/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 220 V is applied to two differen... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

220 V is applied to two different conductors made of the same material. One conductor is twice as long and twice the diameter of the second. What is the ratio of the power transformed in the first relative to the second?

Short Answer

Expert verified
The power ratio is 2:1.

Step by step solution

01

Understand the Problem

We have two conductors made of the same material, which means they have the same resistivity \( \rho \). The first conductor is twice as long and has twice the diameter as the second conductor. We need to find the ratio of power dissipation for both of these conductors when the same voltage is applied across them.
02

Calculate Resistance of Conductors

Use the formula for resistance: \[ R = \rho \frac{L}{A} \]where \( L \) is the length and \( A \) is the cross-sectional area of the conductor.Conductor 1:- Length \( = 2L \)- Diameter \( = 2d \)- Area \( A_1 = \pi \left(\frac{2d}{2}\right)^2 = \pi d^2 \)- Resistance \( R_1 = \rho \frac{2L}{\pi d^2} \)Conductor 2:- Length \( = L \)- Diameter \( = d \)- Area \( A_2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \)- Resistance \( R_2 = \rho \frac{L}{\left(\frac{\pi d^2}{4}\right)} = \frac{4\rho L}{\pi d^2} \)
03

Determine Ratio of Resistances

Using the resistances calculated for each conductor: \[ R_1 = \rho \frac{2L}{\pi d^2} \text{ and } R_2 = \frac{4\rho L}{\pi d^2} \]The ratio \( \frac{R_1}{R_2} = \frac{\rho \frac{2L}{\pi d^2}}{\frac{4\rho L}{\pi d^2}} = \frac{2}{4} = \frac{1}{2} \).
04

Calculate Power Dissipated

Power dissipated in a conductor can be calculated using the formula:\[ P = \frac{V^2}{R} \]For Conductor 1:\[ P_1 = \frac{220^2}{R_1} \]For Conductor 2:\[ P_2 = \frac{220^2}{R_2} \]
05

Determine Ratio of Powers

The ratio of power dissipated in the first conductor to the second conductor is given by:\[ \frac{P_1}{P_2} = \frac{\frac{220^2}{R_1}}{\frac{220^2}{R_2}} = \frac{R_2}{R_1} \]Using the resistance ratio from Step 3, we have:\[ \frac{P_1}{P_2} = \frac{1}{\frac{1}{2}} = 2 \]
06

Conclude the Ratio

Therefore, the ratio of the power transformed in the first conductor to the second is 2:1.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistance Calculation
The resistance of a conductor determines how difficult it is for electric current to flow through it. If you imagine electricity as water flowing through a pipe, resistance is similar to the narrowness of the pipe, which slows down the water flow. The formula to calculate resistance, denoted as \( R \), is given by:\[ R = \rho \frac{L}{A} \]where:- \( \rho \) is the resistivity of the material, a constant depending on what the conductor is made of.- \( L \) is the length of the conductor. A longer pipe offers more resistance, similar to how a longer conductor does.- \( A \) is the cross-sectional area, akin to the thickness of a pipe, and a larger area means less resistance.Understanding these relationships helps in calculating the resistance of any conductor, such as in our exercise with two different yet similar conductors. The longer and wider a conductor, typically the change in resistance needs thorough evaluation as done in the provided solution.
Voltage Applied
Voltage is like the force pushing electricity through a conductor. It's the potential difference that causes electric current to flow along the conductor's length. In simpler terms, it's like the pressure at the start of a water pipe that ensures water flows to the other end. In the given exercise, a voltage of 220 V is applied to both conductors.When voltage is applied, the rate of energy transfer or electric power dissipation in the conductor is dependent on both the resistance and the applied voltage. The formula used to determine the power dissipation \( P \) is:\[ P = \frac{V^2}{R} \]where:- \( V \) is the voltage applied.- \( R \) is the resistance of the conductor.This formula highlights why voltage is crucial. Even with identical material and conditions, different resistances result in variable power dissipations, particularly when the same voltage is applied across different conductors.
Conductors in Series
In electrical circuits, conductors can be arranged in many configurations, one of which is in series. When conductors are connected in series, they are arranged end to end in a line, making the total resistance the sum of the individual resistances. In the current exercise, although we have two distinct conductors, they aren't exactly in series but are considered separately for comparison of power dissipation.However, understanding series arrangements is crucial for comprehending such exercises. In series:- The same current flows through each conductor.- The total voltage across the series combination is the sum of the voltages across each conductor.- The total or equivalent resistance is the sum of the individual resistances: \( R_{\text{total}} = R_1 + R_2 \).Knowing how conductors behave in series helps in grasping more complex circuit analyses and ensures a solid foundation for understanding power dissipation across several components when applied in more extensive network systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(II) A neuron is stimulated with an electric pulse. The action potential is detected at a point 3.70 cm down the axon 0.0052 s later.When the action potential is detected 7.20 cm from the point of stimulation, the time required is 0.0063 s. What is the speed of the electric pulse along the axon? (Why are two measurements needed instead of only one?)

(II) Can a 2.2-mm-diameter copper wire have the same resistance as a tungsten wire of the same length? Give numerical details.

(I) An electric car uses a 45-kW (160-hp) motor. If the battery pack is designed for 340V, what current would the motor need to draw from the battery? Neglect any energy losses in getting energy from the battery to the motor.

(II) An ordinary flashlight uses two D-cell 1.5-V batteries connected in series to provide 3.0 V across the bulb, as in Fig. 18-4b (Fig. 18-36). The bulb draws 380 mA when turned on. (\(a\)) Calculate the resistance of the bulb and the power dissipated. (\(b\)) By what factor would the power increase if four D cells in series (total 6.0 V) were used with the same bulb? (Neglect heating effects of the filament.)Why shouldn't you try this?

An air conditioner draws 18 A at 220-V ac. The connecting cord is copper wire with a diameter of 1.628 mm. (\(a\)) How much power does the air conditioner draw? (\(b\)) If the length of the cord (containing two wires) is 3.5 m, how much power is dissipated in the wiring? (\(c\)) If no. 12 wire, with a diameter of 2.053 mm, was used instead, how much power would be dissipated in the wiring? (\(d\)) Assuming that the air conditioner is run 12 h per day, how much money per month (30 days) would be saved by using no. 12 wire? Assume that the cost of electricity is 12 cents per kWh.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.