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How long would it take, following the removal of the battery, for the potential difference across the resistor in an \(R L\) circuit (with \(L=2.00 \mathrm{H}, R=3.00 \Omega\) ) to decay to \(10.0 \%\) of its initial value?

Short Answer

Expert verified
The potential difference decays to 10% of its initial value in approximately 1.535 seconds.

Step by step solution

01

Understand the Problem Statement

We need to find how long it takes for the potential difference across a resistor in an \(RL\) circuit to reduce to 10% of its initial value after the battery is removed. Given are the inductance \(L = 2.00\, \text{H}\) and resistance \(R = 3.00\, \Omega\).
02

Recall the Formula for Decay in an RL Circuit

The potential difference across the resistor in an \(RL\) circuit follows the equation \(V(t) = V_0 e^{-\frac{R}{L}t}\), where \(V_0\) is the initial voltage. We want \(V(t) = 0.1 V_0\).
03

Substitute and Rearrange the Equation

Substitute \(V(t) = 0.1 V_0\) into the decay formula: \[0.1 V_0 = V_0 e^{-\frac{R}{L}t}\]. Simplifying, we have \[0.1 = e^{-\frac{R}{L}t}\].
04

Solve for Time \(t\)

Take the natural logarithm on both sides to solve for \(t\): \(-\ln(0.1) = \frac{R}{L}t\). Substituting \(R = 3.00\, \Omega\) and \(L = 2.00\, \text{H}\), we get \[t = \frac{-\ln(0.1) \cdot L}{R}\].
05

Calculate the Time

Calculate the time \(t\) using values: \(t = \frac{-\ln(0.1) \cdot 2.00}{3.00}\). The natural logarithm \(-\ln(0.1) = 2.302\), so \[t = \frac{2.302 \cdot 2.00}{3.00} = \frac{4.604}{3.00} \approx 1.535\, \text{seconds}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Difference
In the context of RL circuits, the potential difference refers to the voltage across the resistor. When a battery is connected to an RL circuit, it provides an initial voltage. Upon removal of the battery, the potential difference begins to decrease over time.
This reduction in the potential difference is due to the energy stored in the inductor, which attempts to keep the current flowing even after the power source is disconnected.
As a result, the voltage across the resistor decreases exponentially with time. This is crucial to understand because our goal here is to determine how quickly this potential difference drops to a certain percentage of its original value.
To sum up, the potential difference changes dynamically based on circuit conditions and time. Understanding this fluctuation is key in analyzing RL circuits.
Inductance
Inductance in an RL circuit is a property of the coil or inductor, symbolized by the letter 'L'. It is measured in Henrys (H) and it reflects the ability of the circuit to store energy in a magnetic field.
Inductance causes a delay in the change of current, which essentially influences how the potential difference evolves over time when the circuit is opened.
  • A higher inductance value means more energy storage capability, hence a slower decay of current.
  • In our exercise, the inductance is given as 2.00 H, indicating that the inductor is quite capable of storing energy.
Understandably, inductance plays a significant role in how quickly the potential difference decays, as it opposes changes in the current within the circuit.
Resistance
Resistance in an RL circuit is the opposition to the flow of current and is symbolized by 'R'. It is measured in Ohms (Ω). In our particular problem, the resistance is 3.00 Ω.
The resistor is pivotal because it determines how quickly the potential difference decreases after the battery is removed.
The interplay between resistance and inductance dictates the rate of decay of voltage in the circuit. Simply put:
  • A higher resistance results in a faster decay of potential difference.
  • A lower resistance leads to a slower decay.
Thus, in the decay formula, both resistance and inductance must be considered to evaluate the time required to reach a specified potential difference.
Exponential Decay
Exponential decay in an RL circuit describes how the potential difference decreases over time, following an exponential curve. This phenomenon is mathematically represented by the equation: \[V(t) = V_0 e^{-\frac{R}{L}t}\]
Here:
  • \(V(t)\) is the voltage at a time \(t\).
  • \(V_0\) is the initial voltage.
  • \(R\) is the resistance.
  • \(L\) is the inductance.
In our specific problem, the aim is to find out 't', where 90% of the voltage has decayed. We did this by transforming and solving the equation to isolate 't'.
Exponential decay happens because the current and potential difference decrease at a rate proportional to their current value, which is a hallmark of exponential functions. By understanding this decay, it is possible to predict how long it will take to reach a certain voltage level in an RL circuit.

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Most popular questions from this chapter

Figure \(30-56\) shows a copper strip of width \(W=16.0 \mathrm{~cm}\) that has been bent to form a shape that consists of a tube of radius \(R=1.8 \mathrm{~cm}\) plus two parallel flat extensions. Current \(i=35 \mathrm{~mA}\) is distributed uniformly across the width so that the tube is effectively a one-turn solenoid. Assume that the magnetic field outside the tube is negligible and the field inside the tube is uniform. What are (a) the magnetic field magnitude inside the tube and (b) the inductance of the tube (excluding the flat extensions)?

A solenoid having an inductance of \(6.30 \mu \mathrm{H}\) is connected in series with a \(1.20 \mathrm{k} \Omega\) resistor. (a) If a \(14.0 \mathrm{~V}\) battery is connected across the pair, how long will it take for the current through the resistor to reach \(80.0 \%\) of its final value? (b) What is the current through the resistor at time \(t=1.0 \tau_{L} ?\)

Coil 1 has \(L_{1}=25 \mathrm{mH}\) and \(N_{1}=100\) turns. Coil 2 has \(L_{2}=40\) \(\mathrm{mH}\) and \(N_{2}=200\) turns. The coils are fixed in place; their mutual inductance \(M\) is \(3.0 \mathrm{mH}\). A \(6.0 \mathrm{~mA}\) current in coil 1 is changing at the rate of \(4.0 \mathrm{~A} / \mathrm{s}\). (a) What magnetic flux \(\Phi_{12}\) links coil 1, and \((\mathrm{b})\) what self-induced emf appears in that coil? (c) What magnetic flux \(\Phi_{21}\) links coil 2 , and \((\mathrm{d})\) what mutually induced emf appears in that coil?

A coil with an inductance of \(2.0 \mathrm{H}\) and a resistance of \(10 \Omega\) is suddenly connected to an ideal battery with \(\mathscr{E}=100 \mathrm{~V}\). (a) What is the equilibrium current? (b) How much energy is stored in the magnetic field when this current exists in the coil?

The inductance of a closely packed coil of 400 turns is \(8.0\) \(\mathrm{mH}\). Calculate the magnetic flux through the coil when the current is \(5.0 \mathrm{~mA}\).

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