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Chapter 18: Problem 64

Show that \(A^{2} \times \Omega=W\) (amperes squared times ohms \(=\) watts).

Short Answer

Expert verified
Answer: Yes, the equation \(A^2 \times \Omega = W\) holds true by substituting Ohm's law into the power formula and simplifying, as demonstrated in the provided solution.

Step by step solution

01

Write down Ohm's law and the power formula

Ohm's law: \(V = I \times R\) Power formula: \(P = I \times V\)
02

Substitute Ohm's law into the power formula

Replace V in the power formula with \(I \times R\) from Ohm's Law: \(P = I \times (I \times R)\)
03

Simplify the equation

Multiply I by I × R: \(P = I^2 \times R\)
04

Compare the equation with the given equation

We now have: \(P = I^2 \times R\) Comparing this to the given equation, \(A^2 \times \Omega = W\), we see that: \(I^2 \times R = A^2 \times \Omega\) Since I is current in amperes \(A\) and R is resistance in ohms \(\Omega\), and P is power in watts \(W\), the equation holds true. Therefore, we have proved that \(A^2 \times \Omega = W\).

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Most popular questions from this chapter

Chelsea inadvertently bumps into a set of batteries with an emf of $100.0 \mathrm{V}\( that can supply a maximum power of \)5.0 \mathrm{W} .$ If the resistance between the points where she contacts the batteries is $1.0 \mathrm{k} \Omega,$ how much current passes through her?
Find the maximum current that a fully charged D-cell can supply - if only briefly-such that its terminal voltage is at least \(1.0 \mathrm{V}\). Assume an emf of \(1.5 \mathrm{V}\) and an internal resistance of \(0.10 \Omega .\)
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?
A parallel plate capacitor is constructed from two square conducting plates of length \(L=0.10 \mathrm{m}\) on a side. There is air between the plates, which are separated by a distance \(d=89 \mu \mathrm{m} .\) The capacitor is connected to a \(10.0-\mathrm{V}\) battery. (a) After the capacitor is fully charged, what is the charge on the upper plate? (b) The battery is disconnected from the plates and the capacitor is discharged through a resistor \(R=0.100 \mathrm{M} \Omega .\) Sketch the current through the resistor as a function of time \(t(t=0\) corresponds to the time when \(R\) is connected to the capacitor). (c) How much energy is dissipated in \(R\) over the whole discharging process?
A string of 25 decorative lights has bulbs rated at \(9.0 \mathrm{W}\) and the bulbs are connected in parallel. The string is connected to a \(120-\mathrm{V}\) power supply. (a) What is the resistance of each of these lights? (b) What is the current through each bulb? (c) What is the total current coming from the power supply? (d) The string of bulbs has a fuse that will blow if the current is greater than 2.0 A. How many of the bulbs can you replace with 10.4-W bulbs without blowing the fuse?
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