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Chapter 25: Problem 25

Two conductors of the same length and radius are connected to the same emf device. If the resistance of one is twice that of the other, to which conductor is more power delivered?

Short Answer

Expert verified
Answer: The conductor with lower resistance has more power delivered to it.

Step by step solution

01

Recall Ohm's Law and the expression for power

According to Ohm's Law, the electric current (I) flowing through a conductor is inversely proportional to its resistance (R) and directly proportional to the voltage (V) across the conductor: I = V / R Also, the power (P) delivered to a conductor is given by the expression: P = I^2 * R
02

Express the currents in terms of resistance

Let's denote the resistance of the first conductor as R1 and the resistance of the second conductor as R2. Given that the resistance of one conductor is twice that of the other: R2 = 2 * R1 Since both conductors are connected to the same emf device, the voltage across both conductors is the same. Let's denote the voltage as V. Using Ohm's Law, we can express the current flowing through each conductor in terms of its resistance: I1 = V / R1 I2 = V / (2 * R1)
03

Calculate the power delivered to each conductor

Now we can calculate the power delivered to each conductor using the expression for power: P1 = I1^2 * R1 P2 = I2^2 * R2 Substitute the expressions for I1 and I2: P1 = (V / R1)^2 * R1 P2 = (V / (2 * R1))^2 * (2 * R1)
04

Compare the power delivered to each conductor

To compare the power delivered to each conductor, we can simplify the expressions for P1 and P2: P1 = V^2 / R1 P2 = V^2 / (4 * R1) * (2 * R1) = V^2 / (2 * R1) Since P1 = V^2 / R1 and P2 = V^2 / (2 * R1), it's clear that P1 > P2. Therefore, more power is delivered to the conductor with the lower resistance (R1).

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Most popular questions from this chapter

Two cylindrical wires of identical length are made of copper and aluminum. If they carry the same current and have the same potential difference across their length, what is the ratio of their radii?

A modern house is wired for \(115 \mathrm{~V}\), and the current is limited by circuit breakers to a maximum of \(200 .\) A. (For the purpose of this problem, treat these as DC quantities.) a) Calculate the minimum total resistance the circuitry in the house can have at any time. b) Calculate the maximum electrical power the house can consume.

A light bulb is connected to a source of emf. There is a \(6.20 \mathrm{~V}\) drop across the light bulb, and a current of 4.1 A flowing through the light bulb. a) What is the resistance of the light bulb? b) A second light bulb, identical to the first, is connected in series with the first bulb. The potential drop across the bulbs is now \(6.29 \mathrm{~V},\) and the current through the bulbs is \(2.9 \mathrm{~A}\). Calculate the resistance of each light bulb. c) Why are your answers to parts (a) and (b) not the same?

If the current through a resistor is increased by a factor of \(2,\) how does this affect the power that is dissipated? a) It decreases by a factor of 4 . b) It increases by a factor of 2 . c) It decreases by a factor of 8 . d) It increases by a factor of 4 .

What is the current density in an aluminum wire having a radius of \(1.00 \mathrm{~mm}\) and carrying a current of \(1.00 \mathrm{~mA}\) ? What is the drift speed of the electrons carrying this current? The density of aluminum is \(2.70 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3},\) and 1 mole of aluminum has a mass of \(26.98 \mathrm{~g}\). There is one conduction electron per atom in aluminum.
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