Question

Using the exact exponential treatment, find the time required for the current through a 2.00 H inductor in series with a 0.500 Ω resistor to be reduced to 0.100% of its original value.

a) 13.86 seconds
b) 18.42 seconds
c) 24.03 seconds
d) 31.52 seconds

Answer 1

To find the time required for the current through the inductor to be reduced to 0.100% of its original value, we can use the formula for the current through an RL circuit. Plug in the values of the inductance and resistance to calculate the time constant. Set up the equation using the given values and solve for time.

Step-by-step explanation:

To find the time required for the current through the inductor to be reduced to 0.100% of its original value, we can use the formula for the current through an inductor in an RL circuit:

I(t) = I0 * e^(-t/τ)

Where I(t) is the current at time t, I0 is the initial current, e is the base of the natural logarithm, t is the time, and τ is the time constant given by the formula τ = L/R.

Substituting the given values of L = 2.00 H and R = 0.500 Ω into the formula, we have:

τ = 2.00 H / 0.500 Ω = 4.00 seconds

Next, plug the values of I(t) = 0.001 * I0 (0.100% of I0) and τ into the equation and solve for t:

0.001 * I0 = I0 * e^(-t/4.00)

Simplifying the equation, we get:

e^(-t/4.00) = 0.001

Taking the natural logarithm of both sides, we get:

-t/4.00 = ln(0.001)

Finally, solving for t:

t = -4.00 * ln(0.001) = 18.42 seconds

Therefore, the time required for the current through the inductor to be reduced to 0.100% of its original value is approximately 18.42 seconds (option b).