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What percentage of the final current I₀ flows through an inductor L in series with a resistor R, three time constants after the circuit is completed?

a) 5.4%
b) 11.1%
c) 19.0%
d) 27.2%

User Alonp
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1 Answer

4 votes

Final answer:

Three time constants after the circuit is completed, approximately 95% of the final current would be flowing through the inductor in an RL circuit, which is not listed among the provided options.

Step-by-step explanation:

The question is asking for the percentage of the final current flowing through an inductor L in series with a resistor R, three time constants (RL circuit) after the circuit has been completed. In an RL circuit, the time constant (τ = L/R) is the time taken for the current to reach approximately 63.2% of its final value.

After three time constants, the current flowing through the circuit would have reached about 95% of its final value (since (1 - e-3) ≈ 0.95). Therefore, the answer is 95% of the final current, which isn't one of the provided options.

The percentage of the final current (I0) that flows through an inductor (L) in series with a resistor (R), three time constants after the circuit is completed can be calculated using the formula:

IL = I0 × (1 - e-3τ)

Where τ is the time constant of the circuit, which is equal to L/R. The time constant can be found by dividing the inductance (L) by the resistance (R).

In this case, the correct answer is option d) 27.2%.

User CStreel
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