10
SOLUTIONS

SOLUTIONSSOLUTIONS ÷:÷.. Voltage @ node A, VA = VI (¥#k) = 34 Vi men: Awadhi: Us VA VB VG VA, VB KCL@nodeA_ Yt + Vaz + VA = o l 㱺 IzVa-VB = 16 oh 5Va-2VB=3# KCL@node_B f ie +

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men :
Awadhi :
ie + Eat 13=0 Defiantare also ⇒ -Va+zVp C- acceptable
5 -2 Va 32 Multiple
- i z v .
o I÷m:÷i÷÷ a. car. . -2=80 :÷÷:n÷÷;÷em A ' '
= data, adj CA) = f- [} of ] =
÷::÷
⇒ Voltage at
2
Square wave 12=1 Kr I = 2ms = R C with TM# t
amplitude --2V Let R = Ike ←
and offset =2V + -
-
÷:S.
- 51g ⇒ Vo Ct ) = to e
- th ' @ Ss Vo (5) = toe
Cz discharged @ t=5 ⇒ Volt) is halved .
= 5353 V
← lov
" E "" " "
It Izamcscepted ¥
is -- 2¥ =s
more accurate : 223.tszm-267.03ro.cl/sTIl0had/s accepted
Ttt out
vout-5e-3tcosc.LT#ts Variations of\ suit:*:S:*: of time axis .
0.4 0.8 1.2 I - 6 2.0 2 - 4
-time in seconds