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2 years ago

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

1 answer:

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.

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1. Springs____________ energy when compressed And _________energy when they rebound.

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Answer:

Springs store energy when compressed and release energy when they rebound

Explanation:

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3 years ago

You are considering purchasing a compact washing machine, and you have the following information: The Energy Guide claims an est

RSB [31]

Answer: $15.34

Explanation: see image below

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3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago

Which one of the following activities is not an example of incident coordination

Lady bird [3.3K]

Directing, ordering, or controlling

7 0

4 years ago

A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 c

gayaneshka [121]

Answer:

V1=5ft3

V2=2ft3

n=1.377

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=4lb*1.25ft3/lb=5ft3

state 2

V2=m.v2

V2=4lb*0.5ft3/lb= 2ft3

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/53)/ ln (2/5)

n=1.377

3 0

4 years ago