Answer: The current in both circuit instances are equal (the current is unchanged).
Step-by-step 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.