All categories

3 years ago

Electric fuse and a Circuit breaker contains?

2 answers:

7 0

Fuses and Circuit Breakers (CB) are used for the protection of power system. These along with other protective equipment like relays, isolater, switches, are collectively called as switchgear(switching equipment used in power systems).

The basic function is to break the circuit in case of faulty conditions so as to protect the power system equipment and auxiliaries.

FUSE is a low resistance device which is placed in the circuit for protection. Under faulty conditions when the current becomes more than the desired value, then due to the increase in temperature the fuse wire melts and breaks, thus breaking the circuit. These are used for lower power ratings and can be used only once, after that it has to be replaced with a new one.

CIRCUIT BREAKER also solves the same purpose i.e. it breaks the circuit when the fault occurs. These are used for large ratings and in the power systems and auxiliaries. CB also has the capability to re-close after the fault is through i.e. when the system gets healthy. The CB has two parts/contacts one is static and the other is moving(the one responsible for making or breaking the contacts).When the contacts of CB separate, then it said to have tripped. Tripping of the circuit breaker has to be very fast so as to avoid the arc formation between the two contacts, otherwise it will defeat the purpose of installing the CB. The trip signal is given to the CB by an other protective device called as an ELECTRICAL RELAY. CB’s are categorized on the basis of the power ratings and the medium used to extinguish the arc formed during the breaking of the contacts.

mark brainliest :)

Fiesta28 [93]3 years ago

4 0

Electric fuse and a Circuit breaker both contain ways to interrupt the flow of electricity.

Hope this helps!

Davinia.

You might be interested in

In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal

SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0

3 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

11 months ago

What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?

Assoli18 [71]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

m is the mass

f is the force

From the question we have

$a = \frac{70}{7} = 10 \\$

We have the final answer as

Hope this helps you

4 0

2 years ago

How much charge is on each plate of a 3.00-Î¼F capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c

Sauron [17]

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

6 0

3 years ago

Many planets in our solar system (Saturn, Uranus, Neptune) have rings. If these rings are made of chunks of debris, why do they

Liono4ka [1.6K]

Generally, rings form from moons, asteroids, or comets that have disintegrated due to a collision or because they got too close to their planet (Roche Limit). ... Most of the debris, however, will not have enough energy to overcome the planet's gravity and will remain in orbit around the planet.

5 0

3 years ago