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Chapter 10: Problem 4

A battery with voltage \(V\) sends a current \(I\) through a resistor \(R\). If \(V\) is doubled and \(R\) halved, the power dissipated by the resistor (A) goes up by a factor of eight. (B) quadruples. (C) halves. (D) doubles. (E) goes down by a factor of four.

Short Answer

Expert verified
Question: When the voltage is doubled and the resistance is halved, the power dissipated by a resistor: A) goes up by a factor of eight. B) goes down by a factor of eight. C) goes up by a factor of four. D) goes down by a factor of four. E) remains the same. Answer: A) goes up by a factor of eight.

Step by step solution

01

Write down the original power formula

The original formula for power is given as: \(P=VI\), where P is the power, V is the voltage, and I is the current.
02

Write down the Ohm's Law formula

We will need to use Ohm's Law as well to relate V, I, and R. Ohm's Law formula is given by: \(V = IR\), where V is the voltage, I is the current, and R is the resistance.
03

Relate power formula to Ohm's Law

To find the power in terms of V and R, we can substitute the Ohm's Law formula into the power formula. So, we get \(P = V(IR) = VI = \frac{V^2}{R}\).
04

Apply the given changes to the formula

Now we need to apply the changes given in the exercise: the voltage is doubled, and the resistance is halved. So, \(V\) will turn to \(2V\) and \(R\) will turn to \(R/2\).
05

Find the new power

Now we need to find the new power using the modified formula: \(P_{new} = \frac{(2V)^2}{R/2} = \frac{4V^2}{R/2} = 8\frac{V^2}{R}\).
06

Compare the new power to the original power

Now we need to compare \(P_{new}\) and \(P_{old}\) to find the answer. We have \(P_{new} = 8\frac{V^2}{R}\) and \(P_{old} = \frac{V^2}{R}\). Divide \(P_{new}\) by \(P_{old}\) to find the factor by which power changed: \(\frac{P_{new}}{P_{old}} = \frac{8\frac{V^2}{R}}{\frac{V^2}{R}} = 8\).
07

Conclusion

So, the power dissipates by a factor of 8. The correct answer is (A) goes up by a factor of eight.

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Most popular questions from this chapter

A battery \(V\) sends a current \(I\) through two resistors in series, \(R\) and \(2 R\). If the first resistor is doubled to \(2 R\), the current \(I\) is now (A) the same as it was initially. (B) half what it was initially. (C) three-quarters what it was initially. (D) four-thirds what it was initially. (E) twice what is was initially.

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