Solving a Resistance-Inductance Circuit Problem

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SUMMARY

The discussion focuses on solving a resistance-inductance circuit problem involving a coil with a resistance of 8.0 ohms and a self-inductance of 4.0 mH connected to a 100 V potential difference. The steady-state current (If) is calculated to be 12.5 A, and the time constant (T) is determined to be 0.500 ms. The current I as a function of time is expressed as I = (12.5A)(1 - e^(-t/0.500ms)), while the rate of change of current (dI/dt) is derived as dI/dt = 12.5A(-e^(-t/0.500ms)). The discussion highlights the need for clarification on the units and numerical factors in the expression for dI/dt.

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Homework Statement


A coil of resistance 8.0 ohms and self-inductance 4.0 mH is suddenly connected across a constant potential difference of 100 V. Let t = 0 be the time of connection, at which the current is zero. Find the current I and its rate of change dI/dt at times (a) t = 0, (b) t = 0.1 s, (c) t = 0.5 s, and (d) t = 1.0 s.


Homework Equations



If=E_o / R = 100V/8.00ohms = 12.5A

T(time constant)= L/R = 4.0 mH/8ohms = .500 ms

The Attempt at a Solution


Substitute If and T to obtain
I = (12.5A) (1-e^(-t/.500ms))

Express dI/dt :

dI/dt = 12.5A(-e^(-t/.500ms)) <--- here is where I am stuck. Seems like there should be a numer in the thousands to multiply with units s^-1 to cancel out the other units

which in return would yiels aan anser like:
= (##kA)e^(-t/.500ms)

Then substitute all the reset of the values in.

Not sure what i am missing here Please help me!
 
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