Question

An inductor with an inductance of 2.50 H and a resistance of 8.00 \(\Omega\) is connected to the terminals of a battery with an emf of 6.00 V and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is 0.500 A; (c) the current 0.250 s after the circuit is closed; (d) the final steady-state current.

Short answer

(a) 2.4 A/s, (b) 0.8 A/s, (c) 0.464 A, (d) 0.75 A.

Step-by-step solution

Understanding the Circuit Configuration

The circuit consists of an inductor with inductance \(L = 2.50\, \text{H}\) and internal resistance \(R = 8.00\, \Omega\). It is connected to a battery with an emf \(\mathcal{E} = 6.00\, \text{V}\). We are required to find various current-related quantities in this RL circuit upon closure of the switch.

Finding the Initial Rate of Increase of Current

Initially, the current \(I = 0\) because the switch was just closed. The initial rate of increase of current is determined by the formula \(\left(\frac{di}{dt}\right)_{t=0} = \frac{\mathcal{E}}{L}\). Plugging in the values, we get \[\left(\frac{di}{dt}\right)_{t=0} = \frac{6.00}{2.50} \]. Calculate to find the initial rate of increase of current is \(2.4\, \text{A/s}\).

Calculating Rate of Increase of Current at 0.5 A

When the current is \(0.500\, \text{A}\), use the following formula: \(\frac{di}{dt} = \frac{\mathcal{E} - I \cdot R}{L}\). Substitute \(\mathcal{E} = 6.00\), \(I = 0.500\), \(R = 8.00\), and \(L = 2.50\) into the equation: \[\frac{di}{dt} = \frac{6.00 - 0.500 \times 8.00}{2.50} = \frac{2.00}{2.50}\]. Calculate to find the rate of increase \(\frac{di}{dt} = 0.8\, \text{A/s}\).

Current 0.25 Seconds After Circuit is Closed

The current as a function of time \(t\) in an RL circuit is \(I(t) = \frac{\mathcal{E}}{R}(1 - e^{-Rt/L})\). Plug in \(\mathcal{E} = 6.00\), \(R = 8.00\), \(L = 2.50\), and \(t = 0.25\) to get \[I(0.25) = \frac{6.00}{8.00}(1 - e^{-8.00 \cdot 0.25/2.50}) = 0.75(1 - e^{-0.8})\]. Calculate to find \(I(0.25) \approx 0.464\, \text{A}\).

Finding the Final Steady-State Current

In the steady state, the inductor acts as a short circuit, and the current \(I\) achieves a maximum value determined by Ohm's Law: \(I = \frac{\mathcal{E}}{R}\). Substitute \(\mathcal{E} = 6.00\) and \(R = 8.00\) into the equation: \[I = \frac{6.00}{8.00} = 0.75\, \text{A}\]. The final steady-state current is \(0.75\, \text{A}\).

Key concepts

Inductance

Inductance is a fundamental concept in circuit analysis, especially within RL circuits, which are circuits containing resistors (R) and inductors (L). An inductor, typically a coil of wire, has the property of inductance measured in henrys (H). This property is crucial because it opposes changes in current flow through the circuit.
The greater the inductance, the more it opposes sudden changes in current. In our example circuit, the inductance is 2.50 H. This relatively high inductance means that when the circuit is first powered, it will resist the initial increase of current significantly. Understanding inductance is key to explaining why the current doesn't reach its maximum instantaneously when a voltage is applied.
The energy storage aspect of inductance is unique. It stores energy in a magnetic field when the current flows through the inductor. This magnetic field can, in turn, affect the current and voltage in the circuit by maintaining the flow even when the power source is removed for a while. Understanding these interactions is essential for mastering RL circuit analysis.

Initial Rate of Current Increase

The initial rate of current increase in an RL circuit is a crucial parameter that helps us to understand how quickly the current starts to rise just after the switch is closed.
When a voltage is first applied to an RL circuit, there is initially no current because inductors resist changes in current. At this moment, the voltage across the inductor is maximum, and the current begins to increase.
Using the formula \[\left(\frac{di}{dt}\right)_{t=0} = \frac{\mathcal{E}}{L} \]where \(\mathcal{E}\) is the electromotive force (emf), and \(L\) is the inductance, we determine how fast the current is initially rising. In our example, with \(\mathcal{E} = 6.00\) V and \(L = 2.50\) H, the initial rate is calculated to be 2.4 A/s.
This value shows that the current increase is dependent on the emf and inversely on the inductance. The larger the inductance, the slower the initial rate of current increase.

Steady-State Current

Steady-state current refers to the current level in an RL circuit once it has stabilized or reached equilibrium after a certain period, as the effects of the inductor become negligible. In this condition, the inductor behaves like a short circuit.
In steady-state, the current is determined solely by the resistance of the circuit and the applied voltage. According to Ohm's Law, the steady-state current \(I\) can be expressed as: \[ I = \frac{\mathcal{E}}{R} \]where \(\mathcal{E}\) is the applied voltage and \(R\) is the resistance. In the given example, with \(\mathcal{E} = 6.00\) V and \(R = 8.00\) Ω, the steady-state current is found to be 0.75 A.
The fact that this value is solely dependent on resistance and voltage highlights how the inductor's effects diminish as time progresses and perfectly illustrates the role resistance plays in limiting current in circuits.

Ohm's Law

Ohm's Law is a fundamental principle in electronics and electrical engineering. It describes the relationship between voltage, current, and resistance in an electrical circuit. The formula for Ohm's Law is: \[ V = I \cdot R \]where \(V\) is the voltage across the circuit, \(I\) is the current flowing through it, and \(R\) is the resistance.
This principle allows us to calculate one of these values if the other two are known. In the context of RL circuits, Ohm's Law becomes invaluable in determining the steady-state current.
For the given exercise, the steady-state current is calculated using \[ I = \frac{\mathcal{E}}{R} \]showing its direct application. This is a rearrangement of Ohm's Law, indicating how crucial it is for solving not only steady state but also various other conditions in electrical circuits.