IRWIN 9e 6_17

IRWIN 9e 6_17 - Irwin, Basic Engineering Circuit Analysis....

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis. 9/E 1 6.1}!r The H-‘awet'nrm fur the current firming thrmigh the HI-pJ: mpuciinr in Fig. Fri. [Ta is Hiram-'11 in Fig. Ph. 1Tb. lt' 1L": [r = D) = 1V, determine air] at I = 1 ms. 3 m5.4m.~:.ur1il 5 ms. irr] {mic-r} I {ma} (3.] “32' Figure PIS-.11 SOLUTION: I'm Vcflms} : vcr O) r E’— [(igmku 0 lm \IcUmg): 1+ I J(i5m)d~t IoplO \/C ( ,m) -; i—i ioo Xlo3(15Xl63)£—jjolm \/c(ims) = i+i€§ooat \/C(Ims) = i+i500<iw chflms) : 2-5V Chapter 6: Capacitance and Inductance Problem 6.17 2 Irwin. Basic Engineering Circuit Analysis, 9/E at VC(3m5)=\/C(Ims) + ._’_ 15mm IOM hm t VCRWU :2'5 +IOOXI03WS‘X10’3) [ill Vétlms)= 25 +600 Ci-fmj VCCShns) : 2'5 "r '500 [Sm-[m] Vc (37m) = 5'5 V y it \/C ((4)715) :: VCC3m)+_L_ f‘lo'rndi (0H3 ,t Vc (Wm): 55' X(OOX‘O3(‘IOX/O_3)[i 33m Vcflnhs): 5'5 - IOOOCi‘Smj Va (Hms) : 55' )000CHm-3wat Vc (Hms) 3 LI'SV Problem 6.17 - Chapter 6: Capacitance and Inductance Irwin, Basic Engineering Circuit Analysis, 9/E 3 VC(5ms) : L4'5 ’Iooo [i’ W”) : (4'5 " IOOO [5m~ Um] VCCCmg) : 3’5V _ Chapter 6: Capacitance and inductance Problem 6.17 ...
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This note was uploaded on 11/08/2011 for the course EE 2120 taught by Professor Aravena during the Fall '08 term at LSU.

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IRWIN 9e 6_17 - Irwin, Basic Engineering Circuit Analysis....

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