Engineering LR circuit current calculation

[PhysicsTest]

Homework Statement

A 120-V dc generator energizes a motor whose coil has an inductance of 50 H and a resistance of
A field discharge resistor of is connected in parallel with the motor to avoid damage to the
motor, as shown in Fig. 7.147. The system is at steady state. Find the current through the discharge resistor 100 ms after the breaker is tripped.

Relevant Equations

I=I0(1 - e^-t/tau)

In the steady state the current through 100 Ohm path and inductor acts as a short circuit, inductor stores the current equal to as below
I = 120/100 = 1.2Amps
once the circuit breaker is removed the inductor works as current source the 100 Ohm and 400 Ohm are in series and the current after 100ms is

But somehow the answer is not matching which is 441mA. I am not sure why.

 

[PhysicsTest]

Yes i think i have made wrong calculations it is e^-1 and it comes correctly as 441

 

[DaveE]

Because 500*0.1/50 ≠ 0.1. Your method was correct.

Also, yet another example of a sloppy/bad/confusing HW question. Motors have mechanical loads, maybe just the rotor inertia. This shows up on the oscilloscope. Why do they have to be cute? Why can't they just draw a schematic with an inductor if they want you to just use the inductor model?