This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: AP 7.1 [a] The circuit for t <: 0 is shown below. Note that the inductor behaves like a short
circuit, effectively eliminating the 2 Q resistor from the circuit. First combine the 30 Q and 6 ﬂ resistors in parallel: 306 = 5 9
Use vol5tage division to ﬁnd the voltage drop across the parallel resistors:
= — 120 = '
I v 5 + 3( ) 75V Now ﬁnd the currernt using Ohm's law: ' v 5
' ‘ = — = ——— = 12.5A
em ) _ 6 6 1 2 1 #3 2
[b] rum) = 5L: (0) = 2—(8 x 10 )(12.5) = 6251111 [c] To ﬁnd the time constant, we need to ﬁnd the equivalent resistance seen by the inductor fort .‘> 0. When the switch opens. only the 2 Q resistor remains connected to the inductor. Thus,
L 8 x 10'3
1" = — = —————
R 2
_ [d] i(t) = t(0*)e*f'r = —12.5e“/°m4 = —12.5e"25°‘A, [e] 3*(‘5 ms) = —12.5e250<m5) = 4253125 = —3.58A =4ms t>0 So to (5 ms) = §Lt2(5 mm) = %(3) x 103(3.53)2 = 51.3mJ
to (dis) = 625 ~ 51.3 = 573.7mJ 573.7
d' ‘ = —— = _
ﬂ—ﬂ % 1551pated ( 625 ) 100 91 8%
AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves like an open circuit. ‘ Find the voltage drop across the open circuit by ﬁnding the voltage drop across the 50 k9 resistor. First use current division to ﬁnd the current through the
50 k9 resistor: ' 80x103 3 ' .
—“——"‘“——* . " =4
80x103+20><103+50><103(75‘x m ) Use Ohm’s law to ﬁnd the voltage drop: _
1.1(0') = (50 x 103nm = (50 x 103)(o.004) = 200v 3set: = [b] To ﬁnd the time constant, we need to ﬁnd the equivalent resistance seen by the capacitor for t > 0. When the switch opens, only the 50 k9 resistor remains
connected to the capacitor. Thus, T = R0 = (50 x 103)(0.4 x 105) = 20:11.5
[c1 v(t) = v(0")e"m = 2OOe‘t/D'02 = 20:06"50¢ v, t >‘ o [(1] 111(0) = 13—0122 = %(0.4 x 104x200)2 = 8m] 1
[e] w(t) = $0020) = 5(04 >< 106)(zoos50m)2 = sea1W m]
The initial energy is 8 In], so when 75% is dissipated, 2 11]] remains:
8 x 10*3310‘3‘ = 2 x 103, em = 4, t = (In 4)/100 = 13.361115 ...
View
Full
Document
This note was uploaded on 08/12/2010 for the course ENGR 117 taught by Professor Nickarnold during the Fall '08 term at Santa Barbara City.
 Fall '08
 NickArnold

Click to edit the document details