Problem 48 - College Physics

Question

A \(12.0 \mathrm{~V}\) dc battery having no appreciable internal resistance, a \(150.0 \Omega\) resistor, an \(11.0 \mathrm{mH}\) inductor, and an open switch are all connected in series. After the switch is closed, what are (a) the time constant for this circuit, (b) the maximum current that flows through it, (c) the current \(73.3 \mu\) s after the switch is closed, and (d) the maximum energy stored in the inductor?

Step-by-Step Solution

Answer

(a) 73.3 μs, (b) 0.080 A, (c) 0.0506 A, (d) 0.0352 mJ

Step 1: Calculate the Time Constant

The time constant \( \tau \) for an RL circuit is calculated using the formula \( \tau = \frac{L}{R} \), where \( L \) is the inductance and \( R \) is the resistance.Given:\[ L = 11.0 \text{ mH} = 11.0 \times 10^{-3} \text{ H} \]\[ R = 150.0 \Omega \]Substituting these into the equation:\[ \tau = \frac{11.0 \times 10^{-3}}{150.0} \approx 7.33 \times 10^{-5} \text{ s} \]

Step 2: Determine the Maximum Current

The maximum current \( I_\text{max} \) is the current when the inductive effects have diminished. It is calculated using Ohm's Law: \( I_\text{max} = \frac{E}{R} \), where \( E \) is the battery voltage.Given:\[ E = 12.0 \text{ V} \]\[ R = 150.0 \Omega \]Substituting these into the equation:\[ I_\text{max} = \frac{12.0}{150.0} = 0.080 \text{ A} \]

Step 3: Calculate the Current After 73.3 µs

The current \( I(t) \) at time \( t \) can be found using the formula \( I(t) = I_\text{max} \left( 1 - e^{-\frac{t}{\tau}} \right) \).Given:\[ t = 73.3 \mu\text{s} = 73.3 \times 10^{-6} \text{s} \]\[ \tau = 7.33 \times 10^{-5} \text{s} \]\[ I_\text{max} = 0.080 \text{ A} \]Substituting these into the equation:\[ I(73.3 \times 10^{-6}) = 0.080 \left( 1 - e^{-\frac{73.3 \times 10^{-6}}{7.33 \times 10^{-5}}} \right) \]Calculating:\[ I(73.3 \times 10^{-6}) \approx 0.080 (1 - e^{-1}) \approx 0.080 \times 0.632 = 0.0506 \text{ A} \]

Step 4: Determine Maximum Energy Stored in the Inductor

The maximum energy \( U \) stored in an inductor is given by the formula \( U = \frac{1}{2} L I_\text{max}^2 \).Given:\[ L = 11.0 \times 10^{-3} \text{ H} \]\[ I_\text{max} = 0.080 \text{ A} \]Substituting into the equation:\[ U = \frac{1}{2} \times 11.0 \times 10^{-3} \times (0.080)^2 \]\[ U = 0.5 \times 11.0 \times 10^{-3} \times 0.0064 = 0.0352 \times 10^{-3} \text{ J} = 0.0352 \text{ mJ} \]

Key Concepts

Time Constant

The time constant in an RL circuit is a crucial parameter that tells us how quickly the circuit responds to changes in voltage. In this circuit, the time constant \( \tau \) is given by the formula \( \tau = \frac{L}{R} \), where \( L \) represents the inductance and \( R \) is the resistance. This time constant determines how swiftly the current can reach its maximum value after the switch is closed.
For this exercise, with \( L = 11.0 \text{ mH} = 11.0 \times 10^{-3} \text{ H} \) and \( R = 150.0 \Omega \), the calculation is as follows:
\[ \tau = \frac{11.0 \times 10^{-3}}{150.0} \approx 7.33 \times 10^{-5} \text{ s} \]
This means that it takes about 73.3 microseconds for the current to reach approximately 63.2% of its maximum value in this circuit. This parameter is essential for analyzing how quickly a circuit can respond to changes.

Maximum Current

In an RL circuit, the maximum current \( I_{\text{max}} \) is the current that eventually flows through the circuit after the effects of the inductor have subsided. The maximum current can be calculated using Ohm's Law, which states that \( I = \frac{V}{R} \), where \( V \) is the voltage and \( R \) is the resistance.
For this exercise:

• The voltage \( E = 12.0 \text{ V} \)
• The resistance \( R = 150.0 \Omega \)

Substituting these values:
\[ I_{\text{max}} = \frac{12.0}{150.0} = 0.080 \text{ A} \]
Thus, the maximum current that will flow through the circuit is 0.080 A, or 80 mA, after the inductor has become effectively transparent to steady DC current.

Energy in Inductor

Energy stored in an inductor is part of what makes RL circuits interesting. When current flows through the inductor, energy is stored in its magnetic field. The maximum energy \( U \) stored can be computed using the formula \( U = \frac{1}{2} L I_{\text{max}}^2 \), where \( L \) is the inductance and \( I_{\text{max}} \) is the maximum current.
For the present exercise:

• The inductance \( L = 11.0 \times 10^{-3} \text{ H} \)
• The maximum current \( I_{\text{max}} = 0.080 \text{ A} \)

Inserting these values into the equation:
\[ U = \frac{1}{2} \times 11.0 \times 10^{-3} \times (0.080)^2 \]
\[ U = 0.5 \times 11.0 \times 10^{-3} \times 0.0064 = 0.0352 \times 10^{-3} \text{ J} = 0.0352 \text{ mJ} \]
This represents the energy stored in the inductor when the current reaches its maximum.

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits, describing the relationship between voltage, current, and resistance. It asserts \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
In our circuit analysis, Ohm's Law was pivotal in determining the maximum current. By knowing the voltage across the circuit and its resistance, we derived the current that flows through the circuit using:

• Voltage \( E = 12.0 \text{ V} \)
• Resistance \( R = 150.0 \Omega \)

The calculation follows as:
\[ I_{\text{max}} = \frac{E}{R} = \frac{12.0}{150.0} = 0.080 \text{ A} \]
This foundational law enables us to connect the dots between how much electrical 'pressure' is applied to push the current through a resistor and how much of that current actually flows.