Answer:
Compared with the current in the first coil, the current in the second coil is unchanged.
Explanation:
All coils, inductors, chokes and transformers create a magnetic field around themselves consist of an Inductance in series with a Resistance forming an LR Series Circuit.
The steady state of current in the LR circuit is:
I= V/R (1 - e^-Rt/L)
Where I= current
R= Resistance
V= Voltage
Where R/L is the time constant.
For a conducting wire, it has a very small resistance. The time constant will be in microseconds. The current will be in a steady state after few second. The current is independent on the inductance and dependent on the resistance. The length of wire and the resistance here are the same. Therefore, the current remains unchanged.
Answer:
E. The period of oscillation increases.
Explanation:
The period of oscillation is:
T = 2π√(m/k)
Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.
Increasing the mass will increase the period and decrease the frequency.
Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
Explanation:
Current output at the battery will be current of entire circuit, while the current through each bulb in the parallel circuit is the total current circuit.
So, current output through power supply is i and current through each component be 
considering only three component.
Then in a parallel circuit
