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Basile [38]
3 years ago
12

Electric fuse and a Circuit breaker contains?

Physics
2 answers:
Orlov [11]3 years ago
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.

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Answer:

45200J

Explanation:

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Heat of vaporization of water  = 2260J/g

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Temperature = 100°C

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Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

         H  = mL

m is the mass

L is the latent heat of vaporization

Insert the parameters and solve;

         H  = 20g x 2260J/g

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2 years ago
Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu
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Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)²  = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424  x 10²⁵

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4 0
2 years ago
The time constant for RC circuit with the values of R1 and C1 is 5ms. What will be the time constant for a new RC circuit with t
lions [1.4K]

Answer:

d. 25 ms

Explanation:

  • In a RC circuit we call time constant to the product of the resistance times the capacitance, which represents the time when the charge reaches to the 63% of the final value, as follows:

       \tau_{1} = R_{1} *C_{1}  = 5 ms (1)

  • If we have a new circuit with new values for R and C, the time constant will be defined in the same way, as follows:

       \tau_{2} =10* R_{1} *0.5*C_{1}  = 5*(R_{1}* C_{1}) = 5* \tau_{1} = 5* 5 ms = 25 ms (2)

6 0
2 years ago
A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m
anzhelika [568]

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

Given,

The centripetal acceleration of the car, a = 13.33 m/s²

The radius of the curve, r = 30 m

The centripetal force acting on the car is given by the formula

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Where    v²/r is the acceleration component of the force

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Substituting the values in the above equation

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Hence, the speed of the car, v = 19.997 m/s

3 0
2 years ago
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