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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

If a chandelier has a label stating \(120 \mathrm{V}, 5.0 \mathrm{A},\) can its power rating be determined? If so, what is it?
A gold wire and an aluminum wire have the same dimensions and carry the same current. The electron density (in electrons/cm \(^{3}\) ) in aluminum is three times larger than the density in gold. How do the drift speeds of the electrons in the two wires, \(v_{\mathrm{Au}}\) and \(v_{\mathrm{Al}},\) compare?
Two electrodes are placed in a calcium chloride solution and a potential difference is maintained between them. If $3.8 \times 10^{16} \mathrm{Ca}^{2+}\( ions and \)6.2 \times 10^{16} \mathrm{Cl}^{-}$ ions per second move in opposite directions through an imaginary area between the electrodes, what is the current in the solution?
A \(500-\) W electric heater unit is designed to operate with an applied potential difference of \(120 \mathrm{V}\). (a) If the local power company imposes a voltage reduction to lighten its load, dropping the voltage to $110 \mathrm{V}$, by what percentage does the heat output of the heater drop? (Assume the resistance does not change.) (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than calculated in part (a)?
(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?
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