An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th
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Answer:
Incomplete question, no circuit diagram.
Check attachment for further explanation and circuit diagram
Explanation:
We want to find Vo/Vs and Io/Ix and also power delivered to the 2 kΩ resistor
Given that,
g= 0.0025 S (i.e conductance )
Vs= 4 V
R1 = 1 kΩ = 1000 Ω
R2 = 3 kΩ = 3000 Ω
R3 = 10 kΩ = 10000 Ω
R4 = 500Ω = 0.5 kΩ
R5 = 2 kΩ = 2000 Ω
a. At Loop 1: let use voltage divider rule to get Vx
Then, Vx = R2/(R1+R2) • Vs
Vx=3/(1+3) •Vs
Vx=¾ Vs.
The small signal current is given as
Is=g•Vx, since Vx=¾Vs
Then, Is= 0.025×¾Vs
Is=3/1600 Vs. Equation 1
Note: Ressistor R4 and R5 are in series, then the equivalent resistance of R4 and R5 is given as
Req= R4+R5
Req=2+0.5=2.5 kΩ
So, using current divider rule between R3 and the equivalent resistance of R4 and R5.
Therefore, Io= R3/(R3+Req) • Is
Io= R3/(R3+Req) • Is equation 2
Note : using ohms law on resistor R5,
V=iR. , R5=2 kΩ
Vo=IoR5
Vo=2Io
Io=Vo/2. Equation 3
Substitute equation 1 and 3 into 2
Io= R3/(R3+Req) • Is
½Vo = 10/(10+2.5) • 3/1600 Vs
½Vo = 10/12.5 • 3/1600 Vs
½Vo = 3/2000 Vs
Vo/Vs = 3/2000 × 2
Vo/Vs = 1 / 1000
The voltage output gain is
Vo/Vs = 1 / 1000
b. From equation 2
Io= R3/(R3+Req) • Is
Also, applying ohms law to resistor R2,
Vx = Ix• R2, R2=3kΩ
Vx = 3•Ix
Given that, Is= g•Vx
Is=0.0025(3•Ix)
Is= 3/400 Ix
Then, Io= R3/(R3+Req) • Is
Io= 10/(10+2.5) • 3/400 Ix
Io= 10/12.5 • 3/400 Ix
Io= 3/500 Ix
Io/Ix= 3/500
The current gain is
Io/Ix= 3/500
c. Output power
Power is given as
P=I²R
Then, output power at Resistor 5 is
Po = Io²•R5
R5=2000 Ω
From loop 1: using KVL, sum of voltage in a loop is zero
-Vs+1000Ix+3000Ix=0
4000Ix=Vs
Since Vs=4
Then, 4000Ix=4
Ix =4/4000
Ix = 1/1000 A
Since, Io/Ix = 3/500
Then, Io = 3/500 • Ix
Io=3/500 × 1/1000
Io= 6×10^-6 A
Therefore,
Po=Io²•R5. ,R5=2000
Po= (6×10^-6l² × 2000
Po=7.2×10^-8 W
Po=72×10^-9 W
Po=72 nW
The output power at resistor R5 is
72 nW
