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Midterm I [12:00]
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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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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 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)= 0.005e 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
=121−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
=0.005e 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)forthecaseoft=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 5i =5 i =1.111 22
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 1i2 −2i1 5 5i=5+2i
Matrix 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 −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
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