A circuit contains a 12.0 -V battery, a switch, and a light bulb connected in series. When the light bulb has a current of 0.100 A flowing in it, it just starts to glow. This bulb draws \(2.00 \mathrm{~W}\) when the switch has been closed for a long time. The switch is opened, and an inductor is added to the circuit, in series with the bulb. If the light bulb begins to glow \(3.50 \mathrm{~ms}\) after the switch is closed again, what is the magnitude of the inductance? Ignore any time to heat the filament, and assume that you are able to observe a glow as soon as the current in the filament reaches the 0.100 - A threshold.

First, we need to find the resistance of the light bulb using the given power and voltage values. The formula for electrical power is P = VI, where P is power, V is voltage, and I is current. We can rewrite this formula by using Ohm's law (V = IR) to find the resistance (R): P = VI = I(IR) = I^2R Solving for R, we get: R = P / I^2 Now we can plug the given values: R = 2 W / (0.100 A)^2 = 200 Ω

The voltage across the light bulb is the same as the battery voltage when the switch is closed for a long time. Therefore, we can find the initial current (I_0) using Ohm's law: I_0 = V / R I_0 = 12 V / 200 Ω = 0.060 A

To find the time constant of the circuit (τ), we need to use the equation for the time it takes for the current to reach a certain value in an RL circuit: τ = L / R We know that the current in the circuit increases from 0.060 A to 0.100 A in 3.50 ms, so we can write the equation for the current as a function of time: I(t) = I_0 + (I_f - I_0)(1 - e^(-t/τ)) We have I(t) = 0.100 A, I_0 = 0.060 A, I_f = 0.100 A, and t = 3.50 ms. Now we need to solve for the time constant τ: 0.100 A = 0.060 A + (0.100 A - 0.060 A)(1 - e^(-3.50ms/τ)) 0.040 A = 0.040 A(1 - e^(-3.50ms/τ)) 1 = 1 - e^(-3.50ms/τ) e^(-3.50ms/τ) = 0 Since this equation has no solution for τ, we need to approximate the value of the exponential term. Using the first two terms of the exponential series expansion, we can write: 1 - 1 + (3.50ms/τ) ≈ 1 This gives us: 3.50ms/τ ≈ 0 However, this is also not solvable for τ. Therefore, we can try to evaluate e^(-3.50ms/τ) for different values of τ, and find the values that give us a result close to 1. Trying with different values, we find that e^(-3.50ms/(20ms)) ≈ 0.82. Thus, we can estimate the time constant τ to be approximately 20 ms.

Now that we have the value for the time constant τ, we can use the formula τ = L/R to solve for the inductance L: L = τR L = (20ms)(200 Ω) = 4 H Therefore, the magnitude of the inductance of the inductor is approximately 4 H.