Answer:
Technician A
Explanation:
Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.
Answer:
There are six conditions
1. Poles should contain some residual flux.
2. Field and armature winding must be correctly connected so that initial mmm adds residual flux.
3. Resistance of field winding must be less than critical resistance.
4. Speed of prime mover of generator must be above critical speed.
5. Generator must be on load.
6. Brushes must have proper contact with commutators.
Explanation:
Answer:
V₂ = 20 V
Vt = 20 V
V₁ = 20 V
V₃ = 20 V
I₁ = 10 mA
I₃ = 3.33 mA
It = 18.33 mA
Rt = 1090.91 Ω
Pt = 0.367 W
P₁ = 0.2 W
P₂ = 0.1 W
P₃ = 0.067 W
Explanation:
Part of the picture is cut off. I assume there is a voltage source Vt there?
First, use Ohm's law to find V₂.
V = IR
V₂ = (0.005 A) (4000 Ω)
V₂ = 20 V
R₁ and R₃ are in parallel with R₂ and the voltage source Vt. That means V₁ = V₂ = V₃ = Vt.
V₁ = 20 V
V₃ = 20 V
Vt = 20 V
Now we can use Ohm's law again to find I₁ and I₃.
V = IR
I = V/R
I₁ = (20 V) / (2000 Ω)
I₁ = 0.01 A = 10 mA
I₃ = (20 V) / (6000 Ω)
I₃ = 0.00333 A = 3.33 mA
The current It passing through Vt is the sum of the currents in each branch.
It = I₁ + I₂ + I₃
It = 10 mA + 5 mA + 3.33 mA
It = 18.33 mA
The total resistance is the resistance of the parallel resistors:
1/Rt = 1/R₁ + 1/R₂ + 1/R₃
1/Rt = 1/2000 + 1/4000 + 1/6000
Rt = 1090.91 Ω
Finally, the power is simply each voltage times the corresponding current.
P = IV
Pt = (0.01833 A) (20 V)
Pt = 0.367 W
P₁ = (0.010 A) (20 V)
P₁ = 0.2 W
P₂ = (0.005 A) (20 V)
P₂ = 0.1 W
P₃ = (0.00333 A) (20 V)
P₃ = 0.067 W
Answer:
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