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Draw a box around the final answers otherwise you will NOT get any credits And move your solutions to the given boxes.
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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 ≥ 0a) 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<0b) FindvoltageoverCandRwhent≥0c) Find current through C and R when t ≥ 0d) 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)=12VAnd 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 responsev (t)=v t eτ =R⋅C where i (t)=C d v(t)+ −t =C vt0 e ⋅Rd −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 τv−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-upwhen t ≥ 0a) @ t ≥ 0 , show the detail procedure that voltage across the capacitor and plot it.b) @ t ≥ 0 , show the detail procedure that howyou got the voltage over the resistor (no needc) @ t ≥ 0 , show the detail procedure that howyou got the current through the resistor andd) @ t = 10ms , how much power is delivered tothe 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)=3.73mwThe 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 isvR (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 isp(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 = 1.2mA4. [15 pts] Plot voltage and powera) 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=3.333vL :i =0.277 11L :4⋅i =−2i +5i +5i wherei =i +i −i 2×123 x231L3 :5=−2i1 +5i2 +5i3L :4⋅(i +i −i)=−2i +5i +5i ↔i +i =−2i2231123231 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 −4v(t∞ )= 0.5385vy(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 =−0.525vv1Ω =v7v +v5.25v =0.7v−0.525v=0.175v The value of v (t ) right after the flipping the switch is v(t0+ )=0.175vv(t=∞)= 1(7v)=0.5385v 13 The equation for the voltage over the resistor isy(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: assignmentchef QQ: 1823890830 Email: [email protected]