Answer:
0.6 A
Explanation:
As the motor gears towards full speed, the voltage of the circuit tends to become the difference that exists between the line voltage and the that of the back emf. Remembering Ohm's law, we then apply it to get the final, lower current that is based on the reduced voltage Ef. We use the provided resistance in the question, that is 3 ohms.
Ef = (120 V) - (117 V) = 3 V
If = Ef/R = (3 V) / (5.0 Ω) = 0.6 A
It helps you understand the concept of what you are predicting
The resonant frequency of a circuit is the frequency 
at which the equivalent impedance of a circuit is purely real (the imaginary part is null).
Mathematically this frequency is described as
Where
L = Inductance
C = Capacitance
Our values are given as
Replacing we have,
From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.
For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.
Answer:
Explanation:
When an moving electric charge passes through a uniform magnetic field
its motion becomes circular .
If m be the mass v be the velocity , q be the charge on the mass B be the magnetic field and R be the radius of circular path
force on the moving charge created by magnetic field
= B q v
Centripetal force required for circular motion
= m v² / R
For balancing
B q v = m v² / R
v = B q R / m
Time period of rotation
T = 2π R / v
= 2 π R m / B q R
= 2 π m / B q
For first particle
T₁ = 2 π m₁ / B q₁
For second particle
T₂ = 2 π m₂ / B q₂
q₁ = q₂ and 10 m₁ = m₂ ( given )
Putting the values in second equation
T₂ = 2 π 10 m₁ / B q₁
= 10 x 2 π m₁ / B q₁
= 10 T₁
Given T₁ = T
T₂ = 10 T
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