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IRWIN 9e 7_40

# IRWIN 9e 7_40 - InNin Basic Engineering Circuit Analysis...

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Unformatted text preview: InNin, Basic Engineering Circuit Analysis, 9/E 1 1.40 Liar: the step-hy-stcp melhnd [u ﬁnd Hair) [111' r [J in the circuit 1'11F1'g. PT-‘AIIL Figure P140 SOLUTION: Chapter 7: First— and Second- Order Transient Circuits Problem 7.40 2 lwvin, Basic Engineering Circuit Analysis, 9/E KCL', £3154, L": ILL-L3 KVL'. R‘L' * Kris +RSLB+QRI=D R‘i' * VH1 * (Ru +25%; =0 ‘ c011 «- «15 .. o KVLi \uinzﬁo 1.: 'ﬁ; 2‘; I“ : -—(oD~ KVL‘. Problem 7.40 Chapter 7: First- and Second- Order Transient Circuits Irwin, Basic Engineering Circuit Analysis, 9/E 3 QL1’5L3=O 1 :Li ,7 3 3 611 +AC3£3)=\2 _________________—___—__———————————————- Chapter 7: First- and Second- Order Transient Circuits Problem 7.40 L 4 Irwin. Basic Engineering Circuit Analysis, 9/E KVL', . \2; (25+ RH+25)53-Rsil—(Rq+zs)x+ ~ KVL'. 0: 12¢, +?;£;+(Rq’ri€s)£5—(Ru95)::4 rsgji—i— 512+4i3—41+ :0 Problem 7.40 Chapter 7: First- and Second- Order Transient Circuits Irwin, Basic Engineering Circuit Analysis, 9/E 5 £5 = 0.10% VOW):- (13—42,)?5 V0 (0*): (0.70L'+2)C2.) Vow") = 5.1m: K‘ -+ K,,= ’5.“ t;0° I. Chapter 7: First- and Second- Order Transient Circuits ' Problem 7.40 i 6 mm, Basic Engineering Circuit Analysis, 9/E Problem 7.40 Chapter 7: First- and Second- Order Transient Circuits Irwin, Basic Engineering Circuit Analysis, 9/E 7 "/ ~11 ,LO =10 +,r0 A0: .35— . «to 3A Vo(oo);LDR5 Chapter 7: First- and Second- Order Transient Circuits Problem 7.40 ‘ 8 V Irwin. Basic Engineering Circuit Analysis, 9/E ?%:|.5_{1 ’1”: Kr’ l5 7 ﬁgs -i VtD (by: VH1” K26. Vo(\>3= 5-55t0-08507S-t V) ’00 Problem 7.40 Chapter 7: First- and Second- Order Transient Circuits ...
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