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Chapter 10: Problem 24

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

Short Answer

Expert verified
Answer: (B) Sometimes, if \(A_1\) is small enough.

Step by step solution

01

Recall Ohm's Law and the resistivity formula

Ohm's Law states that the resistance (R) is equal to the voltage (V) divided by the current (I): \(R = \frac{V}{I}\). The resistivity formula relates the resistance of an object to its geometry and the material's resistivity (\(ρ\)): \(R = ρ \frac{L}{A}\), where L is the length of the conductor, and A is its cross-sectional area.
02

Combine and rearrange formulas

We can combine the two formulas into one expression: \(ρ \frac{L}{A} = \frac{V}{I}\). Rearranging terms, we can express the resistance as \(R = ρ \frac{L}{A}\). Therefore, resistance is directly proportional to the length of the conductor and inversely proportional to the cross-sectional area.
03

Compare resistance with current through L1 and L2

Now, let's compare the resistance with the current going through the length L1 (R1) and the resistance with the current going through the length L2 (R2): 1. R1 = \(ρ \frac{L_1}{A_1}\) 2. R2 = \(ρ \frac{L_2}{A_2}\) Since \(L_1 > L_2\), the resistance will be higher if the length is more significant, all other factors being constant. However, let's consider the condition when the resistance with current flowing through L2 is more significant than the resistance with current flowing through L1.
04

Evaluate the conditions for R2>R1

For R2 to be greater than R1, the following inequality must be true: \(ρ \frac{L_2}{A_2} > ρ \frac{L_1}{A_1}\). Canceling out the resistivity constant (ρ) and multiplying each side by \(A_1 A_2\), we get the inequality \(L_2 A_1 > L_1 A_2\). If \(L_2\) is shorter than \(L_1\), the resistance R2 can be greater than R1 if the factor \(A_1/A_2\) is large enough. In other words, we need to satisfy the condition that \(A_1 > A_2 * \frac{L_1}{L_2}\). Since we have found a condition when the shorter length can have greater resistance, the answer to this problem is: (B) Sometimes, if \(A_1\) is small enough

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Most popular questions from this chapter

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.

A battery \(V\) sends a current \(I\) through two resistors in series, \(R\) and \(2 R\). If the second resistor is halved to \(R\), the current dissipated power (A) stays the same. (B) goes up by a factor of two. (C) goes down by a factor of two. (D) is two-thirds what it was initially. (E) is three-halves what it was initially.

A camera flash releases \(3 \mathrm{~J}\) of energy, which is supplied by a capacitor. If the voltage of the flash unit is \(200 \mathrm{~V}\), the capacitance of the capacitor is (A) \(200 \mathrm{~F}\) (B) \(200 \mu \mathrm{F}\) (C) \(150 \mu \mathrm{F}\) (D) \(10^{-8} \mathrm{~F}\) (E) \(150 \mathrm{nF}\)

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) \(7.1 \times 10^{-6} \Omega\)

If bulb \(B\) is removed from the circuit, the brightness of bulb \(C\) (A) remains the same. (B) goes up by a factor of four. (C) goes down by a factor of four. (D) goes down by a factor of \(4 / 9\). (E) goes up by a factor of \(9 / 4\).
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