Question

Using the exact exponential treatment, find the time required to bring the current through an 80.0 mH inductor in series with a 15.0 Ω resistor to 99.0% of its final value, starting from zero.

a) 2.56 seconds
b) 4.32 seconds
c) 6.21 seconds
d) 8.09 seconds

Answer 1

Using exact exponential treatment for an RL circuit with an 80.0 mH inductor and a 15.0 Ω resistor, the time constant is 5.33 × 10^{-3} s. The time required to reach 99.0% of the final current value is approximately 6.21 seconds, corresponding to choice (c).

Step-by-step explanation:

To find the time required to bring the current through an 80.0 mH inductor in series with a 15.0 Ω resistor to 99.0% of its final value, we use the exact exponential treatment which is based on the formula for the current in an RL circuit:

I(t) = I_{final}(1 - e^{-t/τ})

where τ is the time constant of the circuit and is equal to L/R. For a circuit with an 80.0 mH inductor and a 15.0 Ω resistor, the time constant τ is:

τ = L/R = (80.0 × 10^{-3} H)/(15.0 Ω) = 5.33 × 10^{-3} s

To find the time when the current reaches 99.0% of its final value, we set I(t)/I_{final} to 0.99 and solve for t:

0.99 = 1 - e^{-t/τ}

e^{-t/τ} = 0.01

-t/τ = ln(0.01)

t = -τ × ln(0.01) = -5.33 × 10^{-3} s × ln(0.01) ≈ 6.21 seconds

Therefore, the time required to bring the current to 99.0% of the final value is approximately 6.21 seconds, which corresponds to answer choice (c).