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A Light-Dimmer Circuit The intensity of a lightbulb with

Chapter 24, Problem 95GP

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QUESTION:

A Light-Dimmer Circuit The intensity of a lightbulb with a resistance of \(120\ \Omega\) is controlled by connecting it in series with an inductor whose inductance can be varied from \(L=0\) to \(L=L_\max\). This “light dimmer” circuit is connected to an ac generator with a frequency of 60.0 Hz and an rms voltage of 110 V. (a) What is the average power dissipated in the light bulb when \(L=0\)? (b) The inductor is now adjusted so that \(L=L_\max\). In this case, the average power dissipated in the light bulb is one-fourth the value found in part (a). What is the value of \(L_\max\)?

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QUESTION:

A Light-Dimmer Circuit The intensity of a lightbulb with a resistance of \(120\ \Omega\) is controlled by connecting it in series with an inductor whose inductance can be varied from \(L=0\) to \(L=L_\max\). This “light dimmer” circuit is connected to an ac generator with a frequency of 60.0 Hz and an rms voltage of 110 V. (a) What is the average power dissipated in the light bulb when \(L=0\)? (b) The inductor is now adjusted so that \(L=L_\max\). In this case, the average power dissipated in the light bulb is one-fourth the value found in part (a). What is the value of \(L_\max\)?

ANSWER:

Step 1 of 2

a)

When \(L=0\) we have a circuit containing only the resistor so the average power dissipated is given by

\(P_{\mathrm{av}}=\frac{V_{\mathrm{rms}}^{2}}{R}\)

Inserting the numbers we find

\(\begin{array}{l}P_{\mathrm{av}}=\frac{(110\mathrm{~V})^2}{120\Omega}\\ \Longrightarrow P_{\mathrm{av}}=100.8\mathrm{\ W}\\ \Longrightarrow P_{\mathrm{av}}=0.10\ \rm{kW}\end{array}\)

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