ¿ A ¿& b¿ p=(250 ) (−8 )=−2000! 2000! ¿& ¿ A
c ¿ p=(−150 ) (16 )=−2400! 2400! ¿& ¿ A d ¿ p=(−480 ) (−10 )=4800! 4800! ¿ A¿ &
a` p x _*(ƒN`_*1N` x (ƒNN [ (ƒNN [ fro! O to ‚
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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ا
ـنع
ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
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! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
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• \uestion 1#1):[hen a car has a dead batter" it can often be started b connecting the batter fro!
another car across its ter!inals# Lhe positi%e ter!inals are connected together as are the
negati%e ter!inals# Lhe connection is illustrated in ]ig# •1#1)# Ossu!e the current i
in ]ig# •1#1) is !easured and found to be 'N O#a` [hich car has the dead batter
b` +f this connection is !aintained for 1 !in" how !uch energ is transferred to the
dead batter
-o7=t6o;Pa` Current i=30 A flows fro! +ve %olt to % ve %olt in car O" $o car O is
absorbing power# Car O has the dead batter#
b¿ p=v i ¿12#30=360!
w=∫0
60
p dt =∫0
60
360dt =360 [t ]600
¿21,600 $ =21.6 "$
Œ ــW ــœ ا ب_ ــ, T ــN Œ ــW ا ا Ž _ــ & 3يت Ÿ Tـ &’4 ــ * –¡لZŒ ي
X&’+ ت
( ا
X
Y+ ا
ة
+
a ا
+k’
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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ا
ـنع
ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
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.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
• \uestion 1#1X:
Lhe !anufacturer of a Z V dr*cell flashlight batter sas that the batter will
deli%er &N !O for ƒN continuous hours# „uring that ti!e the %oltage will drop
fro! Z V to X V# Ossu!e the drop in %oltage is linear with ti!e# €ow !uchenerg does the batter deli%er in this ƒN h inter%al
-o7=t6o;P
p=vi
pt =0=9# (20#10−3 )=0.18!
pt =80h=6# (20#10−3
)=0.12!
w=∫0
80h
p dt = Area o' ( p−t )diagram
¿( 0.12+0.182 )# (80#60#60)
¿43,200 $ =43.2 "$
’ Xــ / ــa ا Xد ـ ـ* بــ—N ا Uــ Vبة +اـــ ـ ـ ـg! ة ـ ـ يـ¢& ا ) # تـــaN ل ـ ـ ـ ا [“£ ل
( ــت
,بــ
¤ Œ ــن
X
a ا
ا
4
n—N
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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ـنع
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& ك
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,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
Lhe %oltage and current at the ter!inals of the circuit ele!ent in ]ig# 1#) are shown in
]ig# •1#1Z#
a` $-etch the power %ersus t" plot for N … t … )N s#
b` Calculate the energ deli%ered to the circuit ele!ent at t x (" 1&" 'X" and )N s#
• \uestion 1#1Z:
• "olution*
a` p=vi
t (sec)0−48−1620−3646−50
v (V )2.5 t 30−2.5 t t −3075−1.5 t
i ((A )1.0−1.00.4−0.6
p((! )2.5 t 2.5 t −300.4 t −120.9 t −45
N 3نـ[ ا ! 4ا Y [+ تى
ا % ـ ـ ـ ـ ’ ى ـ ـ ـ ـع ا ـ ـ ـ ـ¤
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
s ?Z^
2Z]
s]Z0
s10
10 10
• 5o;t6;7eC x}7e8t6o; 1Z1\yP
b)Calculate the area uner the curve from +ero up to 50 s*
w (4)=1
2(4 ) (10)=20 μ $
w (12 )=w (4 )+1
2(4 ) (−10 )=0 $
w (36 )=w (12)+1
2(4 ) (10 )+
1
2(10 ) (−4 )+
1
2(6 ) (2.4 )=7.2 μ $
w (50)=w (36)+ 12
(4 ) (−3.6 )=0 $
ـ ¥ـ
_ ب
N ــ
ــ
ــة
(_ ا
ــع
ـ ¥ ـ
_ ا
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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عى
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ا
ـنع
ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
• \uestion 1#&&:
Lhe %oltage and current at the ter!inals of an auto*!obile batter during a charge ccle
are shown in ]ig# •1#&&#
a` Calculate the total charge transferred to the batter#
b` Calculate the total energ transferred to the batter#
•
-o7=t6o;Pa) q=areaunder (i−t ) plot
¿[ 12 (4 ) (5 )+ (4 ) (10 )+1
2(8 ) (4 )+(8 ) (6 )+
1
2(3 ) (6 )]#10
3
¿ [10+40+16+48+9 ]103=123,000C
ŒV ــŽ + ــ ي 4 ــ(( ا ــ ــ4% ـ ـ ـ ـ
ا
ــــ
n ن
ــــل
m’4 ــــ
_ ا
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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ـ " !
#$ % ا
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3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
5o;t6;7e x}7e8t6o; 1Z22yPb) w=∫ p dt =∫ vidt
v=0.2#10−3
t +9
• 0)t ) 4000 s
i=15−1.25#10−3
t
p=135−8.25#10−3 t −0.25#10−6 t 2
w1=∫
0
4000
(135−8.25#10−3
t −0.25#10−6
t 2 ) dt
¿ (540−66−5.3333 )103=468.667 "$
• 4,000) t )12,000
i=12−0.5#10−3
t
p=108−2.1#10−3
t −0.1#10−6
t 2
108−2.1#10−3
t −0.1
(¿#10−6
t 2)dt
w2= ∫4000
12,000
¿
¿ (864−134.4−55.467 )103=674.133"$
• 12,000)t )15,000
i=30−2#10−3
t
p=270−12#10−3−0.4#10
−6t 2
w3= ∫
12,000
15,000
(270−12#10−3
t −0.4#10−6
t 2 ) dt
¿ (810−486−219.6 )103=104.4 "$
w* =w1+w
2+w
3=468.667+674.133+104.4=1247.2 "$
Œ & $ X& œŸ V(W ‡/mNة 4“¦
ا
V
N § ( ع
n ن
N
4
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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كـل
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ا
ـنع
ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
• \uestion 1#&':
Lhe %oltage and current at the ter!inals of the circuit ele!ent in ]ig# 1#) are ^ero for t |
N" and for t † N the are:
v=(16,000t +20) e−800t
V
i=(128 t +0.16 ) e−800 t
A
a` Ot what instant of ti!e is !a,i!u! power deli%ered to the ele!ent b` ]ind the !a,i!u! power in watts#c` ]ind the total energ deli%ered to the ele!ent in !illiyoules#
ŒV ــ
Ž
ـ +ـ
ي
4 ــ
(( ا
ـــ
ــ
4ا ــــ % ــــn ن ا _ــــ m’4 ــــل
• -o7=t6o;Pa` p=vi=[ (16,000 t +20 )e−8000 t ] [ (128 t +0.16 ) e
−800t ]¿2048#10
3t 2
e−1600 t +5120 t e
−1600 t +3.2e−1600 t
¿3.2e−1600 t [640,000 t
2+1600 t +1 ]
dp
dt =3.2 {e−1600 t [1280#10
3t +1600 ]−1600e
−1600 t [640,000 t 2+1600t +1 ] }
¿−3.2e−1600 t [128#10
4 (800 t 2+t ) ]
¿−409.6#104
e−1600t
t (800 t +1)
*here'ore dp
dt =0whent =0
so p ma+ occurs at t =0
b¿ pma+=3.2e−0 [0+0+1 ]=3.2!
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October 2010
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بنت
ا
ـنع
ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
5o;t6;7eC x}7e8t6o; 1Z2?yP
c ¿w=∫0
t
p d+
640,000 t 2
e−1600 t
dt +¿∫0
t
1600 t e−1600t
dt +∫0
t
e−1600t
dt
w3.2
=∫0
t
¿
¿640,000 e
−1600 t
−4096#106 [256#10
4t 2+3200t +2 ] t
0+1600 e
−1600t
256#104
(−1600 t −1 ) t
0+[ e
−1600 t
−1600 ] t
0
!hent → ∞ all theupper limits evaluate ¿ ,ero hence
w3.2
= (640,000 )(2)4096#10
6 + 1600256#10
4+ 11600
w=10−3+2#10
−3+2#10−3=5m$
’ /` ن & ¤ … Wا ,ت ‘ (VN Tل ̃ا ™Z د V ا ‡ — ا ىع 3™ U_Nل
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October 2010
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كـل
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بنت
ا
ـنع
ـ " !
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& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
5FF9 GBHB9 .IJKIL9 DBtB -tr7ct7re89 I=<or6t>A89 D68crete .Bt>9 D6<6tB= Ko<6c9 5o;ceMt8 &N+ ب( ك دا+P
.ec>B;6cB= De86<; QRQQ9 -tr7ct7rB= I;B=E868 QRQQ9 .B;B<eAe;t9 5ID9 S=76C .ec>B;6c89 -E8teA DE;BA6c8 T (UN +Pاد
بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل
• \uestion 1#&):
Lhe %oltage and current at the ter!inals of the circuit ele!ent in ]ig# 1#) are:
^ero for t | N" and for t † N the are:
v=(10,000t +5 ) e−400t
V t ≥0
i=(40 t +0.05) e−400t
A t ≥0
a` ]ind the ti!e _in !illiseconds` when the power deli%ered to the circuit ele!ent is
!a,i!u!# b` ]ind the !a,i!u! %alue of p in !illiwatts#c` ]ind the total energ deli%ered to the circuit ele!ent in !illiyoules#
• -o7=t6o;Pa` p=vi=400#10
3t 2
e−800 t +700 t e
−800t +0.25 e−800 t
¿e−800 t [400,000 t
2+700t +0.25 ]dp
dt ={e
−800 t [800#103
t +700 ]−800e−800 t [400,000 t
2+700 t +0.25 ] }
¿ [−3,200,000 t 2+2400t +5 ] 100e
−800 t
*here'ore dp
dt =0when3,200,000 t
2−2400 t −5=0
so p ma+ occurs at t =1.68ms
b` pma+=[400,000 (0.00168 )2+700 (0.00168 )+0.25 ] e−800(0.00168)
¿666m!
Œ & $ X& œŸ V(W ‡/mNة 4“¦
ا
V
N § ( ع
n ن
N
4
7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
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October 2010
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كـل
عى
بنت
ا
ـنع
ـ " !
#$ % ا
& ك
'( )*+( ةا
,&! -.-/& ت
3ا
ي
& ب
! 4
مو
: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
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c` w=∫0
t
p dt
w=∫0
t
400,000 t 2
e−800t
dt +∫0
t
700 t e−800 t
dt +∫0
t
0.25e−800t
dt
¿ 400,000e
−800 t
−512#106 [64 #10
4t 2+1600t +2 ] t
0+700e
−800 t
64#104 [−800t −1 ] t
0+0.25 [ e
−800t
−800 ] t
0
whent =∞ all theupper limits evaluate¿ ,er hence
w=(400,000 )(2)
512#106 +
700
64#104+0.25
800 =2.97m$
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7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010
http://slidepdf.com/reader/full/circuits-sol-ch-01-circuit-variables-oct-2010 24/24
October 2010
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: 56rc76t89 :;<=68> 12?9عا @7Aer6cB= .et>oC89 DE;BA6c89 -tre;<t>9 -tBt6c8
• \uestion 1#&X:
The numerical values for the currents an voltages in the circuit in Fig.
,1.2% are given in Table ,1.2%.Fin the total po$er evelope in the
circuit.
• -o7=t6o;P
p=vi pa=vaia=(150) (0.6)=90m! pb=vbib=(150) (−1.4 )=−210m!
pc=−vc ic=−(100) (−0.8)=80m! pd=vd id=(250) (−0.8 )=−200m!
pe=−ve ie=−(300) (−2)=600m! p' =v ' i ' =(−300) (1.2 )=−360m!
• +f the power is positi%e" the circuit ele!ent is absorbing" otherwise de%eloping#
∑ pabsorbing=90+80+600=770m!
∑ pdeveloping=210+200+360=770m!
• Lhe power balances _j‡`
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