Question

A typical light dimmer used to dim the stage lights in a theater consists of a variable inductor L (whose inductance is adjustable between zero and L_max) connected in series with a lightbulb B, as shown in Fig. 31-34. The electrical supply is ${\mathbf{V}}_{\mathbf{rms}}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{V}$at $\mathbf{f}\mathbf{=}\mathbf{60}\mathbf{}\mathbf{Hz}$; the light bulb is rated at 120 V, $\mathbf{p}\mathbf{=}\mathbf{1000}\mathbf{}\mathbf{W}$. (a) What ${\mathbf{L}}_{\mathbf{max}}$ is required if the rate of energy dissipation in the lightbulb is to be varied by a factor of 5 from its upper limit of 1000W? Assume that the resistance of the lightbulb is independent of its temperature. (b) Could one use a variable resistor (adjustable between zero and R_max ) instead of an inductor? (c) If so, what R_max is required? (d) Why isn’t this done?

Short answer

(a) The maximum value of the inductor is ${\mathrm{L}}_{\max }=7.64\times {10}^{-2}\mathrm{m}$.

(b)Yes; one could use a variable resistor.

(c) The maximum value of the resistance is data-custom-editor="chemistry" ${\mathrm{R}}_{\max }=17.8\mathrm{\Omega }$.

(d) This is not done because the resistors would consume, rather than temporarily store, electromagnetic energy.

Step-by-step solution

Listing the given quantities:

RMS voltage,${\mathrm{V}}_{\mathrm{rms}}=120\mathrm{V}$

Frequency, $\mathrm{f}=60\mathrm{Hz}$

Voltage, $\mathrm{V}=120\mathrm{V}$

Power,$\mathrm{P}=1000\mathrm{W}$

Understanding the concepts of power and inductance =:

Define the relation between inductance, voltage, and power from the formula for power consumed by light bulb. Inserting the given values will give us the value of L_max.Using it,R_maxcan be found. From this, we can determinewhether inductor can be replaced by a variable resistor.

Formulas:

$\mathrm{P}=\frac{{\mathrm{I}}^2\mathrm{R}}{2}$

Here, I is the current and R is the resistance.

Hence, the power is,

$\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}$

(a) Calculation of the Lmax  is required if the rate of energy dissipation in the lightbulb is independent of its temperature

Write the equation for power as below.

$\mathrm{P}=\frac{{\mathrm{I}}^2\mathrm{R}}{2}$

From the above equation you can say that, the power the directly proportional to the square of the current. Therefore,

data-custom-editor="chemistry" $\begin{array}{rcl}\mathrm{P}& \propto & {\mathrm{I}}^2\\ \frac{{\mathrm{P}}_{\min }}{{\mathrm{P}}_{\max }}& =& {\left(\frac{{\mathrm{I}}_{\min }}{\mathrm{I}}\right)}^2\\ & =& {\left(\frac{{\displaystyle \frac{\mathrm{\epsilon }}{{\mathrm{Z}}_{\min }}}}{{\displaystyle \frac{\mathrm{\epsilon }}{{\mathrm{Z}}_{\max }}}}\right)}^2\\ & =& \left[\frac{\mathrm{\epsilon }}{{\mathrm{Z}}_{\min }}\times \frac{{\mathrm{Z}}_{\max }}{\mathrm{\epsilon }}\right]\end{array}$

Here, Z is the impedance.

The maximum impedance is,

${\mathrm{Z}}_{\max }=\sqrt{{\mathrm{R}}^2+{\left({\mathrm{\omega L}}_{\max }\right)}^2}$

The minimum impedance is,

data-custom-editor="chemistry" ${\mathrm{Z}}_{\min }=\mathrm{R}$

The rate of energy dissipation in the light bulb is to be varied by a factor of 5. Therefore,

$\begin{array}{rcl}5& =& {\left(\frac{\sqrt{{\mathrm{R}}^2+{\left({\mathrm{\omega L}}_{\max }\right)}^2}}{\mathrm{R}}\right)}^2\\ & =& \left(\frac{{\mathrm{R}}^2+{\left({\mathrm{\omega L}}_{\max }\right)}^2}{{\mathrm{R}}^2}\right)\\ 5{\mathrm{R}}^2& =& {\mathrm{R}}^2+{\left({\mathrm{\omega L}}_{\max }\right)}^2\\ {\left({\mathrm{L}}_{\max }\right)}^2& =& \frac{4{\mathrm{R}}^2}{{\mathrm{\omega }}^2}\\ {\mathrm{L}}_{\max }& =& \frac{2\mathrm{R}}{\mathrm{\omega }}\end{array}$

The equation of power is,

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}} \\ \mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}} \end{gathered}$$

Hence, the maximuminductor is,

$\begin{array}{rcl}{\mathrm{L}}_{\max }& =& \frac{{\displaystyle \frac{2{\mathrm{V}}^2}{\mathrm{P}}}}{\mathrm{\omega }}\\ & =& \frac{2{\mathrm{V}}^2}{\mathrm{P}\times 2\mathrm{\pi f}}\\ & =& \frac{{\mathrm{V}}^2}{\mathrm{P}\times \mathrm{\pi f}}\end{array}$

Substitute known numerical values in the above equation.

$\begin{array}{rcl}{\mathrm{L}}_{\max }& =& \frac{{\left(120\right)}^2}{1000\times 3.14\times 60}\\ & =& 7.64\times {10}^{-2}\mathrm{H}\end{array}$

Hence, the maximum value of the inductor is $7.64\times {10}^{-2}\mathrm{H}$.

(b) Explanation

Since,${\mathrm{L}}_{\max }=\frac{2\mathrm{R}}{\mathrm{\omega }}$

Then, one could use the variable resistor instead of an inductor.

(c) Calculation of the  :

As given,

${\left(\frac{{\mathrm{R}}_{\max }+{\mathrm{R}}_{\mathrm{bulb}}}{{\mathrm{R}}_{\mathrm{bulb}}}\right)}^2=5$

Simplifyingtheabove equation gives,

$\begin{array}{rcl}{\left(\frac{{\mathrm{R}}_{\max }+{\mathrm{R}}_{\mathrm{bulb}}}{{\mathrm{R}}_{\mathrm{bulb}}}\right)}^2& =& 5\\ {\mathrm{R}}_{\max }+{\mathrm{R}}_{\mathrm{bulb}}& =& \sqrt{5{\mathrm{R}}_{\mathrm{bulb}}}\\ {\mathrm{R}}_{\max }& =& \sqrt{5{\mathrm{R}}_{\mathrm{bulb}}}-{\mathrm{R}}_{\mathrm{bulb}}\\ {\mathrm{R}}_{\max }& =& \left(\sqrt{5}-1\right){\mathrm{R}}_{\mathrm{bulb}}\\ & =& \left(\sqrt{5}-1\right)\times \frac{{\mathrm{V}}^2}{\mathrm{P}}\\ & =& \left(\sqrt{5}-1\right)\times \frac{{120}^2}{1000}\\ & =& 17.8\mathrm{\Omega }\end{array}$

Hence, the maximum value of resistance is data-custom-editor="chemistry" ${\mathrm{R}}_{\max }=17.8\mathrm{\Omega }$.

(d) Explanation:

This is not done because the resistors would consume, rather than temporarily store, electromagnetic energy.