For reference, here is your circuit properly drawn with component designators:

The question is what happens when the inductor current starts at 0, then the switch is closed.

The first observation should be that the current source and the two resistors can be reduced to the equivalent Norton or Thevenin circuits. That is a current source with a single resistor in parallel, or a voltage source with a single resistor in series. The two are equivalent from the inductor's point of view. I'll pick the Thevenin version:

The current function should now be obvious from inspection because it's a simple L-R decay:

    I = I_final⋅(1 - e^-t/Τ)

    I = (V1/R3)⋅(1 - e^-t(R3/L1))

We can easily see that:

• The initial current is 0.
• The "final" current is V1/R3 = (5 V)/(15 Ω) = 333 mA.
• The time constant is L1/R3 = (1 H)/(15 Ω) = 66.7 ms.

the time costant is L1/R so the t/T must have units of Ohms*seconds/Henries

I'm not sure what you are getting at. Resistance divided by inductance yields units of time. Specifically, Henrys / Ohms = seconds.

Ah, I just noticed that I accidentally flipped the time constant in the equation above. The exponential is supposed to be e^-t/Τ, with Τ being L1/R3. I'll go fix that now.

posted over 3 years ago

CC BY-SA 4.0

3y ago

Olin Lathrop‭

8059 reputation 18 332 961 469