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riadik2000 [5.3K]
2 years ago
14

When a technician is ohming out a load in a circuit, what reading should the technician get if the load is good

Engineering
1 answer:
xz_007 [3.2K]2 years ago
6 0

When a technician is ohming out a load in a circuit, the reading that the technician should get if the load is good is some numbers.

<h3>What does Ohming out a circuit mean?</h3>

The term “Ohming out a motor” is known to be the act of measuring the electrical resistance that is present in the motor windings and comparing it with the normal values.

Note that When a technician is ohming out a load in a circuit, the reading that the technician should get if the load is good is some numbers as it will tell if there is a measure of comparison or not.

Learn more about technician from

brainly.com/question/2328085

#SPJ1

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Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

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3 years ago
10. In CCS, Northing and Easting are
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True or False: Stress can effectively be relieved through physical activity, getting enough rest and sleep, and relaxation techn
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Answer:

True

Explanation:

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3 years ago
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What does Clay say will happen if the system is rejected?
juin [17]

Answer:

the nation will suffer terrible consequences

Explanation:

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Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage cha
alexira [117]

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α =  15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ +  β - α  = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}

r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}

r=0.88

New coefficient of friction ,μ'  = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' +  β' - α  = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}

r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}

r'=0.70

Percentage change

\Delta r=\dfrac{r-r'}{r}\times 100

\Delta r=\dfrac{0.88-0.70}{0.88}\times 100

Δr=20.45 %

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3 years ago
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