Which statement correctly describes how increased voltage affects brake decay time?

Which statement correctly describes how increased voltage affects brake decay time?

• A. It decreases decay time
• B. It does not affect decay time
• C. It increases decay time
• D. It makes decay time unpredictable

Answer: (C)

When braking current is released, the current decays through the braking resistor like an RL circuit. The current follows i(t) = I0 e^{-R t / L}, where the time constant is L/R. The rate of decay is governed by this constant, which doesn’t change with voltage. What changes with voltage is the starting current I0, since I0 = V/R. A higher voltage means a higher initial braking current, so to reach a fixed lower current level (the point where braking is effectively finished), more time is needed. In other words, while the decay rate (the exponential curve) stays the same, starting from a larger current takes longer to drop to the same final value, so the braking current remains in the braking range longer.

When braking current is released, the current decays through the braking resistor like an RL circuit. The current follows i(t) = I0 e^{-R t / L}, where the time constant is L/R. The rate of decay is governed by this constant, which doesn’t change with voltage. What changes with voltage is the starting current I0, since I0 = V/R. A higher voltage means a higher initial braking current, so to reach a fixed lower current level (the point where braking is effectively finished), more time is needed. In other words, while the decay rate (the exponential curve) stays the same, starting from a larger current takes longer to drop to the same final value, so the braking current remains in the braking range longer.