In an RL circuit with a DC source Vs, after the switch is closed, the current i(t) evolves toward its steady-state value with a time constant τ = L/R. Which expression correctly represents i(t)?

In an RL circuit with a DC source Vs, after the switch is closed, the current i(t) evolves toward its steady-state value with a time constant τ = L/R. Which expression correctly represents i(t)?

• A. I(t) = (Vs/R) (1 - e^{-t/τ})
• B. I(t) = (Vs/R) e^{-t/τ}
• C. I(t) = (Vs/R) (1 + e^{-t/τ})
• D. I(t) = (Vs/R) (1 - e^{t/τ})

Answer: (A)

The current in an RL circuit with a DC source starts at zero (if the inductor is initially unenergized) and rises toward the steady-state value Vs/R, with a time constant τ = L/R. This happens because the inductor resists sudden changes in current, so the loop equation Vs = iR + L di/dt leads to L di/dt + R i = Vs. Solving with the initial condition i(0) = 0 gives i(t) = (Vs/R) [1 − e^(−t/τ)]. This expression yields i(0) = 0 and i(∞) = Vs/R, matching the physical behavior of the circuit. The other forms would imply starting from a nonzero current or diverging, which contradicts the initial condition and the DC steady-state behavior.

The current in an RL circuit with a DC source starts at zero (if the inductor is initially unenergized) and rises toward the steady-state value Vs/R, with a time constant τ = L/R. This happens because the inductor resists sudden changes in current, so the loop equation Vs = iR + L di/dt leads to L di/dt + R i = Vs. Solving with the initial condition i(0) = 0 gives i(t) = (Vs/R) [1 − e^(−t/τ)]. This expression yields i(0) = 0 and i(∞) = Vs/R, matching the physical behavior of the circuit. The other forms would imply starting from a nonzero current or diverging, which contradicts the initial condition and the DC steady-state behavior.