An RL series circuit has: 5.3-mH coil,2.2-Ω resistor,15-V battery,ammeter and switch.After the switch is closed for a long time,find the final value of the current.
A) 1300 A
B) 6.8 A
C) 0.8 mA
D) 33 A
E) 0.17 A

Question

An RL series circuit has: 5.3-mH coil,2.2-Ω resistor,15-V battery,ammeter and switch.After the switch is closed for a long time,find the final value of the current.
A) 1300 A
B) 6.8 A
C) 0.8 mA
D) 33 A
E) ​0.17 A

Quizplus · Accepted Answer

First, we can calculate the time constant of the circuit using the formula:

τ = L/R = 5.3 mH / 2.2 Ω = 2.41 μs

Since the switch is closed for a long time, we can assume that the circuit has reached steady state, and the capacitor has been fully charged.

At steady state, the inductor will act like a short circuit, and the current will only be limited by the resistance of the circuit.

Using Ohm's law, we can calculate the steady-state current as:

I = V/R = 15 V / 2.2 Ω = 6.8 A

Therefore, the final value of the current is 6.8 A, which is option B.

An RL Series Circuit Has: 5