Problem# 3. Determine v, in the circuit below anyway you like: Draw the frequency-domain network and calculate vo(t) in the circuit shown in Fig below if vs(t) is 4 sin (500t +45°) V and is(t) is 1 cos (500t + 45°) A. 90 Vs(t) 333uF 20mH + vo (t) is(t)

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Problem# 3. Determine v, in the circuit below anyway you like:
Draw the frequency-domain network and calculate vo(t)
in the circuit shown in Fig below if vs(t) is
4 sin (500t +45°) V and is(t) is 1 cos (500t + 45°) A.
90
Vs(t) (S
333uF
20mH
vo (t)
is(t)
Transcribed Image Text:Problem# 3. Determine v, in the circuit below anyway you like: Draw the frequency-domain network and calculate vo(t) in the circuit shown in Fig below if vs(t) is 4 sin (500t +45°) V and is(t) is 1 cos (500t + 45°) A. 90 Vs(t) (S 333uF 20mH vo (t) is(t)
6:39 PM Wed Apr 26
<=
S BOQOT*5*
MC Engr M20
+ Bookmark + Qui...kmark
MC Engr M20
X
4
Draw the frequency domain network & cabalate
wit) in the ckt shown below if 4(4)
40h (500€ +45) € ilt) is leas (s00t +45²) A
40
Velt) Ⓒ
400/5001-457
j ques
up
w=500
4299" (
381
+H
4(1001-95% 20
203401 bit)
201
303
4299² Ⓒ
(500)
↓
-j6006
40
-j6.006
310 31
15
j(500) (2010)
40
40
4/45²-v
-16.006
MC Engr M20 X
i=1/45°
Ⓒlas(500-45²)
103401 Diár
4/45
283-ja-V V-O
ca(500-45)
DIG
48 + 168
283-283
-j6006 JID + 1/45°
+283 +1283-ju= +1645²
-6.006
-283-j2.83+ju
60(-283 6006 ) = (50 +124²) 60
-29-3-j283 + jou=600+60 (45°
-29-3-j283 + jov=6+60 (45°
-283-j2B3+jlv-60132²=0
-283-ja83.jkv-42-43-4243
Simply
-29-3-j28·3+j10v=-jov +60 (45°
tijibu +jbu
-70.73 -√70.73 +jlbv=D
+70.73 +70-73
jolast
2.4.4.
jlbv = 70.73 +370-43
j16
j16
+7073 107
V= 70.73-j 70-73
V₂
J16
V= 10003/45°
16 (90
V-6-25/45-40
V-625-45
V-6-25 cos (600t-45″)
O
MC Engr M20
62%
(+) 8 1
B
15
15
Transcribed Image Text:6:39 PM Wed Apr 26 <= S BOQOT*5* MC Engr M20 + Bookmark + Qui...kmark MC Engr M20 X 4 Draw the frequency domain network & cabalate wit) in the ckt shown below if 4(4) 40h (500€ +45) € ilt) is leas (s00t +45²) A 40 Velt) Ⓒ 400/5001-457 j ques up w=500 4299" ( 381 +H 4(1001-95% 20 203401 bit) 201 303 4299² Ⓒ (500) ↓ -j6006 40 -j6.006 310 31 15 j(500) (2010) 40 40 4/45²-v -16.006 MC Engr M20 X i=1/45° Ⓒlas(500-45²) 103401 Diár 4/45 283-ja-V V-O ca(500-45) DIG 48 + 168 283-283 -j6006 JID + 1/45° +283 +1283-ju= +1645² -6.006 -283-j2.83+ju 60(-283 6006 ) = (50 +124²) 60 -29-3-j283 + jou=600+60 (45° -29-3-j283 + jov=6+60 (45° -283-j2B3+jlv-60132²=0 -283-ja83.jkv-42-43-4243 Simply -29-3-j28·3+j10v=-jov +60 (45° tijibu +jbu -70.73 -√70.73 +jlbv=D +70.73 +70-73 jolast 2.4.4. jlbv = 70.73 +370-43 j16 j16 +7073 107 V= 70.73-j 70-73 V₂ J16 V= 10003/45° 16 (90 V-6-25/45-40 V-625-45 V-6-25 cos (600t-45″) O MC Engr M20 62% (+) 8 1 B 15 15
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