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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

During a "brownout," which occurs when the power companies cannot keep up with high demand, the voltage of the household circuits drops below its normal $120 \mathrm{V} .\( (a) If the voltage drops to \)108 \mathrm{V},$ what would be the power consumed by a "100-W" light-bulb (that is, a light-bulb that consumes \(100.0 \mathrm{W}\) when connected to \(120 \mathrm{V}\) )? Ignore (for now) changes in the resistance of the light-bulb filament. (b) More realistically, the light-bulb filament will not be as hot as usual during the brownout. Does this make the power drop more or less than that you calculated in part (a)? Explain.
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