An inductor has an inductance of 0.025 H and a wire resistance of 3 V. How long will it take the current to reach its full Ohm’s law value?
The correct answer and explanation is :
The time constant ((\tau)) for an RL circuit is given by:
[
\tau = \frac{L}{R}
]
where:
- (L = 0.025) H (inductance),
- (R = 3) Ω (resistance).
Substituting the values:
[
\tau = \frac{0.025}{3} = 0.0083 \text{ seconds} \text{ (or 8.3 milliseconds)}.
]
The current in an RL circuit follows the equation:
[
I(t) = I_{\text{max}} \left(1 – e^{-t/\tau} \right)
]
where (I_{\text{max}}) is the steady-state current given by Ohm’s law:
[
I_{\text{max}} = \frac{V}{R}
]
The current reaches approximately 99% of its final value after 5 time constants:
[
t_{99\%} = 5\tau = 5 \times 0.0083 = 0.0415 \text{ seconds} \text{ (or 41.5 milliseconds)}.
]
Explanation:
An inductor resists changes in current due to the back EMF it generates. When voltage is applied to an RL circuit, the current does not reach its final Ohm’s law value instantaneously. Instead, it follows an exponential growth pattern. The time constant (\tau) determines how quickly the current approaches its steady-state value.
Initially, when the switch is closed, the current starts at zero. As time progresses, the current gradually increases. After one time constant ((\tau)), the current reaches about 63% of its maximum value. After 5τ, the current is effectively at 99%, which is considered “full” for practical purposes.
This behavior is crucial in applications like electrical circuits, motor controls, and signal processing, where inductors influence the timing and response of systems.