Question

V\_s mV\_pp 1.0 kHz V\_cc +12V R\_1 56 k\Omega R\_c 4.7 k\Omega C\_3 R\_3 5.6 k\Omega C\_1 1.0 \muF R\_2 10 k\Omega Q\_1 0.22 \muF R\_E1 68 \Omega Q\_2 R\_4 22 k\Omega V\_o\_u\_t R\_E2 560 \Omega C\_2 100 \muF C\_4 100 \muF R\_9 Speaker 8 \Omega

          V\_s
mV\_pp
1.0 kHz
V\_cc
+12V
R\_1
56 k\Omega
R\_c
4.7 k\Omega
C\_3
R\_3
5.6 k\Omega
C\_1
1.0 \muF
R\_2
10 k\Omega
Q\_1
0.22 \muF
R\_E1
68 \Omega
Q\_2
R\_4
22 k\Omega
V\_o\_u\_t
R\_E2
560 \Omega
C\_2
100 \muF
C\_4
100 \muF
R\_9
Speaker
8 \Omega
        
Show more…
VmV1.0 kHz
V+12V
R
56 kΩR4̧.7 kΩC
R
5.6 kΩC
1.0 R
10 kΩQ
0.22 R1
68 ΩQ
R
22 kΩVøR2
560 ΩC
100 C
100 R
Speaker
8 Ω

Added by David M.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Vs = 200mVpp Solver for Vout2. Please show work on how you solved for this. R = 6.0kΩ RC = 4.7kΩ R = 5.61Ω C = 0.22μF 1.0μF R = 10Ω 68mVpp at 1.0kHz 22Ω ZR = 0.9Ω C = 100μF eake
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Transcript

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00:01 This all description we have at the key migration of current process l.
00:07 I1 -dash plus r1 into i1 minus i2 plus e equals to 0.
00:14 This is equation of 1.
00:16 We have i2r2 plus 1 of c integral i1dt plus r1 into i1d plus r1 into i1 equals to 0.
00:28 This is a equation moving and further solving one of these things by putting we at i .1 dash it is to minus 2 i1 plus 2 i2 minus 4 for this and d equation 3 8 i 2 dash plus 2 i1 plus 2 i 2 i 2 dash equals to 0 this is equation for solving out of these equations and things we get that that x -dash equals to the a x plus f of t first we will find back such t -bass on the i -dash equals to minus 2 -1 2...
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