/*! 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 83 A variable inductor can be place... [FREE SOLUTION] | 91影视

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Chapter 21: Problem 83

A variable inductor can be placed in series with a lightbulb to act as a dimmer. (a) What inductance would reduce the current through a \(100-\) W lightbulb to \(75 \%\) of its maximum value? Assume a \(120-\mathrm{V}\) rms, \(60-\mathrm{Hz}\) source. (b) Could a variable resistor be used in place of the variable inductor to reduce the current? Why is the inductor a much better choice for a dimmer?

Short Answer

Expert verified
Answer: The value of the inductance needed to reduce the current to 75% of its maximum value is 0.5304 H. A variable resistor can be used as a dimmer, but a variable inductor is a more energy-efficient and safer choice.

Step by step solution

01

Find the maximum current (I_max) through the lightbulb

We are given the power rating of the lightbulb (100 W) and the RMS voltage (120 V). Let's use the formula for power in an AC circuit to find the maximum current. Power (P) = V_rms * I_max Rearrange the formula for I_max: I_max = Power / V_rms Plug in the given values: I_max = 100 W / 120 V = 0.8333 A
02

Find the current when reduced to 75%

Now, we need to find the current when it's reduced to 75% of its maximum value. I_reduced = 0.75 * I_max = 0.75 * 0.8333 A = 0.625 A
03

Calculate the impedance needed to achieve the reduced current

We will now calculate the impedance needed to achieve the reduced current in the circuit. Using Ohm's law: V = I * Z Where V is the RMS voltage, I is the current, and Z is the impedance. Rearrange the formula for Z: Z = V / I Plug in the known values for the reduced current: Z = 120 V / 0.625 A = 192 惟
04

Find the required inductance (L)

We know the impedance required, and we are assuming that it's purely inductive (ignoring any resistance in the coil), so we will calculate the required inductance for the circuit at a frequency of 60 Hz. The formula for the inductive reactance (X_L) is: X_L = 2 * pi * f * L Where f is the frequency, and L is the inductance. To find the inductance, we will rearrange the formula: L = X_L / (2 * pi * f) Since we found that the impedance (Z) required is 192 惟 and we assume that it's purely inductive, X_L = Z: L = 192 惟 / (2 * pi * 60 Hz) = 0.5304 H (a) The inductance needed to reduce the current to 75% of its maximum value is 0.5304 H.
05

Part (b): Comparing variable resistor and variable inductor as a dimmer

A variable resistor could also be used to reduce the current in the circuit. However, the inductor is a more efficient choice than a resistor as a dimmer, mainly for two reasons: 1. A resistor dissipates the energy in the form of heat as it reduces the current flow. This could make the dimmer quite hot and waste energy. 2. An inductor, on the other hand, stores energy in its magnetic field when it restricts the flow of current. It releases that energy back into the circuit when the current is allowed to increase. This makes the inductor-based dimmer more energy-efficient and safer than a resistor-based dimmer. In conclusion, while a variable resistor could potentially be used as a dimmer, a variable inductor is a better choice due to its energy efficiency and safety.

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Most popular questions from this chapter

A variable capacitor is connected in series to an inductor with negligible internal resistance and of inductance \(2.4 \times 10^{-4} \mathrm{H} .\) The combination is used as a tuner for a radio. If the lowest frequency to be tuned in is \(0.52 \mathrm{MHz}\). what is the maximum capacitance required?
An RLC series circuit is driven by a sinusoidal emf at the circuit's resonant frequency. (a) What is the phase difference between the voltages across the capacitor and inductor? [Hint: since they are in series, the same current \(i(t) \text { flows through them. }]\) (b) At resonance, the rms current in the circuit is \(120 \mathrm{mA}\). The resistance in the circuit is \(20 \Omega .\) What is the rms value of the applied emf? (c) If the frequency of the emf is changed without changing its rms value, what happens to the rms current? (W) tutorial: resonance)
An ac series circuit containing a capacitor, inductor, and resistance is found to have a current of amplitude \(0.50 \mathrm{A}\) for a source voltage of amplitude \(10.0 \mathrm{V}\) at an angular frequency of $200.0 \mathrm{rad} / \mathrm{s} .\( The total resistance in the circuit is \)15.0 \Omega$ (a) What are the power factor and the phase angle for the circuit? (b) Can you determine whether the current leads or lags the source voltage? Explain.
A series \(R L C\) circuit has \(R=500.0 \Omega, L=35.0 \mathrm{mH},\) and $C=87.0 \mathrm{pF} .$ What is the impedance of the circuit at resonance? Explain.
The voltage across an inductor and the current through the inductor are related by \(v_{\mathrm{L}}=L \Delta i / \Delta t .\) Suppose that $i(t)=I \sin \omega t .\( (a) Write an expression for \)v_{\mathrm{L}}(t) .$ [Hint: Use one of the relationships of Eq. \((20-7) .]\) (b) From your expression for \(v_{\mathrm{L}}(t),\) show that the reactance of the inductor is \(X_{\mathrm{L}}=\omega L_{\mathrm{r}}\) (c) Sketch graphs of \(i(t)\) and \(v_{\mathrm{L}}(t)\) on the same axes. What is the phase difference? Which one leads?
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