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Chapter 10: Problem 3

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

Short Answer

Expert verified
Answer: (B) quadruples.

Step by step solution

01

Recall Ohm's law and Power formula

Remember that Ohm's law states that \(V=IR\), which means the voltage across a resistor is equal to the product of the current through the resistor and the resistance itself. Also, recall the power formula, which is \(P=IV\). This formula states that the power dissipated by the resistor is equal to the product of the current through the resistor and the voltage across the resistor.
02

Equation for power in terms of resistance and voltage

We want to find the power dissipated by the resistor in terms of \(R\) and \(V\). From Ohm's law, we have \(I=\frac{V}{R}\). Substitute this expression for \(I\) into the power formula, we get \(P = (\frac{V}{R})V = \frac{V^2}{R}\).
03

Analyze the effect of doubling the voltage

Now, let's see what happens to the power when the voltage is doubled. Let the initial voltage be \(V\), and the doubled voltage be \(2V\). Using the equation found in Step 2, we can find the new power. The new power, \(P' = \frac{(2V)^2}{R} = \frac{4V^2}{R}\).
04

Compare the initial power and the new power

To understand how the power changes when the voltage is doubled, we compare the two power equations. Recall that the initial power, \(P = \frac{V^2}{R}\). The new power, \(P' = \frac{4V^2}{R}\). Notice that \(P' = 4P\), which means the new power is four times greater than the initial power.
05

Conclusion

Since the power dissipated by the resistor quadruples when the voltage is doubled, the correct answer is (B) quadruples.

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Most popular questions from this chapter

A bar magnet of \(0.1 \mathrm{~T}\) is thrust into a coil of wire from a large distance in \(0.5 \mathrm{~s}\). If the coil has 1000 turns, an area of \(0.01 \mathrm{~m}^2\) and its resistance is \(0.5 \Omega\), what is the approximate current going through the wire? (A) \(4 \mathrm{~A}\) (B) \(2 \mathrm{~A}\) (C) \(0.5 \mathrm{~A}\) (D) \(0.05 \mathrm{~A}\) (E) \(0.01 \mathrm{~A}\)

A rectangular copper block has a resistivity \(1.7 \times 10^{-8} \Omega \mathrm{m}\). \(L_1=15 \mathrm{~cm} ; L_2=3 \mathrm{~cm}\). The area of the faces are \(A_1=18 \mathrm{~cm}^2\) and \(A_2=90 \mathrm{~cm}^2\). A current is passed through the length \(L_1\). The resistance of the block is (A) \(1.4 \times 10^{-6} \Omega\) (B) \(1.4 \times 10^{-10} \Omega\) (C) \(5.7 \times 10^{-8} \Omega\) (D) \(5.7 \times 10^{-4} \Omega\) (E) \(7.1 \times 10^{-6} \Omega\)

For a rectangular block with \(L_1>L_2\), as drawn, can the resistance with the current through the length \(L_2\) be greater than the resistance with the current going through the length \(L_1\) ? (A) Always (B) Sometimes, if \(L_2\) is long enough (C) Sometimes, if \(A_2\) is large enough (D) Sometimes, if \(A_1\) is small enough (E) Never

If \(V=480 \mathrm{~V} ; C_1=10 \mu \mathrm{F} ; C_2=2 \mu \mathrm{F}\) and \(C_3=4 \mu \mathrm{F}\), rank the capacitors in terms of stored energy, from highest to lowest. (A) \(C_3, C_1=C_2\) (B) \(C_3, C_1, C_2\) (C) \(C_1, C_2, C_3\) (D) \(C_1=C_3, C_2\) (E) \(C_2, C_3, C_1\)

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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