Question

What is the inductance in an \(\mathrm{RL}\) circuit with \(R=17.88 \Omega\) if the time required for the current to reach \(75 \%\) of its maximum value is \(3.450 \mathrm{~ms} ?\)

Short answer

Answer: The inductance in the RL circuit is approximately 0.1776 H.

Step-by-step solution

1. Obtain the time constant formula for an RL circuit

The time constant of an RL circuit, denoted as \(\tau\), is given by the formula: $$\tau = \frac{L}{R}$$ Where \(L\) is the inductance of the coil in henries (H) and \(R\) is the total resistance of the circuit in ohms (\(\Omega\)).

2. Calculate the time constant from the given time

We are given the time \(t\) to reach \(75\%\) of its maximum value. Since the time constant, \(\tau\), is the time required for the current to reach \(1 - 1/e \approx 63.2\%\) of its maximum value in an RL circuit, we need to find a way to relate this given time with the time constant. For an RL circuit, the current as a function of time is given by: $$i(t) = I_\max (1 - e^{(-t/\tau)})$$ Now, we know $$0.75 I_\max = I_\max (1 - e^{(-3.450\mathrm{ms}/\tau)})$$ Dividing both sides by \(I_\max\), we have: $$0.75 = 1 - e^{(-3.450\mathrm{ms}/\tau)}$$ From here, we can solve for \(\tau\).

3. Solve for the time constant

Rearranging last equation: $$e^{(-3.450\mathrm{ms}/\tau)} = 1 - 0.75$$ $$e^{(-3.450\mathrm{ms}/\tau)} = 0.25$$ Taking the natural logarithm (ln) of both sides, we get: $$-\frac{3.450\mathrm{ms}}{\tau} = \ln{0.25}$$ Now, we can find the value of \(\tau\): $$\tau = -\frac{3.450\mathrm{ms}}{\ln{0.25}}$$

4. Calculate the time constant

Plugging the given values into the expression for \(\tau\), we get: $$\tau = -\frac{3.450 \times 10^{-3}\mathrm{s}}{\ln{0.25}}$$ $$\tau \approx 0.00993\,\mathrm{s}$$

5. Use the time constant formula to find the inductance

Now, we have the time constant \(\tau\) and the given resistance \(R = 17.88\,\Omega\). We can rearrange the time constant formula to solve for the inductance \(L\): $$L = \tau \cdot R$$ Plugging in the values of \(\tau\) and \(R\), we get: $$L = 0.00993\,\mathrm{s} \cdot 17.88\,\Omega$$ $$L\approx 0.1776\,\mathrm{H}$$ Therefore, the inductance in the RL circuit is approximately \(0.1776\,\mathrm{H}\).

Key concepts

Time Constant Formula

The time constant, denoted as \(\tau\), is a crucial parameter in an RL (resistor-inductor) circuit. It helps to describe how quickly the current reaches a certain percentage of its maximum value after the power is switched on. Specifically, the formula for calculating the time constant in an RL circuit is given by:
\[\tau = \frac{L}{R}\]
where \(L\) represents the inductance in henries (H), and \(R\) is the resistance in ohms (\(\Omega\)). The time constant \(\tau\) has units of seconds and it signifies the time it takes for the current to reach approximately 63.2% of its final value in an ideal RL circuit.

Understanding the time constant is fundamental because it ultimately determines the transient response speed of the circuit. When solving for inductance, as in the given exercise, it is vital to relate the time constant to the given time at which a certain percentage of maximum current is achieved, as the relationship between them is not direct. Therefore, additional steps involving exponential functions and natural logarithms were taken to establish \(\tau\) before proceeding with the inductance calculation.

Current in RL Circuit

In an RL circuit, the behavior of the current over time is characterized by exponential growth. When the switch completes the circuit, the current doesn't immediately reach its maximum value; instead, it gradually increases and eventually levels off. This response can be mathematically represented by:
\[i(t) = I_{\max} (1 - e^{(-t/\tau)})\]

In this expression, \(i(t)\) is the current at any time \(t\), \(I_{\max}\) is the maximum current possible in the circuit, and \(e\) is the base of the natural logarithm. As shown in the solution steps, to find the time \(t\) when the current reaches a specific percentage of \(I_{\max}\), we need to work with this relationship and apply understanding of exponential decay. This process heavily relies on the time constant \(\tau\) and involves taking the natural logarithm to isolate \(\tau\) for further calculations.

Calculating Inductance

Once the time constant (\(\tau\)) of an RL circuit has been established, calculating inductance is a straightforward task. Inductance quantifies the opposition of an inductor to changes in current and is a fundamental aspect of RL circuits. Using the time constant formula,
\[L = \tau \cdot R\]
and substituting the previously determined values of \(\tau\) and the resistance \(R\), the inductance \(L\) can be calculated, as demonstrated in the given exercise.

It's important to note that the unit of inductance is the henry (H), which accounts for the conversion factors when using time in milliseconds (\(ms\)) or resistance in ohms. In the exercise, we first converted milliseconds to seconds before multiplying by the resistance to get the correct unit of inductance. The concept of inductance is pivotal in understanding how energy is stored in a magnetic field around a coil when a current flows through it. Hence, the ability to calculate inductance is critical for designing circuits that rely on controlled magnetic interactions.