Question

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 A after the resistor is connected to the battery?

Short answer

Answer: It takes approximately \(2.58 \times 10^{-8}s\) for the current to build up to within \(0.500\%\) of its final value.

Step-by-step solution

Identify the given values

We are given the following values: - Battery Voltage: \(V=100V\) - Resistance: \(R=500\Omega\) - Inductance: \(L=0.200\mu H=0.200\times10^{-6}H\) - Final current: \(I_\text{final} = 0.200A\) - Percentage of the final current: \(0.500\%\)

Compute the time constant

In an RL circuit, the time constant is given by the ratio \(τ=\frac{L}{R}\). So, let's compute the time constant: $$ \tau = \frac{L}{R} = \frac{0.200 \times 10^{-6}}{500} = 4\times10^{-10}s. $$

Determine the current in terms of time

The current in an RL circuit can be expressed as a function of time: $$ I(t) = I_\text{final}(1 - e^{-t/\tau}). $$

Calculate the time for the current to reach \(0.500\%\) of its final value

We want to find the time it takes for the current to build up to within \(0.500\%\) of its final value. So we have: $$ I(t) = 0.995 I_\text{final}, $$ where \(0.995\) comes from \(1 - 0.500\%\). Substitute the expression for \(I(t)\) from Step 3: $$ 0.995 I_\text{final} = I_\text{final} (1 - e^{-t/\tau}). $$ Divide by \(I_\text{final}\): $$ 0.995 = 1 - e^{-t/\tau}. $$ Subtract \(0.995\) from both sides and multiply by -1: $$ e^{-t/\tau} = 0.005. $$ Now, take the natural logarithm of both sides: $$ \ln{0.005} = - \frac{t}{\tau}. $$ Multiply by \(-\tau\) to solve for \(t\): $$ t = -\tau \ln{0.005} = (4\times10^{-10}s) \ln{0.005} \approx 2.58 \times 10^{-8}s. $$ It will take approximately \(2.58 \times 10^{-8}s\) for the current to build up to within \(0.500\%\) of its final value of \(0.200A\).

Key concepts

Faraday's Law of Induction

Michael Faraday introduced the concept of electromagnetic induction, demonstrating how a change in magnetic flux can induce an electromotive force (EMF) in a conductor. This principle is mathematically defined by Faraday's Law of Induction, which states that the induced EMF in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit.

The mathematical representation of Faraday's Law is \( emf = -\frac{d\Phi_B}{dt} \), where \( emf \) is the induced electromotive force, and \( \Phi_B \) represents the magnetic flux. In the context of the textbook exercise, the stray inductance in the RL circuit, which is due to a coiled wire or other conductor, causes the magnetic flux to change as the current changes, resulting in an induced EMF. This induced EMF opposes changes in the current, thereby causing the current to increase over time rather than instantaneously.

Inductive Reactance

Inductive reactance arises when an inductor is placed in an alternating current (AC) circuit. Unlike resistance, which hinders electron flow, inductive reactance temporarily stores energy in a magnetic field and releases it back into the circuit. Specifically, inductive reactance is the opposition that an inductor presents to the current in an AC circuit, and its value changes with the frequency of the alternating current.

The formula for inductive reactance is given by \( X_L = 2\pi f L \), where \( X_L \) is the inductive reactance, \( f \) is the frequency of the AC current, and \( L \) is the inductance. In the exercise, the focus is on a direct current (DC) system, where reactance is not directly considered. However, understanding inductive reactance can be valuable in more complex circuits that involve AC.

Exponential Decay in RL Circuits

In an RL circuit, the time it takes for the current to change is characterized by an exponential decay function due to the inductor's behavior. The time constant \( \tau \) of an RL circuit is the time required for the current to reach approximately 63.2% of its final steady-state value after a sudden change, such as opening or closing a switch.

An RL circuit's current response over time can be expressed as \( I(t) = I_\text{final}(1 - e^{-t/\tau}) \), where \( I(t) \) is the current at time \( t \) and \( I_\text{final} \) is the final steady-state current. As demonstrated in the exercise solution, to find how long it will take for the current to reach a specified percentage of the final value, one can manipulate this formula, ultimately leading to solving for \( t \) using logarithms. An inductor thus affects how quickly the current builds up to its final value, causing an exponential approach rather than a linear one.