[SOLVED] IT Name and ID:

Name and ID:
Midterm I [12:00]
Student ID #:
Email: signature:

Copyright By Assignmentchef assignmentchef

Draw a box around the final answers otherwise you will NOT get any credits And move your solutions to the given boxes.
Your phone must be turned off and kept in your bag. One (8.511) cheat sheet is allowed.
Calculator is ok.
No cell phone for calculator
Only pencil and eraser

1. [87 15pts] Assume that the op-amp is ideal amp.
Find output voltage, vo
Find the current flow over the 9k with direction?
a) 129v b) 5mA
i9k =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 vR (t)=vs e vt vt c) iR (t)= s e iC (t)= s e a) FindvoltageovertheCandRwhent<0b) FindvoltageoverCandRwhent0c) 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 e =RC where i (t)=C d v(t)+ t =C vt0 e Rd t vs e R=Cvs eR=CRvs e=v e t s1t 1t RC iR (t)=iC (t)d t =C vs e dt 1t 1t =Cvs e =Cvs e vt = s e 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) 1e 3 =(12v) 1e138.88t 7.210 ( ) i (t)=(12v)e 7.2103 C i(t)= 0.005e 7.2103 d) p(t =10ms)=3.73mwThe general equation for the DC input ist t y(t0)=y(t+)e +y()1e The voltage over the capacitor is fully discharged t < 0 since it is disconnected from the source,y(t0)=y()1e where=RC=(2.4k)(3F)=0.0072 =121e 7.2103 The voltage over the resistor isvR (t)=iR=C d v(t)R=Cy()d 1e t R dt 1 t =(3F)(12v) e 7.2103 (2.4k)=(12v)e 7.2103 7.2103 The current through the resistor is controlled by this equation i (t ) = C d v (t )i(t)=Cy()d 1e t dt 1 t =(3F)(12v) e 7.2103 7.2103 =0.005e 7.2103 p = (i2 )R = 0.00373w=3.73mw The power delivered to the resistor at the given time isp(t =10ms)=i(t =10ms)R =Cy()d 1e t (2.4k)dt 1 10103 =(3F)(12v) e 7.2103 (2.4k) 7.2103 = 0.0012A = 1.2mA4. [15 pts] Plot voltage and powera) Findv(t)forthecaseoft=0,0+& b) Write v (t ) and plot it. c) Findv(t=0.1ms)v(0 ) v(t+ )=L :(k i )5v=i (2)i (2)(4)i 5v=i (2)i (2) 1Rx12 x12 L 2 : 5 = i 1 ( 2 ) + i 2 ( 2 + 3 )wherei =i i x21L :(4)(i i )5v=i (2)i (2) 1 21 1 2 L 2 : 5 = i 1 ( 2 ) + i 2 ( 5 )6 6 i1 5 i1 0.2772 5i =5 i =1.111 22 v(t0 )=i2 3=(1.111)3=3.333vL :i =0.277 11L :4i =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 + 2i1 1i2 2i1 5 5i=5+2iMatrix is sin gular to working precision[7 6 2 pts] a) Findv(t)forthecaseoft=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(t0)=0.3635e +0.5385=0.0185 0.1103 y (t = 0.1ms ) = 0.3635e + 0.5385 = 0.177v @t(0):v(t0 )=17=1.75 4@t(0+ ) 5.25V Using superposition property,v = 1 7=0.7v 7v 9+1 v5.25v =0.525vv1 =v7v +v5.25v =0.7v0.525v=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(t0)=(y(0+)y())e +y() =RC=(3||10)8mF=0.0185= (0.175 0.5385)e + 0.5385= 0.3635e + 0.5385 = 0.0185 0.1103 y (t = 0.1ms ) = 0.3635e + 0.5385 = 0.177v CS: assignmentchef QQ: 1823890830 Email: [email protected]