Answer:
a) the point at which the transistor enters the saturation region is 1.5 v
b) the value of ID is obtained in saturation is 1.125 mA
Explanation:
Given the data in the question;
for an NMOS, the condition for the saturation is;
V_DS ≥ V_GS - V_t
V_GS is 2.5 v and V_t is 1 v
so we substitute
V_DS ≥ 2.5 - 1
V_DS = 1.5 v
so the point at which the transistor enters the saturation region is 1.5 v
The drain current I_d in the saturation region;
I_d = 1/2×μₙ×Cₐₓ×W/L×( V_GS - V_t)²
= 1/2Kₙ ( V_GS - V_t)²
our Kₙ is 1 mA/V², V_GS is 2.5v and V_t is 1 v
so we substitute
I_d = 1/2(1 mA/V²)( 2.5 - 1 )²
= 1/2(1 mA/V²)( 2.25)
= 1.125 mA
therefore, the value of ID is obtained in saturation is 1.125 mA
Which option identifies the type of engineer described in the following scenario?
Sean is an engineer whose current project is a skyscraper in Richmond, VA. He relies heavily on geometry in his research of building design.
Material
Civil
Mechanical
Chemical
Answer:
civil
Explanation:
mark be branilist
Answer:
Civil
Explanation:
civil engineering – the application of planning, designing, constructing, maintaining, and operating infrastructure while protecting the public and environmental health, as well as improving existing infrastructure that may
A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven.
Answer:
a) the moisture content before it was placed in the oven is 18.18%
b) degree of saturation for soil is 72.19%
Explanation:
Given the data in the question;
Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100
so we substitute
Moisture content = [(53.3 - 45.1) / 45.1 ] × 100
= (8.2/45.1) × 100
= 18.18%
Therefore the moisture content before it was placed in the oven is 18.18%
Dry Unit Weight = dry weight / volume
Dry Unit Weight = 45.1 lb / 0.45 ft³
Dry Unit Weight = 100.22 lb/ft³
we know that;
dry unit weight = (Specific gravity × unit weight of water) / (1 + e)
we also know that; unit weight of water is 62.43 lbf/ft³
so we substitute
e = (2.70×62.43 / 100.22) - 1
e = 1.68 - 1
e = 0.68
so void ratio e = 0.68
Now we determine the degree of saturation using the equation;
degree of saturation = (Moisture content × specific gravity) / void ratio
we substitute
degree of saturation = ( 18.18% × 2.7) / 0.68
= 0.49086 / 0.68
= 0.7219 ≈ 72.19%
Therefore degree of saturation for soil is 72.19%
