Final answer:
The correct expression for the current i(t) through an inductor in an RL circuit after a switch is opened is i(t) = i₀ · R₁/(R₁ + R₂) · e-(R₁ + R₂)/L · t, which accounts for the initial steady-state current and the exponential decay over time due to the combined resistance and inductance.
Step-by-step explanation:
When an inductor is connected in an RL circuit with resistors R1 and R2 and an inductance L, and if there is a steady-state current is before the circuit is switched, then the expression for the current i(t) through the inductor for t ≥ 0 can be given by one of the expressions provided. Understanding RL circuits and the behavior of inductors in such circuits is crucial in determining the correct expression. Specifically, in a simple RL circuit where an inductor L is in series with a resistor R, the time constant is given by τ = L/R, and the current decay can be expressed as I(t) = I0 · e-(R/L) · t. Considering a circuit that has been initially subjected to a steady current, the current through the inductor after the switch is opened will decay exponentially over time. The term R1/(R1+R2) represents the current division factor between the resistors R1 and R2 in series, while e-(R1+R2)/L · t represents the exponential decay of the inductor current with the combined resistance (R1+R2) and inductance L. Hence, the correct expression for the current i(t) would be i(t) = is · R1/(R1 + R2) · e-(R1 + R2)/L · t, matching answer choice (b).