ac circuit analysis (part 1)
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8/10/2019 AC Circuit Analysis (Part 1)
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8/10/2019 AC Circuit Analysis (Part 1)
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8/10/2019 AC Circuit Analysis (Part 1)
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dt )t ( i
C
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dt
di Lv =
! +=
t
dt )t ( v
L
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8/10/2019 AC Circuit Analysis (Part 1)
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8/10/2019 AC Circuit Analysis (Part 1)
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G/(1F '/% 1'%04<P1'0'% &())%.' -1
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R
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LR
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8/10/2019 AC Circuit Analysis (Part 1)
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dt
di16i16
dt
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di10
dt
idi16
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2
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)t(v2i36dt
di12
dt
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2
2
22
=++
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0i)16D10D(i16 22
1 =+++! AYC(8D+16)
8/10/2019 AC Circuit Analysis (Part 1)
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di
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)tcos(F)t(f m !+"=
@/%)%UB Z 0B=5-'(4% 2) =%0W 805(%
! Z 0.9(50) 3)%H(%.&<F )04R1%&
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#
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Fm
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8/10/2019 AC Circuit Analysis (Part 1)
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tT
2cosFtcosF)t(f mm
!
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8/10/2019 AC Circuit Analysis (Part 1)
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G/% =%0W 805(% 23 '/% 825'09% -1 >BZ"XX 825'1: G/% 0.9(50)
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9):+')8 G/% 3)%H(%.&< 23 '/% 1-.(12-4
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1f =
1%&
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2) &<&5%1R1%& 2) T%)'? AT?C
G/%.F '/% 1-.(12-4 B0< 0512 O% %N=)%11%4 01
ft2cosFtcosF)t(f mm !="=
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V )t377cos(311)t(v !+=
8/10/2019 AC Circuit Analysis (Part 1)
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Note: We say either f 1(t) leads f 2(t) by an angleof 30° or that f 2(t) lags f 1(t) by an angle of 30°.
)30tcos(F)t(f m2 °+!=
72.1-4%) '/% =52' 23 '/% 1-.(12-405 3(.&'-2.1
and)60tcos(F)t(f m1 °+!=
30
60
90 180-30
-60
-90!t, deg
270 360
)t(f 1
)t(f 2
8/10/2019 AC Circuit Analysis (Part 1)
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U)2B I/B1 50@F @% 9%' tcosRIRivmR R
!==
72.1-4%) 0 )%1-1'2): $%' '/% &())%.'
O% 4%1&)-O%4 O<
tcosIimR !=
R
S P 8a
-a
90 180-90!t, deg
270 360
vR
iR
8/10/2019 AC Circuit Analysis (Part 1)
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72.1-4%) 0. -.4(&'2): $%' '/%
&())%.' O% 4%1&)-O%4 O<
tcosIimL !=
L-$
S
P
8$
From vL= , we get tsinLIvmL
!!"=
dt
diL L
90 180-90!t, deg
270 360
iL
vL
!".)8 G/% &())%.' 5091 '/% 825'09% O< b[2:
8/10/2019 AC Circuit Analysis (Part 1)
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-7
87S P
C72.1-4%) 0 &0=0&-'2): $%' '/%
&())%.' O% 4%1&)-O%4 O<
tcosIimC !=
From vC= , we get tsin
C
Iv
m
C !
!
= ! dtiC
1C
90 180-90!t, deg
270 360
iC
!".)8 G/% &())%.' 5%041 '/% 825'09% O< b[2:
vC
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8/10/2019 AC Circuit Analysis (Part 1)
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72.1-4%) 0 +7 A&2.1'0.'C &())%.' ; 0.4 0. 67 A1-.(12-405C
&())%.' -A'CZ;B&21 ':
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M-.&% '/% &())%.' ; -1 &2.1'0.'F '/%. `6>F+7 Z ;Ya:
72.1-4%) a @-'/ '/% +7 &())%.' ;: G/%
=2@%) 4-11-=0'%4 O< a -1
R IP 2=
R
;
8/10/2019 AC Circuit Analysis (Part 1)
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G/% 08%)09% 805(% 23 0.< 1-.(12-405 3(.&'-2. &0. O% 1/2@. '2
O% %H(05 '2 ?%)2: G/(1
RIP 2
m21
AC,AV =
tcosRIRi)t(p 22
m2
!==
R
-A'C
72.1-4%) .%N' a @-'/ '/% 67 &())%.' -A'C:
G/% -.1'0.'0.%2(1 =2@%) 4-11-=0'%4 O< a -1
M-B=5-3<-.9F @% 9%'
!"#$
%& '
+
=
2t2cos1RI)t(p 2
m t2cosR IR I 2m2
12m2
1 !+=
8/10/2019 AC Circuit Analysis (Part 1)
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!".)8 G/% 10B% 4%,-.-'-2. 0==5-%1 '2 0 1-.(12-405 825'09%
8A'CZ>B &21 ':
!H(0'-.9 08%)09% =2@%)F @% 9%'
R IR I 2
m2
12=
or
m
mI707.0
2
II !=
9):+'+.+"'8 G/% %33%&'-8% 805(% 23 0 1-.(12-405 &())%.' @-'/0. 0B=5-'(4% ;B -1 %H(05 '2
2
II
m
EFF =
8/10/2019 AC Circuit Analysis (Part 1)
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G/% %33%&'-8% 805(% 23 0 =%)-24-& 3(.&'-2. -1 0512 &055%4 '/%
a22'PL%0.PMH(0)% AaLMC 805(%:
G/0' -1F 9-8%. 0 =%)-24-& 3(.&'-2. 3A'CF @% 9%'
! ==
T
0
2RMSEFF dt)t(f
T
1FF
G/% %33%&'-8% 2) aLM 805(% 23 '/% 825'09% -1
V220)311(707.0V ==
!".)8 G/% .2B-.05 825'09% -. '/% `/-5-==-.%1 -1 0 1-.(12-44%1&)-O%4 O<
V )t377cos(311)t(v !+=
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