This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem # 11 In the above circuit, the switch closes at t = 0. The source is a steady
DC voltage source equal to 50 volts. The current through the
inductor at t = (0) is 0 and the voltage across the capacitor at t = (0 ) is 0.
1. Use KCL to write a differential equation for the voltage across the capacitor. Put this equation in standard form. 2. What is the form of the particular solution ? 3. What is the form for the complementary solution. State your
reason for choosing this form of complementary solution. KCL 4.0 +11. driz‘fxzvso 1... tjwdkg—io 1’ lg— P ,_——— >0 No 5.9. @ a K 0a” 0 saw: rm! 0161’ \
W—J
ZM— U01,
. _L,_ J,
(a R: 2 [07
watt 4—” 2 ,___l—___r—" .. I , _L—
1» C /o/a° Ions" Io’~/6’°‘ W"
S» ,2 1°?
_ [o7 ’ o2 _" Zotés cut )Zavl «— undc'] Problem # 1 For the series R, L, C circuit A70 V 1 3 31. Develop a differential equation to solve for the voltage across the capacitor. 6 ‘ 2’2. Solve for the particular solution. ' 3. Solve for 0C and W; and determine the correct form for the
complementary solution. You do not have to solve for K1 and K2. 1,
 .Vi__.t/ ' A {j 4. Is this circuit critically damped, under damped, or over damped ? WONG ‘ ' ._ o
"50 l “M '— 5?“ " LL97 Acta 2 C :3U
[C $12.95) + LQ dz 1’2”) a, 02m , )5/0
\p our our" Ifo
J/ cl‘v’oU) 2 Auwr) * J— at’dt [LC
011:1 L 4,?
W Macaw k____,L_,_____ Whig
41’1/ ’0 ' '0 M 10:5)(5) (JDc) 50.]: (3::— r [010 46 +_ 2040 U;
5mm: gaicwc Faye—11¢ ls Cpl1574'U'r
a O
91%” ;/
l0~loj 5
‘ 0&1, 1' +Zoloh’z 3000 /o
353:0 I0
rrz Qo< : lo 10’ 64‘ 5000
M
“37“ ’ 2° “’6 ﬁ/ 5 = 5500 t 5; = 52763.31’ &2
Sb: “ {723406 é ...
View
Full Document
 Spring '08
 GFReid

Click to edit the document details