Problem 23 - University Physics with Modern Physics

Question

An inductor with an inductance of 2.50 \(\mathrm{H}\) and a resistance of 8.00\(\Omega\) is connected to the terminals of a battery with an emf of 6.00 \(\mathrm{V}\) 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 \(0.500 \mathrm{A} ;\) (c) the current 0.250 \(\mathrm{s}\) after the circuit is closed; (d) the final steady-state current.

Step-by-Step Solution

Answer

(a) 2.4 A/s; (b) 0.8 A/s; (c) 0.145 A; (d) 0.750 A.

Step 1: Understanding the circuit

This is an RL circuit with an inductor (L = 2.50 H) and a resistor (R = 8.00 \(\Omega\)) connected to a battery with emf of 6.00 V.

Step 2: Calculate the initial rate of increase of current

At the initial moment when the circuit is turned on, the current is zero. According to the formula \(\frac{di}{dt} = \frac{\varepsilon}{L}\), we have: \(\frac{di}{dt} = \frac{6.00}{2.50} = 2.4 \mathrm{A/s}\).

Step 3: Rate of increase of current when current is 0.5 A

Using the formula \(\frac{di}{dt} = \frac{\varepsilon - iR}{L}\), substitute \(\varepsilon = 6.00\), \(i = 0.500\), \(R = 8.00\), and \(L = 2.50\). \(\frac{di}{dt} = \frac{6.00 - 8.00 \cdot 0.500}{2.50} = 0.8 \mathrm{A/s}\).

Step 4: Calculate the current after 0.250 s

Using the formula \(i(t) = \frac{\varepsilon}{R}(1 - e^{-\frac{R}{L}t})\), substitute \(\varepsilon = 6.00\), \(R = 8.00\), \(L = 2.50\), and \(t = 0.250\). Calculate \(i(0.250) = \frac{6.00}{8.00}(1 - e^{-\frac{8.00}{2.50} \cdot 0.250}) = 0.145 \mathrm{A}\).

Step 5: Calculate the final steady-state current

At steady state, the current is given by Ohm's Law \(i = \frac{\varepsilon}{R}\). So, the steady-state current is \(i = \frac{6.00}{8.00} = 0.750 \mathrm{A}\).

Key Concepts

Inductance

Inductance is a fundamental concept in the study of RL circuits and plays a key role in understanding how currents change over time. Essentially, inductance is the property of an inductor, which is a coil of wire within a circuit, to resist changes in the current flowing through it. This resistance occurs because a change in current generates a change in magnetic flux, thereby inducing a voltage across the inductor that counteracts the current change. This is described by the formula:
\[ V = L \frac{di}{dt} \]
where \( V \) is the voltage induced, \( L \) is the inductance, and \( \frac{di}{dt} \) is the rate of change of current.
- Inductance allows us to predict how quickly a current can rise or fall in an RL circuit. - The greater the inductance, the slower the rate of change of current, making inductors crucial in controlling circuits where soft start or voltage surge protection is desired.
In our exercise, we have an inductance of 2.50 H. This means the inductor will moderate how swiftly the circuit reaches its steady-state current.

Rate of Current Change

The rate of current change is a vital parameter when analyzing the behavior of RL circuits, especially during switches like turning the circuit on or off. It tells us how fast the current increases or decreases within the circuit over time. The expression for the initial rate of increase in an RL circuit, when just connected to a voltage source, is described by
\[ \frac{di}{dt} = \frac{\varepsilon}{L} \]
At the start when the switch is closed, there is no current, and this rate signifies how fast this current begins to flow.
In the discussed exercise:

• The initial rate of change was determined to be 2.4 A/s, meaning the current increases by this rate immediately upon closing the switch.
• When the current reaches 0.5 A, the rate of change diminishes to 0.8 A/s, as the voltage across the inductor breaks the uniformity of current flow according to \( \frac{di}{dt} = \frac{\varepsilon - iR}{L} \).

Understanding this helps in predicting how circuits--especially those starting with zero current--behave over time.

Steady-State Current

In an RL circuit, the steady-state current is the current that eventually flows once all transient effects have settled. This happens after the time constant, which is denoted by \( \tau = \frac{L}{R} \), has passed and the circuit reaches equilibrium. During this state, the effect of the inductor diminishes as it no longer opposes the current change.
- The steady-state current can be found using Ohm's Law: \( i = \frac{\varepsilon}{R} \).- It is typically the maximum current that will flow in the circuit when the inductor is fully magnetized and no longer affecting the current.
In our exercise, this current was calculated to be 0.750 A. This indicates that once the circuit stabilizes, this is the continuous current that will be present.

Ohm's Law

Ohm's Law is one of the fundamental principles used to analyze electrical circuits, providing a simple relationship between voltage, current, and resistance. In the context of our RL circuit, Ohm's Law is essential for calculating both the instantaneous and steady-state currents. The law states that
\[ V = IR \]
where \( V \) is the voltage across the circuit, \( I \) is the current, and \( R \) is the resistance. In scenarios where an emf, akin to a battery, is applied:

• The voltage supplied by the battery sets the pace at which current flows through the resistor, once the transient period is over.
• Ohm's Law directly gives us the steady-state current: \( i = \frac{\varepsilon}{R} \) as mentioned earlier.

This law provides a foundation for circuit analysis by simplifying the understanding of how each component in an electrical circuit affects the flow of electric current. In our specific exercise, it helped in determining that the final steady-state current would reach 0.750 A.