Chapter 27: Q104P (page 802)

A three-way $120V$ lamp bulb that contains two filaments is rated for $100 W, 200 W, 300 W$ . One filament burns out. Afterward, the bulb operates at the same intensity (dissipates energy at the same rate) on its lowest as on its highest switch positions but does not operate at all on the middle position.
(a) How are the two filaments wired to the three switch positions? What are the (b) smaller and
(c) larger values of the filament resistances?

Short Answer

Expert verified

a. The two filaments wired to the three switch positions:

For lower position power, connect the larger resistance, say $R_{1}$ for the middle position power, connect the smaller resistance, say $R_{2}$ and for high position power, connect both resistances parallel.

b. The smaller value of the filament resistance is $R_{2} = 72 Ω$ .

c. The larger value of the filament resistance is $R_{1} = 144 Ω$

Step by step solution

Write the given quantities

The given voltage is $V = 120 V$ .

The power ratings are $P = 100 W, 200 W, 300 W$ .

Determine the formulas and concept of power

Consider equationof power in terms of voltage and current. Forthefirst part, determine the positions of resistances according tothepowers given. Using the equation, find the value of resistance.

Consider the formula for power:

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

(a) Determine the connection of the two filaments to the three switch positions:

The two filaments wired to the three switch positions. Since, the power is inversely proportional to resistance, so, for smaller resistance the higher power is obtained. For lower position power, connect larger resistance, say $R_{1,}$ and for middle position power,connect smaller resistance, say $R_{2.}$ Further, for high position power, connect both resistances parallel.

(b) Determine the smaller value of filament resistance

For smaller value of filament resistance,consider the power and the equivalent resistance is as follows:

$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} R_{e q} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} P = \frac{V^{2}}{R_{e q}}$

Determine the value of $R_{1}$ from the lower power:

$P = \frac{V^{2}}{R_{1}} 100 = \frac{120^{2}}{R_{1}}$

Solve further as:

$\begin{matrix} R_{1} = \frac{120^{2}}{100} \\ = \frac{14400}{100} \\ = 144 Ω \end{matrix}$

Consider the equation for power:

$P = \frac{V^{2}}{R_{e q}} P = \frac{V^{2}}{(\frac{R_{1} R_{2}}{R_{1} + R_{2}})}$

Substitute the values and solve as:

$300 = \frac{120^{2}}{(\frac{144 R_{2}}{144 + R_{2}})} \frac{300 (144 R_{2})}{(144 + R_{2})} = 14400 43200 R_{2} = 14400 (144 + R_{2})$

Solve further as:

$43200 R_{2} = 2073600 + 14400 R_{2} (43200 - 14400) R_{2} = 2073600 28800 R_{2} = 2073600 \begin{array}{l} R_{2} = \frac{2073600}{28800} \\ R_{2} = 72 Ω \end{array}$