An RL series circuit has: 4.8-mH coil,2.8-Ω resistor,1.0-V battery,ammeter and switch.After the switch is closed for a long time,find the final value of the current.
A)7.4E+1 A
B)3.6E-1 A
C)1.3E-2 A.
D)2.8E+2 A
E)1.3E-2 A

Question

Quizplus · Accepted Answer

To find the final value of current, we need to calculate the total impedance and the total current in the circuit. 
The impedance in an RL circuit is given by: Z = sqrt(R^2 + Xl^2) where Xl is the inductive reactance (2πfL)
Given: L = 4.8mH, R = 2.8Ω, V = 1.0V 
Frequency is not given, therefore we can assume it to be a DC circuit 
Xl = 2πfL = 2π * 0 * 4.8E-3 = 0Ω 
Total impedance Z = sqrt(R^2 + Xl^2) = sqrt(2.8^2 + 0^2) = 2.8Ω

Total current I = V/Z = 1.0/2.8 = 0.36A or 3.6E-1 A 
Therefore, the answer is B.

An RL Series Circuit Has: 4