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An \(R L\) circuit with \(L=3.00 \mathrm{H}\) and an \(R C\) circuit with \(C=\) \(3.00 \mu \mathrm{F}\) have the same time constant. If the two circuits have the same resistance \(R\), (a) what is the value of \(R\) and (b) what is this common time constant?

Short Answer

Expert verified
The value of \( R \) is 1000 \( \Omega \) and the common time constant is 0.003 s.

Step by step solution

01

Formulate the equation

Since the time constants are equal, we write: \(\frac{L}{R} = RC\). The unknown here is \(R\), the common resistance. So we can rearrange to solve for \(R\). From the exercise, we know that \(L = 3H\) and \(C = 3 \times 10^{-6} F\).
02

Solve for R

When rearranging for \(R\), the equation becomes \(R = \sqrt{\frac{L}{C}}\). So, we get \(R = \sqrt{\frac{3}{3 \times 10^{-6}}} \, \Omega \), which simplifies to \(R = 1000 \, \Omega \).
03

Calculate the time constant

The time constant in both cases is found by using the equation for the time constant for either circuit type. Using the RL circuit time constant equation as an example, we find that the time constant \(\tau\) is \( \tau = \frac{L}{R} = \frac{3}{1000} = 0.003 \, s\).

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Key Concepts

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

Time Constant
The time constant is a crucial concept in analyzing both RL (resistor-inductor) and RC (resistor-capacitor) circuits. It helps us understand how quickly these circuits respond to changes, such as the introduction of a voltage change.

For an RL circuit, the time constant, denoted as \(\tau\), is defined by the equation \(\tau = \frac{L}{R}\), where \(L\) is the inductance and \(R\) is the resistance. In an RC circuit, the time constant is \(\tau = RC\), where \(C\) represents the capacitance. Essentially, the time constant tells us how fast the current in an RL circuit or the voltage across a capacitor in an RC circuit will reach approximately 63.2% of its final value after a change has occurred.

Understanding the time constant helps in predicting how circuits will behave in real-world applications, such as in filtering, where the speed of response to voltage changes is critical. It provides a time frame to know when circuits are settling into their stable conditions.
Resistance Calculation
Calculating resistance is an essential aspect when working with RL and RC circuits. The resistance in a circuit determines how fast the circuit reaches its steady state, either building up current in an RL circuit or charging/discharging a capacitor in an RC circuit.

In our problem, both circuits share a common resistance \(R\). Given that both circuits have the same time constant, we can use the relationship \(\frac{L}{R} = RC\) to find \(R\). By rearranging for \(R\), we derive \(R = \sqrt{\frac{L}{C}}\). With the provided values of \(L = 3H\) and \(C = 3 \times 10^{-6} F\), our calculation simplifies to \(R = 1000 \Omega\).

The correct calculation and understanding of resistance are crucial for designing circuits with desired characteristics. It ensures the circuits not only function but do so in an efficient and controlled manner.
Inductor and Capacitor Properties
Inductors and capacitors are fundamental components in electrical circuits, each having unique properties and tasks.
  • Inductors: Made typically of coils of wire, they store energy in a magnetic field when electrical current passes through them. Inductance, measured in henries (H), describes how effectively the inductor stores energy. This property plays a key role in determining the time constant in RL circuits. More inductance means a longer time constant, meaning it takes longer for current to build up to its final value.
  • Capacitors: Made of two conductive plates separated by an insulating material, they store energy in an electric field. Capacitance, measured in farads (F), indicates how much electric charge the capacitor can hold at a given voltage. In RC circuits, capacitance is directly proportional to the time constant. The greater the capacitance, the slower the circuit charges or discharges.
Understanding these properties gives insight into circuit behavior, such as what happens when either the inductor or capacitor is changed. This knowledge is vital for designing circuits for filters, oscillators, and transformers, where precise timing and voltage control are necessary.

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Most popular questions from this chapter

An aluminum ring of radius \(5.00 \mathrm{~cm}\) and resistance \(3.00 \times 10^{-4} \Omega\) is placed around the top of a long air-core solenoid with 1000 turns per meter and a smaller radius of \(3.00 \mathrm{~cm}\), as in Figure \(\mathrm{P} 20.64\). If the current in the sole noid is increasing at a constant rate of \(270 \mathrm{~A} / \mathrm{s}\), what is the induced current in the ring? Assume the magnetic field produced by the solenoid over the area at the end of the solenoid is one-half as strong as the field at the center of the solenoid. Assume also the solenoid produces a negligible field outside its cross-sectional area.

A uniform magnetic field of magnitude \(0.50 \mathrm{~T}\) is directed perpendicular to the plane of a rectangular loop having dimensions \(8.0 \mathrm{~cm}\) by \(12 \mathrm{~cm}\). Find the magnetic flux through the loop.

The rolling axle, \(1.50 \mathrm{~m}\) long, is pushed along horizontal rails at a constant speed \(v=3.00 \mathrm{~m} / \mathrm{s}\). A resistor \(R=0.400 \Omega\) is connected to the rails at points \(a\) and \(b_{1}\) directly opposite each other. (The wheels make good electrical contact with the rails, so the axle, rails. and \(R\) form a closed-loop circuit. The only significant resistance in the circuit is \(R\).) A uniform magnetic field \(B=0.800 \mathrm{~T}\) is directed vertically downward. (a) Find the induced current \(I\) in the resistor. (b) What horizontal force \(\overrightarrow{\mathbf{F}}\) is required to keep the axle rolling at constant speed? (c) Which end of the resistor, \(a\) or \(b\), is at the higher electric potential? (d) After the axle rolls past the resistor, does the current in \(R\) reverse direction?

A 300 -turn solenoid with a length of \(20.0 \mathrm{~cm}\) and a radius of \(1.50 \mathrm{~cm}\) carries a current of \(2.00 \mathrm{~A}\). A second coil of four turns is wrapped tightly around this solenoid, so it can be considered to have the same radius as the solenoid. The current in the 300 -turn solenoid increases steadily to \(5.00 \mathrm{~A}\) in \(0.900 \mathrm{~s}\). (a) Use Ampere's law to calculate the initial magnetic field in the middle of the 300 -turn solenoid. (b) Calculate the magnetic field of the 300 -turn solenoid after \(0.900\) s. (c) Calculate the area of the 4 -turn coil. (d) Calculate the change in the magnetic flux through the 4 -turn coil during the same period. (c) Calculate the average induced emf in the 4 -turn coil. Is it equal to the instantaneous induced emf? Explain. (f) Why could contributions to the magnetic field by the current in the 4 -turn coil be neglected in this calculation?

A 500-turn circular-loop coil \(15.0 \mathrm{~cm}\) in diameter is initially aligned so that its axis is parallel to Earth's magnetic field. In \(2.77 \mathrm{~ms}\) the coil is flipped so that its axis is perpendicular to Earth's magnetic field. If an average voltage of \(0.166 \mathrm{~V}\) is thereby induced in the coil, what is the value of the Earth's magnetic field at that location?

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