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Dmitriy789 [7]
3 years ago
12

A(n) _________ is a current greater than the equipment rated current or conductor ampacity, which is confined to the normal cond

uctive paths provided by the conductors and other equipment.
Engineering
1 answer:
exis [7]3 years ago
3 0

Answer:

overcurrent

Explanation:

It is any electrical current in excess of the nominal value indicated in the protection device, in the electrical equipment or in the current carrying capacity of a conductor. The overcurrent can be caused by an overload, a short circuit or a ground fault.

The overcurrent raises the operating temperature in the different elements of the electrical installation where this presentation is.

An overcurrent can be an overload or electric shock current.

<u>Overload</u>: the overload current is an excessive current in relation to the nominal operating current. It occurs in drivers and other components of a distribution system. Overloads are in most cases, more frequent between a range of one to six times the nominal current level. They are caused by temporary increases in current and occur when the motors start or when the transformers are energized.

Electric shock: as the name implies, a electric shock current is one that flows out of the normal conduction pathways. Electric shock or fault currents can be hundreds of times greater than the nominal operating current.

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An AC circuit has a resistor, capacitor and inductor in series with a 120 V, 60 Hz voltage source. The resistance of the resisto
aliina [53]

Answer:

(i) 3.5385 ohm, 3.768 ohm (ii) 39.89 A (III) 4773.857 W (vi) 348 var (vii) 0.9973 (viii) 4.1796°

Explanation:

We have given voltage V =120 volt

Frequency f=60 Hz

Resistance R =3 ohm

Inductance L =0.01 H

Capacitance C =0.00075 farad

(i) reactance of of inductor X_L=\omega L=2\pi fL=2\times 3.14\times 60\times 0.01=3.768ohm

X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 60\times 0.00075}=3.5385ohm

(ii) Total impedance Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{3^2+(3.768-3.5385)^2}=3.008ohm

Current i=\frac{V}{Z}=\frac{120}{3.008}=39.89A

(viii) power factor cos\Phi =\frac{R}{Z}=\frac{3}{3.008}=0.9973

(VII) cos\Phi =0.9973

\Phi =4.1796^{\circ}

So power factor angle is 4.1796°

(iii) Apparent power P=VICOS\Phi =120\times 39.89\times 0.9973=4773.875W

(vi) Reactive power Q=VISIN\Phi =120\times 39.89\times SIN4.17^{\circ}=348var

5 0
3 years ago
How many meters is equivalent to a mile?
Nutka1998 [239]

Answer:

1609.344 metres

I think, correct me if I'm wrong

5 0
3 years ago
Read 2 more answers
A particle travels along a straight line with a velocity v = (12 – 3t2) m/s. When t = 1 s, the particle is located 10 m to the l
arlik [135]

Answer:

The displacement from t = 0 to t = 10 s,  is -880 m

Distance is 912 m

Explanation:

v = (12 - 3t^2) m/s = ds/dt.  .  . . . . . . . .  A

integrate above equation we get

s = 12t - t^3 + C

from information given in the question  we have

t = 1 s, s = -10 m

so distance s will be

-10 = 12 - 1 + C,

C = -21

s(t) = 12t - t^3 - 21

we know that acceleration is given as

a(t) = dv/dt = -6t  

[FROM EQUATION A]

Acceleration at  t = 4 s, a(4) = -24 m/s^2

for the displacement from t = 0 to t = 10 s,

s(10) - s(0) = (12*10 - 10^3 - 21) - (-21) = -880 m

the distance the particle travels during this time period:

let v = 0,

3t^2 = 12

t = 2 s

Distance = [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m

7 0
4 years ago
Which size of impurity atom, smaller impurity atom or larger impurity atom, when located near a dislocation, will nullify some o
svp [43]

Answer:

Smaller impurity atom will nullify some of the compressive strain of a dislocation in a crystal. Because, smaller impurity atoms located near a dislocation creates tensile strain on atoms around it thereby partially nullifying compressive strain at the dislocation.

4 0
3 years ago
Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move re
slava [35]

Answer:

The answer is "15 N".

Explanation:

Please find the complete question in the attached file.

In frame B:

For just slipping:

\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\

        =2 \times 1 \times  g \times \tan 37^{\circ}\\\\ =2 \times 10 \times  \frac{3}{4}\\\\ =15 \ N

4 0
3 years ago
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