Answer:
2.44 mV
Explanation:
This question has to be one of analog quantization size questions and as such, we use the formula
Q = (V₂ - V₁) / 2^n
Where
n = 12
V₂ = higher voltage, 5 V
V₁ = lower voltage, -5 V
Q = is the change in voltage were looking for
On applying the formula and substitutiting the values we have
Q = (5 - -5) / 2^12
Q = 10 / 4096
Q = 0.00244 V, or we say, 2.44 mV
The correct statement is: a higher than a normal voltage drop could indicate high resistance. Technician B is correct.
<h3>Ohm's law</h3>
Ohm's law states that the current flowing through a metallic conductor is directly proportional to the voltage provided all physical conditions are constant. Mathematically, it is expressed as
V = IR
Where
V is the potential difference
I is the current
R is the resistance
<h3>Technician A</h3>
High resistance causes an increase in current flow
V = IR
Divide both side by I
R = V / I
Thus, technician A is wrong as high resistance suggest low current flow
<h3>Technician B</h3>
Higher than normal voltage drop could indicate high resistance
V = IR
Thus, technician B is correct as high voltage indicates high resistance
<h3>Conclusion </h3>
From the above illustration, we can see that technician B is correct
Learn more about Ohm's law:
brainly.com/question/796939
Answer:
The surface area of the primary settling tank is 0.0095 m^2.
The effective theoretical detention time is 0.05 s.
Explanation:
The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.
Volumetric flow rate = 0.570 m^3/s
Overflow rate = 60 m/s
Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2
Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate
Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3
Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s
