Question

What is the inductance in a series \(\mathrm{RL}\) circuit in which \(R=3.00 \mathrm{k} \Omega\) if the current increases to \(\frac{1}{2}\) of its final value in \(20.0 \mu \mathrm{s} ?\)

Short answer

Answer: The inductance of the RL circuit is approximately \(4.14 \times 10^{-5} H\).

Step-by-step solution

Write down the given values

We have been given: Resistance (R) = \(3.00 k\Omega = 3000 \Omega\) Time (t) = \(20.0 \mu s = 20 \times 10^{-6} s\)

Write down the formula for the current in an RL circuit

The formula for the current (i) in an RL circuit at some given time (t) is: \[i = I (1 - e^{-\frac{tR}{L}})\] where I is the final current value, R is the resistance, L is the inductance and t is the time.

Plug in the given values and solve for L

We are given that when \(t = 20 \times 10^{-6}s\), the current reaches half of its final value, which means that \(i = \frac{1}{2}I\). So, we can rewrite the formula as follows: \[\frac{1}{2}I = I (1 - e^{-\frac{(20 \times 10^{-6})(3000)}{L}})\] Now, we can solve for L: 1. Divide both sides by I: \[\frac{1}{2} = 1 - e^{-\frac{(20 \times 10^{-6})(3000)}{L}}\] 2. Subtract \(1 - \frac{1}{2} = \frac{1}{2}\) from both sides: \[\frac{1}{2} = e^{-\frac{(20 \times 10^{-6})(3000)}{L}}\] 3. Take the natural logarithm of both sides: \[ln(\frac{1}{2}) = -\frac{(20 \times 10^{-6})(3000)}{L}\] 4. Multiply both sides by -\(L\): \[-L ln(\frac{1}{2}) = (20 \times 10^{-6})(3000)\] 5. Divide both sides by \(- ln(\frac{1}{2})\): \[L = \frac{(20 \times 10^{-6})(3000)}{- ln(\frac{1}{2})}\] 6. Use a calculator to find the value of L: \[L = \frac{(20 \times 10^{-6})(3000)}{- ln(\frac{1}{2})} \approx 4.14 \times 10^{-5} H\] So, the inductance (L) in the RL circuit is approximately \(4.14 \times 10^{-5} H\).

Key concepts

Understanding Series RL Circuits

In the realm of electronics, a series RL circuit is a fundamental configuration that combines a resistor (R) and an inductor (L) in a series connection. This setup is key in understanding how inductance behaves alongside resistance within an electrical circuit.

When a voltage is applied to a series RL circuit, the current that flows through it does not instantly reach its maximum value. Instead, the current gradually increases, and the rate of this increase is influenced by both the resistance and the inductance present in the circuit. The inductor, in particular, resists changes in current due to its property of inductance, which depends on the physical characteristics of the coil, such as the number of turns in the wire and the core material.

In our exercise, we explore a scenario in which the resistance is given as 3.00 kΩ and we are tasked with deducing the inductance from the time it takes for the current to reach half of its final value. Concepts such as inductive time constant play a pivotal role in unraveling such problems.

Time Constant in RL Circuits

The time constant of an RL circuit, typically denoted by the Greek letter τ (tau), is a measure of the time it takes for the current to rise to approximately 63.2% of its final value, or conversely, for the current flowing through an inductor to decrease to 36.8% of its initial value when the voltage supply is switched off. The time constant τ is defined as the product of the resistance R and the inductance L of the circuit: τ = L/R.

For the particular exercise at hand, we're not directly looking for the time constant, but we utilize its concept to solve for the inductance, given that the current reaches half of its final value within a specific time frame. By understanding the relationship between the time constant and the extent of current change, we can infer the inductance value knowing the resistance and the time. The mathematical approach employed subtly hinges on the natural logarithm function, which is deeply intertwined with the exponential behavior in RL circuits.

Exponential Decay in RL Circuits

Electrical circuits involving resistors and inductors exhibit exponential behavior, particularly visible during the charging and discharging processes of an inductor. Exponential decay is a phenomenon where the current or voltage within an RL circuit decreases over time at a rate proportional to its current value. This property manifests due to the inductor's opposition to changes in current, causing the current to change at a rate that slows down as time progresses.

The formula that embodies this behavior is characterized by an exponent that is negative and proportional to the time, indicating decay, and is captured by the term \(e^{-\frac{tR}{L}}}\). This expression represents the transient response of the circuit, wherein \(e\) is the base of the natural logarithm and the exponent includes the product of time \(t\), resistance \(R\), and the inverse of inductance \(L\).

In the solution we've provided, students can see this exponential relationship in action as we solve for inductance by equating the current at a specific time to a fraction of the final current value and using the natural logarithm to unravel the exponential term. This offers a practical glimpse into how exponential decay governs the dynamic response of series RL circuits.