Initial current in an RL circuit

inductor

I would like to confirm that I am going about this exercise the right way.

"After a long time of being open the switch closes at t=0. Find $_L(s) and i_L(t)." (See picture below)

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Do I need to take any initial conditions into account? There are none given, but as stated, the circuit has been on for a long time. Can the initial current be written as simply V_s/2R?

What I did was sum the voltages together V_S(t) = i_LR + Ldi/dt, then I took the Laplace transform and got I(s) = V(s)/L*1/(s + R) and from that I got i_L(t) = V_s(t)/L*e^(-Rt)

Best Answer

Yes, you should consider that the initial current is V/2R, as inductors behave as short circuits in the steady state.

However, without doing any calculations, there are many things that tell me your answer is incorrect:

  1. The dimension for your current is Voltage/Inductance, which is wrong, should be voltage/resistance.
  2. The dimension of your exponent is R*t, which should be dimensionless, and it is not.
  3. When t goes to infinity, according to your answer, the current goes to zero. By the same logic that allowed you to determine that the initial current before the switch is flipped is V/2R, what should be the current after a long time has passed after the switch is closed?

HINT: Your voltage V is constant. What is the Laplace transform of a constant? You have V(s). Also, what is the Laplace transform of a derivative \$\large \frac{df}{dt}\$ ? Are you sure it's just \$s\cdot F(s)\$?

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