An RL circuit with series L and R is connected to a DC source V at t=0. After the switch closes, which expression gives i(t)?

An RL circuit with series L and R is connected to a DC source V at t=0. After the switch closes, which expression gives i(t)?

• A. i(t) = (V/L) (1 - e^(-tR/L))
• B. i(t) = (V/R) e^(-tR/L)
• C. i(t) = (V/R) (1 - e^(-tR/L))
• D. i(t) = (V/R) (1 - e^(-t/L))

Answer: (C)

When a DC source is applied to a series RL circuit, the inductor resists any instantaneous change in current. The current evolves from 0 toward a steady value of V/R, with a time constant τ = L/R. The governing equation is V = L di/dt + iR, which solves to i(t) = (V/R) [1 − exp(−(R/L) t)]. This form satisfies i(0) = 0 and i(∞) = V/R, and the exponential term decays with the correct time constant. The other expressions don’t fit the physics: starting with i(0) ≠ 0 or having the wrong initial slope would conflict with the inductor’s behavior, and using an exponent like t/L is dimensionally inconsistent. Therefore, i(t) = (V/R) [1 − e^(−tR/L)] is the correct description of the transient current.

When a DC source is applied to a series RL circuit, the inductor resists any instantaneous change in current. The current evolves from 0 toward a steady value of V/R, with a time constant τ = L/R. The governing equation is V = L di/dt + iR, which solves to i(t) = (V/R) [1 − exp(−(R/L) t)]. This form satisfies i(0) = 0 and i(∞) = V/R, and the exponential term decays with the correct time constant.

The other expressions don’t fit the physics: starting with i(0) ≠ 0 or having the wrong initial slope would conflict with the inductor’s behavior, and using an exponent like t/L is dimensionally inconsistent. Therefore, i(t) = (V/R) [1 − e^(−tR/L)] is the correct description of the transient current.