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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 惟.

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?
Consider a \(60.0-\mathrm{W}\) lightbulb and a \(100.0-\mathrm{W}\) lightbulb designed for use in a household lamp socket at \(120 \mathrm{V}\) (a) What are the resistances of these two bulbs? (b) If they are wired together in a series circuit, which bulb shines brighter (dissipates more power)? Explain. (c) If they are connected in parallel in a circuit, which bulb shines brighter? Explain.
Poiseuille's law [Eq. (9-15)] gives the volume flow rate of a viscous fluid through a pipe. (a) Show that Poiseuille's law can be written in the form \(\Delta P=I R\) where \(I=\Delta V / \Delta t\) represents the volume flow rate and \(R\) is a constant of proportionality called the fluid flow resistance. (b) Find \(R\) in terms of the viscosity of the fluid and the length and radius of the pipe. (c) If two or more pipes are connected in series so that the volume flow rate through them is the same, do the resistances of the pipes add as for electrical resistors \(\left(R_{e q}=R_{1}+R_{2}+\cdots\right) ?\) Explain. (d) If two or more pipes are connected in parallel, so the pressure drop across them is the same, do the reciprocals of the resistances add as for electrical resistors \(\left(1 / R_{\mathrm{eq}}=\right.\) $\left.1 / R_{1}+1 / R_{2}+\cdots\right) ?$ Explain.
A gold wire of 0.50 mm diameter has \(5.90 \times 10^{28} \mathrm{con}\) duction electrons/m". If the drift speed is \(6.5 \mu \mathrm{m} / \mathrm{s}\), what is the current in the wire?
The current in a wire is \(0.500 \mathrm{A} .\) (a) How much charge flows through a cross section of the wire in 10.0 s? (b) How many electrons move through the same cross section in $10.0 \mathrm{s} ?$
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