Problem 52 - University Physics with Modern Physics

Question

An inductor is connected to the terminals of a battery that has an emf of \(12.0 \mathrm{~V}\) and negligible internal resistance. The current is \(4.86 \mathrm{~mA}\) at \(0.725 \mathrm{~ms}\) after the connection is completed. After a long time the current is \(6.45 \mathrm{~mA}\). What are (a) the resistance \(R\) of the inductor and (b) the inductance \(L\) of the inductor?

Step-by-Step Solution

Answer

(a) \(R \approx 1860.47 \, \mathrm{\Omega}\); (b) \(L \approx 0.966 \, \mathrm{H}\).

Step 1: Understand the RL Circuit

We have an RL circuit where a resistor and an inductor are in series with a battery. After a long time, the circuit reaches a steady state where the inductor behaves like a regular wire (zero inductive reactance). We need to find the resistance \(R\) and the inductance \(L\) of the inductor.

Step 2: Find the Resistance \( R \)

At steady state, the current through the circuit is \(6.45 \, \mathrm{mA}\). Using Ohm's Law \( V = IR \), where \(V = 12.0 \, \mathrm{V}\), we calculate:\[R = \frac{V}{I} = \frac{12.0}{6.45 \times 10^{-3}} \approx 1860.47 \, \mathrm{\Omega}\]

Step 3: Use the RL Circuit Current Equation

In an RL circuit, the current as a function of time \(t\) is given by:\[ I(t) = I_0 (1 - e^{-t/\tau}) \]where \( \tau = \frac{L}{R} \) is the time constant and \( I_0 \) is the final steady-state current. Here, \(I_0 = 6.45 \, \mathrm{mA} \) and \(I(0.725 \, \mathrm{ms}) = 4.86 \, \mathrm{mA}\).

Step 4: Solve for the Time Constant \( \tau \)

Rearrange the equation for current to solve for \( \tau \):\[ 4.86 = 6.45 (1 - e^{-0.725/\tau}) \, \mathrm{mA} \]\[ e^{-0.725/\tau} = 1 - \frac{4.86}{6.45} \]\[ e^{-0.725/\tau} = 0.2477 \]Take the natural logarithm of both sides and solve for \( \tau \):\[ -0.725/\tau = \ln(0.2477) \]\[ \tau \approx \frac{0.725}{1.3956} \approx 0.519 \times 10^{-3} \, \mathrm{s} \]

Step 5: Find the Inductance \( L \)

Using the relation \( \tau = \frac{L}{R} \) and knowing \( \tau \approx 0.519 \times 10^{-3} \, \mathrm{s} \) and \( R \approx 1860.47 \, \mathrm{\Omega}\):\[ L = R \times \tau = 1860.47 \, \mathrm{\Omega} \times 0.519 \times 10^{-3} \, \mathrm{s} \approx 0.966 \, \, \mathrm{H} \]

Step 6: Final Step: Results

The resistance \( R \) of the inductor is approximately \(1860.47 \, \mathrm{\Omega}\), and the inductance \( L \) is approximately \(0.966 \, \mathrm{H}\).

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle within electrical engineering, crucial for understanding RL circuits. It states that the voltage (\( V \)) across a conductor is directly proportional to the current (\( I \)) flowing through it, with the resistance (\( R \)) being the constant of proportionality. This relationship is expressed in the equation:
\[ V = IR \]
In the context of an RL circuit, like the one in the exercise, Ohm's Law helps determine the resistance of a circuit element. Once the inductor reaches a steady state, it behaves like a regular wire, allowing us to use Ohm's Law straightforwardly.

• For example, with a voltage of 12.0 V and a final steady-state current of 6.45 mA (or 0.00645 A), the resistance can be computed using \( R = \frac{V}{I} \), giving us approximately 1860.47 Ω for the inductor's resistance.

Inductance

Inductance is a property of an electrical circuit that opposes changes in current flow. It is a measure of the tendency of an inductor to resist changes in current and is symbolized by \( L \). This opposition is due to the electromagnetic field generated around the inductor when current passes through it.
The unit of inductance is the Henry (\( ext{H} \)). High inductance means a strong ability to resist changes in current, whereas low inductance indicates less resistance to change.

• In an RL circuit, this property is essential to understand how the current increases over time until it reaches a steady state.
• Calculating inductance can be achieved using the time constant (\( \tau \)) derived from the relationship \( \tau = \frac{L}{R} \). In the exercise, once the time constant is known (approximated as 0.519 ms), and with the resistance \( R \) known, the inductance \( L \) can be calculated to be around 0.966 H.

Time Constant

The time constant, symbolized by \( \tau \), is a critical parameter in understanding RL circuits. It represents the time required for the current to reach approximately 63.2% of its final steady-state value after the circuit is activated.
Given by the formula \( \tau = \frac{L}{R} \), the time constant helps determine how quickly the circuit responds to changes. Here, \( L \) is the inductance, and \( R \) is the resistance.

• For example, in the given exercise, with the time constant calculated as 0.519 ms, it provides insight into how the current ramps up from 0 to the steady-state value.
• A larger time constant indicates a slower response to changes in current, while a smaller \( \tau \) would mean the circuit reaches its steady state faster.
• This concept is particularly useful in timing applications and helps design circuits to achieve desired response times.