Question

Find the voltage gain vO/vS and current gain iO/iX for the circuit for g = 0.0025 S. Then, for vS = 4 V, find the power supplied by the input voltage source and the power delivered to the 2 kΩ resistor.

Answer 1

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