Question

A coil of inductance \(300 \mathrm{mH}\) and resistance \(2 \Omega\) is connected to a source of voltage \(2 \mathrm{~V}\). The current reaches half of its steady state value in \(\ldots \ldots\) (a) \(0.15 \mathrm{sec}\) (b) \(0.3 \mathrm{sec}\) (c) \(0.05 \mathrm{sec}\) (d) \(0.1 \mathrm{sec}\)

Short answer

(d) \(0.1 \mathrm{sec}\)

Step-by-step solution

Calculate the Time Constant τ

To calculate the time constant τ (tau) of the RL circuit, we will use the formula τ = L/R, where L is the inductance (300 mH) and R is the resistance (2 Ω). In order to calculate the time constant in seconds, we need to convert the inductance to henries (H). \( L = 300 mH × \frac{1 H}{1000 mH} = 0.3 H \) Now, let's calculate the time constant τ: \( τ = \frac{L}{R} = \frac{0.3 H}{2 Ω} = 0.15 s \)

Calculate the Steady-State Current

Next, we need to find the steady-state current using Ohm's law (I = V/R), where V is the voltage (2 V) and R is the resistance (2 Ω). \( I_{steady-state} = \frac{V}{R} = \frac{2 V}{2 Ω} = 1 A \)

Find the Time it Takes to Reach Half of the Steady-State Current

Now, our goal is to find the time it takes for the current to reach half of the steady-state value: \( 0.5 × I_{steady-state} = I_{steady-state} × (1 - e^{\frac{-t}{τ}}) \) Dividing both sides by the steady-state current: \( 0.5 = 1 - e^{\frac{-t}{τ}} \) Now solve for t: \( 0.5 = e^{\frac{-t}{τ}} \) Applying the natural logarithm to both sides: \( -ln(2) = \frac{-t}{τ} \) Multiplying both sides by τ: \( -τ ln(2) = t \) Now we can substitute τ (0.15 s) to find the time (t): \( t = -(0.15 s) × ln(2) ≈ 0.104 s \) Since the time is closest to 0.1 s, the correct answer is: (d) \(0.1 \mathrm{sec}\)

Key concepts

Time Constant (τ)

The time constant, represented by the Greek letter \(\tau\), is a crucial factor in analyzing how quickly an RL circuit responds when subjected to a voltage source. In essence, it dictates the time taken for the current in the circuit to rise to approximately 63.2% of its final steady state value after a sudden change in voltage.
In the context of an RL circuit, the time constant \(\tau\) is determined by the inductance \(L\) and the resistance \(R\) using the simple formula:

• \(\tau = \frac{L}{R}\)

Here, \(L\) is measured in henries (H) and \(R\) in ohms (Ω). The resulting time constant is displayed in seconds, providing an idea of how quickly the circuit reaches its stable functioning state. In the given exercise, for instance, an inductance of \(300 \, \mathrm{mH}\) and a resistance of \(2 \, \Omega\) yield a \(\tau\) of \(0.15 \, \mathrm{s}\). This means it would take 0.15 seconds for the circuit to respond significantly to a voltage change initiated at time zero.

Inductance and Resistance

Inductance and resistance are fundamental components of electrical circuits, each contributing to the overall behavior of the system in distinctive ways.
**Inductance (L):** This property, measured in henries (H), is associated with the coil or inductor in the circuit. Inductors store energy in a magnetic field when electrical current flows through them. The ability of an inductor to store this energy is what we refer to as inductance.

• When inductance is high, the circuit's response to changes in current is slower.
• The higher the inductance, the longer it takes for a circuit to reach its steady-state current value.

**Resistance (R):** Resistance, measured in ohms (Ω), refers to the opposition offered by a material to the flow of electric current. Every conductive material has some inherent resistance, expressed as a factor slowing down electron motion through the circuit.
In the exercise, the circuit combines a resistance \(R = 2 \, \Omega\) with an inductance \(L = 0.3 \, \mathrm{H}\) to exhibit behavior characterized by the calculated time constant. Understanding the individual roles of inductance and resistance helps students see how quickly or slowly a circuit can ramp up to its full operating capacity.

Exponential Growth/Decay in Circuits

In RL circuits, the process of current increase or decrease over time is captured by exponential functions. These functions describe how the current approaches its steady-state value.
**Exponential Growth:** When a voltage is applied to an RL circuit initially carrying no current, the current begins to grow exponentially. The governing equation is:

• \(I(t) = I_{steady-state} \times \left(1 - e^{-\frac{t}{\tau}}\right)\)

**Exponential Decay:** If the power source is removed, the current decreases exponentially:

• \(I(t) = I_{initial} \times e^{-\frac{t}{\tau}}\)

The constant \(\tau\) features prominently in both processes, illustrating its control over the circuit's pace of change. This behavior is crucial in timing applications or circuits where gradual charge accumulation or release is required.
Applying this concept to the exercise, we used the exponential growth formula to solve for the time when the current reaches half of its steady-state value. By understanding and manipulating the exponential equations, students can predict how the circuit will behave under varying conditions.