/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 (II) You want to turn on the cur... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) You want to turn on the current through a coil of self- inductance \(L\) in a controlled manner, so you place it in series with a resistor \(R = 2200 \Omega ,\) a switch, and a dc voltage source \(V _ { 0 } = 240 \mathrm { V }\) . After closing the switch, you find that the current through the coil builds up to its steady-state value with a time constant \(\tau .\) You are pleased with the current's steady-state value, but want \(\tau\) to be half as long. What new values should you use for \(R\) and \(V _ { 0 } ?\)

Short Answer

Expert verified
Use \( R' = 4400 \Omega \) and \( V_0' = 480 \text{ V} \).

Step by step solution

01

Identify Given Variables and Formulas

We are given: the initial resistance \( R = 2200 \Omega \), voltage \( V_0 = 240 \text{ V} \), and the need to halve the time constant \( \tau \). The time constant is given by \( \tau = \frac{L}{R} \). In steady state, \( I = \frac{V_0}{R} \).
02

Determine Initial Time Constant and Steady-State Current

Calculate the initial time constant: \( \tau = \frac{L}{R} = \frac{L}{2200} \). Determine the steady-state current: \( I = \frac{240}{2200} \approx 0.109 \text{ A} \).
03

Deduce Condition for Halving Time Constant

We want the new time constant \( \tau' = \frac{L}{R'} = \frac{1}{2}\frac{L}{R} \). Hence, \( R' = 2R = 2 \times 2200 = 4400 \Omega \).
04

Preserve Steady-State Current with New Voltage

To preserve the steady-state current \( I = 0.109 \text{ A} \) with the new resistance \( R' = 4400 \Omega \), solve for the new voltage \( V_0' \) using \( I = \frac{V_0'}{R'} \) which gives \( V_0' = I \times R' = 0.109 \times 4400 \approx 480 \text{ V} \).
05

Conclusion with New Values

The new resistance should be \( 4400 \Omega \) and the new voltage \( 480 \text{ V} \) to halve the time constant while maintaining the steady-state current.

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Key Concepts

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

Time Constant
In an RL circuit, the time constant, denoted by \( \tau \), is a crucial factor that influences how quickly the current reaches its maximum value. It is defined as the ratio between the inductance \( L \) and the resistance \( R \) in the circuit, given by the formula:\[ \tau = \frac{L}{R} \]The time constant represents the time it takes for the current to reach approximately 63.2% of its steady-state value. A smaller time constant means the system responds more quickly, while a larger time constant indicates a slower response. In the problem, we needed to halve \( \tau \), meaning the circuit should respond twice as fast. This change is accomplished by adjusting the resistance, since the inductance \( L \) remains constant. By doubling the resistance from \( 2200 \Omega \) to \( 4400 \Omega \), the new time constant \( \tau' \) becomes:\[ \tau' = \frac{L}{2R} \]
Steady-State Current
Steady-state current is the constant flow of electric charge that occurs after a circuit has reached equilibrium. Once the initial response has settled, and for long time after the switch is closed, the current remains steady, maintained solely by the voltage source overcoming the circuit's resistance. The steady-state current \( I \) in an RL circuit is calculated using Ohm's Law:\[ I = \frac{V_0}{R} \]For the given RL circuit, with resistance \( R = 2200 \Omega \) and voltage \( V_0 = 240 \text{ V} \), the steady-state current is approximately \( 0.109 \text{ A} \). Even with a change in resistance to \( 4400 \Omega \) to modify the time constant, it is important to keep the current constant since that matches the desired outcome, so the voltage must be increased to \( 480 \text{ V} \) to keep \( I \) unchanged according to \( I = \frac{480}{4400} \approx 0.109 \text{ A} \).
Resistance and Voltage
Resistance and voltage play fundamental roles in the behavior of an RL circuit. Resistance \( R \), measured in ohms \( \Omega \), is the opposition that the circuit offers to the flow of current. A higher resistance results in less current flow with the same voltage, and it directly affects the time constant \( \tau \) of the circuit.
  • Resistance: Doubled in this scenario to adjust the time constant. Increasing the resistance makes the circuit reach its steady-state faster.
  • Voltage: The driving force that pushes the current through the circuit. Must be adjusted to keep the steady-state current constant when the resistance changes.
In modifying the original circuit set-up, increasing \( R \) to \( 4400 \Omega \) required us to also adjust the voltage to \( 480 \text{ V} \), ensuring the steady-state current remains consistent with the original configuration. This interplay between resistance and voltage is a classic application of Ohm’s Law, showing the delicate balance needed to achieve specific electrical characteristics in a circuit.

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Most popular questions from this chapter

In a plasma globe, a hollow glass sphere is filled with low-pressure gas and a small spherical metal electrode is located at its center. Assume an ac voltage source of peak voltage \(V_{0}\) and frequency \(f\) is applied between the metal sphere and the ground, and that a person is touching the outer surface of the globe with a fingertip, whose approximate area is \(1.0 \mathrm{~cm}^{2}\). The equivalent circuit for this situation is shown in Fig. \(30-35,\) where \(R_{\mathrm{G}}\) and \(R_{\mathrm{p}}\) are the resistances of the gas and the person, respectively, and \(C\) is the capacitance formed by the gas, glass, and finger. (a) Determine \(C\) assuming it is a parallel-plate capacitor. The conductive gas and the person's fingertip form the opposing plates of area \(A=1.0 \mathrm{~cm}^{2}\). The plates are separated by glass (dielectric constant \(K=5.0\) ) of thickness \(d=2.0 \mathrm{~mm} .\) (b) In a typical plasma globe, \(f=12 \mathrm{kHz}\) Determine the reactance \(X_{C}\) of \(C\) at this frequency in \(\mathrm{M} \Omega\). (c) The voltage may be \(V_{0}=2500 \mathrm{~V}\). With this high voltage, the dielectric strength of the gas is exceeded and the gas becomes ionized. In this "plasma" state, the gas emits light ("sparks") and is highly conductive so that \(R_{\mathrm{G}} \ll X_{C}\). Assuming also that \(R_{\mathrm{p}} \ll X_{C},\) estimate the peak current that flows in the given circuit. Is this level of current dangerous? \((d)\) If the plasma globe operated at \(f=1.0 \mathrm{MHz},\) estimate the peak current that would flow in the given circuit. Is this level of current dangerous?

(III) Two tightly wound solenoids have the same length and circular cross- sectional area. But solenoid 1 uses wire that is 1.5 times as thick as solenoid \(2 .(a)\) What is the ratio of their inductances? (b) What is the ratio of their inductive time constants (assuming no other resistance in the circuits)?

A \(3.5-\mathrm{k} \Omega\) resistor in series with a \(440-\mathrm{mH}\) inductor is driven by an ac power supply. At what frequency is the impedance double that of the impedance at \(60 \mathrm{~Hz} ?\)

(a) For an underdamped \(L R C\) circuit, determine a formula for the energy \(U = U _ { E } + U _ { B }\) stored in the electric and magnetic fields as a function of time. Give answer in terms of the initial charge \(Q _ { 0 }\) on the capacitor. (b) Show how \(d U / d t\) is related to the rate energy is transformed in the resistor, \(I ^ { 2 } R .\)

(I) What is the reactance of a \(9.2 - \mu \mathrm { F }\) capacitor at a frequency of \(( a ) 60.0 \mathrm { Hz } , ( b ) 1.00 \mathrm { MHz }\) ?

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