Problem 57 - College 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) \( R \approx 1860.47 \ \Omega \); (b) \( L \approx 1.769 \ H \).

Step 1: Understand the RL Circuit

An RL circuit consists of a resistor \( R \) and an inductor \( L \) in series with a voltage source. The current \( I(t) \) in the circuit increases over time according to the formula: \[ I(t) = I_0 (1 - e^{-t/\tau}) \] where \( I_0 \) is the steady-state current and \( \tau = \frac{L}{R} \) is the time constant of the circuit. Our goals are to solve for \( R \) and \( L \).

Step 2: Determine the Steady-State Current \( I_0 \)

The steady-state current \( I_0 \) occurs when the circuit has been connected for a long time and is given by Ohm's Law: \[ I_0 = \frac{E}{R} \] where \( E \) is the emf of the battery. Given that \( I_0 = 6.45 \ mA = 0.00645 \ A \) and \( E = 12 \ V \), solve for \( R \).

Step 3: Calculate Resistance \( R \)

From the equation \( I_0 = \frac{E}{R} \), we rearrange to solve for \( R \):\[ R = \frac{E}{I_0} = \frac{12}{0.00645} \approx 1860.47 \ \Omega \]. Thus, the resistance \( R \) is approximately \( 1860.47 \ \Omega \).

Step 4: Solve for Inductance \( L \)

Using the transient current expression, \[ I(t) = I_0 (1 - e^{-t/\tau}) \],plug in the values for \( I(t) = 4.86\ mA = 0.00486 \ A \), \( I_0 = 0.00645 \ A \), and \( t = 0.725 \ ms = 0.000725 \ s \). Solve for \( \tau = \frac{L}{R} \).

Step 5: Substitute Known Values and Rearrange

Substitute in the formula: \[ 0.00486 = 0.00645 (1 - e^{-0.000725/\tau}) \]. Rearrange and solve for \( \tau \): \[ e^{-0.000725/\tau} = 1 - \frac{0.00486}{0.00645} \approx 0.2465 \] \[ -\frac{0.000725}{\tau} = \ln(0.2465) \] \[ \tau \approx \frac{0.000725}{\ln(1/0.2465)} \approx 0.0009502 \ s \].

Step 6: Find the Inductance \( L \)

Now that we have \( \tau \), use the relation \( \tau = \frac{L}{R} \) and our known value of \( R \): \[ L = R \times \tau \approx 1860.47 \times 0.0009502 \approx 1.769 \ H \]. Therefore, the inductance \( L \) is approximately \( 1.769 \ H \).

Key Concepts

Ohm's Law

Ohm's Law is fundamental in understanding electric circuits. It states that the current (\( I \)) through a conductor between two points is directly proportional to the voltage (\( V \)) across the two points. This is mathematically given by the formula:\[I = \frac{V}{R}\]where \( V \) is the voltage measured in volts, \( I \) is the current measured in amperes, and \( R \) is the resistance measured in ohms (\( \Omega \)).In the context of an RL circuit, Ohm's Law helps determine the steady-state current, \( I_0 \), which is the current flowing through the circuit once all changes settle down after the switch has been closed for a long time. Given an electromotive force (emf) of the battery (\( E \)), the steady state current can be calculated by rearranging Ohm's Law as follows:\[I_0 = \frac{E}{R}\]This equation is particularly useful in establishing the resistance of the inductor when the system is at steady state.

Time constant in circuits

The time constant of a circuit, denoted by \( \tau \), is a crucial parameter in analyzing how quickly a circuit responds to changes, e.g., in an RL or RC circuit. In an RL circuit, \( \tau \) is determined by the ratio of inductance (\( L \)) to resistance (\( R \)): \[\tau = \frac{L}{R}\]This constant tells you how fast the current reaches a particular fraction of its final value after the connection is closed. Specifically, in an RL circuit:

• The current reaches about 63.2% of its final steady value after one time constant \( \tau \).
• After five times \( \tau \), the current is considered to have practically reached its steady-state value.

The time constant is essential because it reflects how the inductance and resistance interact to affect the speed of the response to changes in voltage. A small time constant means the circuit responds quickly, while a larger time constant results in a slower response.

Inductance and resistance calculations

Calculating both the inductance (\( L \)) and resistance (\( R \)) of an RL circuit is crucial for understanding its behavior. In this type of circuit analysis, we use both the transient and steady-state characteristics.For the resistance part, once the steady-state current (\( I_0 \)) is known, it can be calculated using Ohm's Law derived formula:\[R = \frac{E}{I_0}\]where \( E \) is the emf of the voltage source.The transient response provides information on how the current changes over time. To find inductance (\( L \)), use the time constant\( \tau \) derived from the exponential growth formula:\[I(t) = I_0 \left(1 - e^{-t/\tau}\right)\]Substituting known values allows solving for \( \tau \), and subsequently for \( L \) using:\[L = R \times \tau\]By understanding these calculations, we gain a deeper insight into how RL circuits function and how they respond to electrical changes over time. This helps in designing circuits with desired dynamic properties.