Chapter 30: Problem 23
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
Step by step solution
Understanding the Circuit Configuration
Finding the Initial Rate of Increase of Current
Calculating Rate of Increase of Current at 0.5 A
Current 0.25 Seconds After Circuit is Closed
Finding the Final Steady-State Current
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inductance
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
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
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
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.