Respuesta :
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: The current in both circuit instances are equal (the current is unchanged).
Explanation:
Provided that a voltage V is the voltage of the battery(capacitor)
The Inductance of the first wound wire, L1 = 5 mH
The Inductance of the second wound wire, L2 = 10 mH
The wire and battery contribute to the resistance of the circuit and consequently, the circuit has a total resistance ,R.
V = I R + L dI/dt
where, I is the current in the circuit and t is the time.
dI/dt which is the change in current with respect to time diminishes after a few seconds, this is due to constant current flowing through the circuit after this time.
Therefore,
The factor dI/dt becomes 0.
The same applies for the second coil with inductance, 10mH
There is no change in the current flowing through the circuit. when the 10mH inductor was just attached in the circuit, a current, I still flows through the circuit.
Based on the explanation above, the current remains unchanged.
