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3 years ago

Infer whether a circuit breaker should be connected in parallel to the circuit that it is protecting.

1 answer:

8 0

A circuit breaker is to be connected in series in the circuit it is to protect.
Explanation:
Let us understand why do we need circuit breakers.
Circuit breaker is a protective device which will break the circuit in case of emergency like the appliance gets short or the wiring may get shorted in between. Also it is provided to cut-off electricity to a specific part of wiring in a house/building etc. for non-emergency situations, say for the purpose of maintenance.
Invariably it is placed in series with the live wire at the switch board or circuit breaker box, so that as soon as circuit breaker trips/switched off, no electrical wiring in the circuit it is protecting should remain live.
The mains circuit breaker is placed in series with the wiring of the dwelling unit and the Electricity incoming point of the Electric Supply Company.
The remaining circuit breakers are connected in parallel to each other to various distribution points/rooms. If connected in series, it will be impossible to isolate a section/part of circuit.
However, each circuit breaker is to be connected in series in the part circuit it is to protect.

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2 years ago

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A 20-kg object is subjected to three forces which produce an acceleration a = -8 m.s^-2 i + 6.0 m.s^-2 j on the object. Two of t

lana66690 [7]

Answer:

F₃ = -151 N i + 96 N j

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Forces acting on the object

F₁= 3.0 N i + 16.0 N j

F₂ = -12.0 N i+ 8.0 N j

F₃ = F₃x N i +F₃ y N j

x component of the net force on the object

Fx=F₁x+F₂x+F₃ x

Fx = 3.0 N-12.0 N +F₃x

Fx = F₃x - 9 N

y component of the net force on the object

Fy=F₁y+F₂y+F₃ y

Fy =16.0 N+ 8.0 N +F₃y

Fy = F₃y + 24 N

Newton's second law to the object:

a = -8 m/s² i + 6.0 m/s² j

∑Fx = m*ax m=20 kg , ax = -8 m/s²

F₃x - 9 = 20 *(-8)

F₃x = -160+9

F₃x = -151 N

∑Fy = m*ay m=20 kg , ay = 6 m/s²

F₃y + 24 =20*( 6 )

F₃y =120 - 24

F₃y = 96 N

F₃ = -151 N i + 96 N j

7 0

3 years ago

Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force

gulaghasi [49]

Answer:

F = 0N

Explanation:

The force between two charges is given by

$F=k\frac{q_1q_2}{r^2}$

where r is the distance between the charges and K is the Coulomb's constant

(k=8-89*10^9Nm^2/C^2)

The force in the first charge is only the sum of the forces due to the other charges. Hence we have

$F_T=F_1+F_2=k\frac{q_2q_1}{r^2}+k\frac{q_3q_1}{r^2}$

$F_T=(8.89*10^9\frac{Nm^2}{C^2})\frac{(3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}+(8.89*10^9\frac{Nm^2}{C^2})\frac{(-3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}\\\\F_T=0.445N-0.445N=0N$