Question

The switch in the circuit of Fig. P5.56 was moved from

position 1 to position 2 at t = 0, after it had been in position 1
for a long time. If L = 80 mH, determine i(t) for t ≥ 0. The answer is supposed to be in the form of: i(t) = i(infinity) + [ i(0) - i(infinity) ] e^(- t/tau)

Answer 1

Without the circuit diagram, it's not possible to provide a detailed solution to this problem. However, based on the information given, we can determine the time constant of the circuit and the initial and final values of the current.

Given that L = 80 mH, the time constant of the circuit is τ = L/R, where R is the total resistance of the circuit. Since the circuit diagram is not provided, we cannot determine R.

At t = 0, the switch is moved from position 1 to position 2, which means that the circuit is now a series RL circuit. At t = 0-, the current through the inductor is i(0-) = i(infinity), where i(infinity) is the steady-state current in the circuit when the switch is in position 2.

At t = 0+, the current through the inductor is i(0+), which is equal to i(infinity) + [i(0) - i(infinity)]e^(-t/τ), where i(0) is the initial current through the inductor just before the switch is moved to position 2.

Therefore, the expression for the current in the circuit for t ≥ 0 is given by:

i(t) = i(infinity) + [i(0) - i(infinity)]e^(-t/τ)

where τ = L/R, i(0) is the initial current through the inductor just before the switch is moved to position 2, and i(infinity) is the steady-state current in the circuit when the switch is in position 2.

Note that the above expression assumes that the circuit is purely a series RL circuit with no other components such as capacitors or voltage sources. If the circuit contains other components, the expression for the current will be more complex.

Step-by-step explanation: