The MCB of a rupas room is tripped and keeps on tripping again and again, and if it is a domestic circuit, what could be the reason for this phenomenon?
The reason could be a short circuit which is resulting in higher level of currents to pass through the MCB which is resulting in trip every time.
OR
The MCB is faulty and might need a replacement.
To Diagnose the problem further more.
Turn off all the switches in rupas room and then try turning on the MCB. If it trips again then MCB is faulty (Subjective to the fact there everything was normal before this issue and no signs of short circuit or spark in wiring were observed)
If MCB does not trip in point 1 then Turn ON all the switches one by one. This shall give you the cause of problem.
Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance

∴ Area of each plate is 7.9060 m²
Voltage

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
The back-and-forth movement of electrons is called alternating current. Electrons go back and forth, the direction of their path alternates from one direction to another.
the movement of electrons in one direction is called direct current. The electrons move in a direct, single path without changing directions.
Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:


DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation

according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero

solving for x we get

x = 0.200 m
b)electric field at half way mean x =0.25

E = 3.84*10^{-4} N/C
Answer:
The work done on the object by the force in the 5.60 s interval is 40.93 J.
Explanation:
Given that,
Force 
Mass of object = 2.00 kg
Initial position 
Final position 
Time = 4.00 sec
We need to calculate the work done on the object by the force in the 5.60 s interval.
Using formula of work done


Put the value into the formula




Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.