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Chapter 18: Problem 60

What is the resistance of a \(40.0-\mathrm{W}, 120-\mathrm{V}\) lightbulb?

Short Answer

Expert verified
Answer: The resistance of the 40.0-W, 120-V lightbulb is approximately 18.97 Ω.

Step by step solution

01

Identify the known variables

We are given the power (P) and voltage (V) of the lightbulb: \(P=40.0\,W\) and \(V=120\,V\).
02

Apply the power formula and Ohm's law

To find the resistance (R), we need to combine the power formula (\(P=IV\)) with Ohm's law (\(V=IR\)). Using Ohm's law, we can express the current (I) as \(I=\frac{V}{R}\). Substituting this into the power formula, we get: \(P = I^2R = \left(\frac{V}{R}\right)^2R\)
03

Rearrange the equation to solve for R

Now, we can rearrange the equation to isolate R: \(P = \left(\frac{V^2}{R^2}\right)R\) Multiply both sides by \(R^2\): \(P \cdot R^2 = V^2\) Divide both sides by P: \(R^2 = \frac{V^2}{P}\) Now, take the square root of both sides: \(R = \sqrt{\frac{V^2}{P}}\)
04

Calculate the resistance

Now that we have the equation to find the resistance, we can plug in the given values for P and V: \(R = \sqrt{\frac{(120\,\mathrm{V})^2}{40.0\,\mathrm{W}}}\) Calculate the square: \(R = \sqrt{\frac{14400\,\mathrm{V^2}}{40.0\,\mathrm{W}}}\) Divide the values: \(R = \sqrt{360\,\mathrm{\Omega^2}}\) Take the square root: \(R = 18.97\,\Omega\)
05

Report the answer

Therefore, the resistance of the \(40.0-\mathrm{W}, 120-\mathrm{V}\) lightbulb is approximately \(18.97\,\Omega\).

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Most popular questions from this chapter

Near Earth's surface the air contains both negative and positive ions, due to radioactivity in the soil and cosmic rays from space. As a simplified model, assume there are 600.0 singly charged positive ions per \(\mathrm{cm}^{3}\) and 500.0 singly charged negative ions per \(\mathrm{cm}^{3} ;\) ignore the presence of multiply charged ions. The electric field is $100.0 \mathrm{V} / \mathrm{m},$ directed downward. (a) In which direction do the positive ions move? The negative ions? (b) What is the direction of the current due to these ions? (c) The measured resistivity of the air in the region is $4.0 \times 10^{13} \Omega \cdot \mathrm{m} .$ Calculate the drift speed of the ions, assuming it to be the same for positive and negative ions. [Hint: Consider a vertical tube of air of length \(L\) and cross-sectional area \(A\) How is the potential difference across the tube related to the electric field strength?] (d) If these conditions existed over the entire surface of the Earth, what is the total current due to the movement of ions in the air?
Two copper wires, one double the diameter of the other, have the same current flowing through them. If the thinner wire has a drift speed \(v_{1},\) and the thicker wire has a drift speed \(v_{2},\) how do the drift speeds of the charge carriers compare?
(a) What is the resistance of the heater element in a \(1500-\mathrm{W}\) hair dryer that plugs into a \(120-\mathrm{V}\) outlet? (b) What is the current through the hair dryer when it is turned on? (c) At a cost of \(\$ 0.10\) per kW .h, how much does it cost to run the hair dryer for 5.00 min? (d) If you were to take the hair dryer to Europe where the voltage is \(240 \mathrm{V},\) how much power would your hair dryer be using in the brief time before it is ruined? (e) What current would be flowing through the hair dryer during this time?
A copper wire and an aluminum wire of the same length have the same resistance. What is the ratio of the diameter of the copper wire to that of the aluminum wire?
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