A \(100 .-V\) battery is connected in series with a \(500 .-\Omega\) resistor. According to Faraday's Law of Induction, current can never change instantaneously, so there is always some "stray" inductance. Suppose the stray inductance is \(0.200 \mu \mathrm{H}\). How long will it take the current to build up to within \(0.500 \%\) of its final value of \(0.200 \mathrm{~A}\) after the resistor is con- nected to the battery?

Since we are given the resistance (500 Ω) and the voltage (100 V) in the circuit, we can use Ohm's Law I = V/R to find the final value of current. Here, I = current, V = voltage, and R = resistance. I = 100V / 500Ω = 0.200 A The final value of the current in the circuit is 0.200 A.

To find the time it takes for the current to build up, we need to know the time constant of the circuit. The time constant (τ) for an inductor-resistor circuit (RL circuit) is given by τ = L/R, where L is the inductance and R is the resistance. Given the stray inductance (0.200 μH = 0.200 x 10^{-6} H) and the resistance (500 Ω), we can find the time constant: τ = (0.200 x 10^{-6} H) / 500Ω ≈ 4 x 10^{-10} s

The equation for the current buildup in an RL circuit is given by: I(t) = I_final * (1 - e^{-t/τ}) Where I(t) is the current at time t, I_final is the final value of the current, e is the base of the natural logarithm, and τ is the time constant. We need to find the time at which the current reaches 0.500% (0.995 times) of its final value. We can set up the equation: 0.995 * I_final = I_final * (1 - e^{-t/τ}) Divide both sides by I_final: 0.995 = 1 - e^{-t/τ} Rearrange the equation to get: e^{-t/τ} = 0.005 Now, take the natural logarithm (ln) of both sides of the equation: -t/τ = ln(0.005) Now isolate t: t = -τ * ln(0.005) We now have an equation for the time at which the current reaches 0.500% of its final value. Plug in the time constant (τ ≈ 4 x 10^{-10} s) we found in Step 2 and calculate: t ≈ -(4 x 10^{-10}s) * ln(0.005) ≈ 2.99 x 10^{-9} s The current build-up time t is approximately 2.99 x 10^{-9} seconds or 2.99 ns.