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1. [87 15pts] Assume that the op-amp is ideal amp.
Find output voltage, vo
Find the current flow over the 9kΩ with direction?
a) −129v b) 5mA↑
i9kΩ =5mA↑

2. [3 pts each 15 pts] First order circuit is shown with DC input Vs and it is called CR circuit. The figure shows that the switch
position is set (switch up) that way when t < 0 and the switch will be flip-down at t ≥ 0 a) vC (t)=vs V vR (t)=0 V −t −t b) vC (t)=vs ⋅e τ vR (t)=−vs ⋅e τ v−t v−t c) iR (t)=− s ⋅e τ iC (t)=− s ⋅e τ a) FindvoltageovertheCandRwhent<0 b) FindvoltageoverCandRwhent≥0 c) Find current through C and R when t ≥ 0 d) PlotvoltageoverCat [t<0,t=0,&t>0]
e) PlotcurrentoverCat [t<0,t=0,&t>0]
The voltage over the C is fully charged @ t < 0 vC (t)=12V And the current flow over the R is none since the capacitor is fully charged. So there is no current flow. Only discharge occurs when t ≥ 0 . It simply is a natural response v (t)=v t e τ =R⋅C where i (t)=C d v(t) + −t =C vt0 e ⋅R d −t vs ⋅e τ ⋅R =C⋅vs⋅− eτ⋅R=C⋅R⋅vs⋅− eτ =−v ⋅e− t s 1−t 1−t  τ  RC iR (t)=iC (t) d −t =C vs ⋅e τ  dt   1−t 1−t =C⋅vs ⋅− e τ =C⋅vs ⋅− e τ v−t =− s ⋅e τ  τ  RC [5 pts each 20 pts] The switch position is set that way when t < 0 (flipped down) and the switch position will be flip-up when t ≥ 0 a) @ t ≥ 0 , show the detail procedure that voltage across the capacitor and plot it. b) @ t ≥ 0 , show the detail procedure that how you got the voltage over the resistor (no need c) @ t ≥ 0 , show the detail procedure that how you got the current through the resistor and d) @ t = 10ms , how much power is delivered to the resistor v (t)=(12v) 1−e −3 =(12v) 1−e−138.88t  7.210  ( ) i (t)=(12v)e 7.2⋅10−3  C  i(t)= 0.005e 7.2⋅10−3    d) p(t =10ms)=3.73mw The general equation for the DC input is −t −t y(t≥0)=y(t+)e τ +y(∞)1−e τ  The voltage over the capacitor is fully discharged t < 0 since it is disconnected from the source, y(t≥0)=y(∞)1−e τ  whereτ=RC=(2.4kΩ)(3μF)=0.0072 =121−e 7.210−3    The voltage over the resistor is vR (t)=i⋅R =C d v(t)⋅R =C⋅y(∞)d 1−e− t ⋅R  τ dt     1 − t  =(3μF)⋅(12v) e 7.2⋅10−3 ⋅(2.4kΩ)=(12v)e 7.2⋅10−3  7.2⋅10−3       The current through the resistor is controlled by this equation i (t ) = C d v (t ) i(t)=C⋅y(∞)d 1−e− t   τ dt     1 − t  =(3μF)⋅(12v) e 7.2⋅10−3  7.2⋅10−3   =0.005e 7.2⋅10−3    p = (i2 )R = 0.00373w=3.73mw The power delivered to the resistor at the given time is p(t =10ms)=i(t =10ms)⋅R =C⋅y(∞)d 1−e− t ⋅(2.4kΩ) dt     1 −10⋅10−3 =(3μF)⋅(12v) e 7.2⋅10−3 ⋅(2.4kΩ) 7.2⋅10−3      = 0.0012A = 1.2mA 4. [15 pts] Plot voltage and power a) Findv(t)forthecaseoft=0−,0+&∞  b) Write v (t ) and plot it. c) Findv(t=0.1ms) v(0− ) v(t+ )= L :(k ⋅i )−5v=i (2Ω)−i (2Ω)⇔(4Ω)⋅i −5v=i (2Ω)−i (2Ω) 1Rx12 x12  L 2 : 5 = − i 1 ( 2 Ω ) + i 2 ( 2 + 3 ) wherei =i −i x21 L :(4)⋅(i −i )−5v=i (2)−i (2Ω) 1 21 1 2  L 2 : 5 = − i 1 ( 2 Ω ) + i 2 ( 5 ) −6 6 i1  5 i1  0.277 −2 5i =5 i =1.111  22  v(t0− )=i2 ⋅3=(1.111)3=3.333v L :i =0.277 11 L :4⋅i =−2i +5i +5i wherei =i +i −i 2x123 x231 L3 :5=−2i1 +5i2 +5i3 L :4⋅(i +i −i)=−2i +5i +5i ↔i +i =−2i 2231123231 L : 5 = −2i + 5i + 5i ↔ 5i + 5i = 5 + 2i 1 1i2 −2i1  5 5i=5+2i Matrix is sin gular to working precision [7 6 2 pts] a) Findv(t)forthecaseoft=0−,0+&∞  b) Write v (t ) and plot it. c) Findv(t=0.1ms) v(t0 )= 1 ⋅7 =1.75V v(t0+ )= 0.175v −4 v(t∞ )= 0.5385v y(t≥0)=−0.3635eτ +0.5385τ=0.0185  − 0.1⋅10−3  y (t = 0.1ms ) =  −0.3635e τ + 0.5385  = 0.177v  @t(0−):v(t0 )=1⋅7=1.75 −4 @t(0+ ) 5.25V Using superposition property, v = 1 7=0.7v 7v 9+1  v5.25v =−0.525v v1Ω =v7v +v5.25v =0.7v−0.525v=0.175v The value of v (t ) right after the flipping the switch is v(t0+ )=0.175v v(t=∞)= 1(7v)=0.5385v 13 The equation for the voltage over the resistor is y(t≥0)=(y(0+)−y(∞))e τ +y(∞) τ=RC=(3||10)⋅8mF=0.0185 = (0.175 − 0.5385)e τ + 0.5385 = −0.3635e τ + 0.5385 τ = 0.0185  − 0.1⋅10−3  y (t = 0.1ms ) =  −0.3635e τ + 0.5385  = 0.177v  程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com