An inductor has an inductance of 0.025 H and a wire resistance of 3 V

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 :

To determine the time it takes for the current to reach its full Ohm’s Law value in an inductor with resistance, we analyze the RL time constant.

Given Data:

  • Inductance: ( L = 0.025 ) H
  • Resistance: ( R = 3 ) Ω

Step 1: RL Circuit Time Constant

The time constant (( \tau )) of an RL circuit is given by:

[
\tau = \frac{L}{R}
]

Substituting the values:

[
\tau = \frac{0.025}{3} = 0.0083 \text{ s} = 8.33 \text{ ms}
]

Step 2: Current Growth in an RL Circuit

In an RL circuit, the current grows according to the equation:

[
I(t) = I_{\text{max}} \left( 1 – e^{-t/\tau} \right)
]

where ( I_{\text{max}} ) is the final current given by Ohm’s Law:

[
I_{\text{max}} = \frac{V}{R}
]

The current approaches 99% of its full value after about 5 time constants:

[
t \approx 5\tau
]

Substituting:

[
t = 5 \times 0.0083 = 0.0417 \text{ s} = 41.7 \text{ ms}
]

Step 3: Explanation

In an RL circuit, an inductor resists changes in current due to the voltage induced across it. When a voltage source is applied, the current does not instantly reach its maximum value but rather increases gradually, following an exponential curve.

The time constant ( \tau ) represents how quickly the current rises. After one ( \tau ), the current reaches about 63% of its maximum value. After about 5τ (41.7 ms in this case), the current reaches 99% of its final value, which is practically considered “full” in electrical engineering.

Thus, in this case, the current will take approximately 41.7 milliseconds to reach its full Ohm’s law value.

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