A measure of how far an object has moved from a starting point
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I have attached the circuit image missing in the question.
Answer:
A) The range of vo is; -6.6V≤ vo ≤-1V
B) σ = 0.1861
Explanation:
A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.
Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0
So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0
Simplifying further; -25 vg - vΔ = 0
From the question, vg = 40mV = 0.04 V
So - 25(0.04) = vΔ
So: vΔ = - 1 V
Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0
So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0
So RΔ = 100kΩ or 100,000Ω from the question.
So, substituting for RΔ, we get,
[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0
Let's put the value of - 1 for vΔ as gotten before.
So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0
Now let's make vo the subject of the equation to get;
-1 - vo = (1 - σ)[2 + (1/σ)]
-1 - vo = 2 - 2σ + (1/σ) - 1
-vo = 1 + 2 - 2σ + (1/σ) - 1
-vo = 2 - 2σ + (1/σ)
vo = - 1 (2 - 2σ + (1/σ))
When σ = 0.2; vo = - 1(2 - 0.4 + 5) =
- 1 x 6.6 = - 6.6V
Also when σ = 1;
vo = - 1(2 - 2 + 1) = - 1V
Therefore, the range of vo is;
- 6.6V ≤ vo ≤ - 1V
B) it will saturate at vo = - 7V
So, from;
vo = - 1 (2 - 2σ + (1/σ))
-7 = - 1 (2 - 2σ + (1/σ))
Divide both sides by (-1)
7 = (2 - 2σ + (1/σ))
Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)
Multiply each term by α to get;
5σ = - 2σ^(2) + 1
So 2σ^(2) + 5σ - 1 = 0
Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861
