Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In Fig. 27-19, a circuit consists of a battery and two uniform resistors, and the section lying along an xaxis is divided into five segments of equal lengths.

(a) Assume thatR1=R2and rank the segments according to the magnitude of the average electric field in them, greatest first.

(b) Now assume thatR1>R2and then again rank the segments.

(c) What is the direction of the electric field along the xaxis?

Short Answer

Expert verified
  1. Rank of electric field for R1=R2is b=d>a=c=e
  2. Rank of electric field for R1>R2isb>d>a=c=e
  3. Direction of electric field is along positive x direction.

Step by step solution

01

Step 1: Given

R1=R2R1>R2

02

Determining the concept

Here, Use the equation of Ohm’s law and the relation between the voltage and the electric field together and make the equation for the electric field relating with the resistance and the length of segment.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

V=IRV=EL

Where, I is current, V is voltage, R is resistance, E is electric field.

03

(a) Determining the rank of segments according to the magnitude of the electric field greatest first forR1=R2

According to Ohm’s law,

V=IR

And relation between voltage and electric field is as follow:

V=EL

So,

EL=IR

E=IRL

Here, the resistance through the wire is less than through the circuits and length of each segment is same.

So, the electric field only depends on R,

Segments a, c and e have same resistance.

So, they have the same electric field. Also, the resistanceR1=R2are greater than the resistance through the segments, So, the electric field through b and d is the same but greater than segments a, c and e.

Hence, the rank will become b=d>a=c=e

04

(b) Determining the rank of segments according to the magnitude of the electric field greatest first for R1>R2

Here, the resistance through the wire is less than through the circuits and the length of each segment is same.

So, the electric field only depends on R,

Segments a, c and e have the same resistance.

So, they have same electric field,

And resistance R1>R2and so, The electric field through b>d

b>d>a=c=e

Hence,rank of electric field for R1>R2isb>d>a=c=e

05

(c) Determining the direction of electric field along the x axis

The direction of the electric field is from more positive end to more negative end in the circuit. So, it is in positive x direction.

Hence, direction of electric field is along positive x direction.

Therefore, use Ohm’s law to find the voltage and the relation between the voltage and the electric field to find the electric field.

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 Vaia!

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

In Fig. 27-54, the resistances areR1=1.0 ΩandR2=2.0 Ω , and the ideal batteries have emf ε1=2.0Vand ε2=ε3=4.0V. What are the (a) size and (b) direction (up or down) of the current in battery 1, the (c) size and (d) direction of the current in battery 2, and the (e) size and (f) direction of the current in battery 3? (g) What is the potential difference VaVb?

In Figure, the ideal batteries have emfsε1=10.0Vandε2=0.500ε1 , and the resistances are each 4.00Ω.

(a) What is the current in resistance 2?

(b) What is the current in resistance 3?

In Fig. 27-55a, resistor 3 is a variable resistor and the ideal battery has emf.ε=12V Figure 27-55b gives the current I through the battery as a function of R3. The horizontal scale is set by.R3s=20ΩThe curve has an asymptote of2.0 mAasR3. What are (a) resistanceR1and (b) resistance R2?

In Fig. 27-25, the ideal batteries have emfs ε1=12vand ε2=6.0v. What are (a) the current, the dissipation rate in (b) resistor 1?(4Ω)And (c) resistor 2 (8Ω), and the energy transfer rate in (d) battery 1 and (e) battery 2? Is energy being supplied or absorbed by (f) battery 1 and (g) battery 2?

A solar cell generates a potential difference of 0.10Vwhen a500 resistor is connected across it, and a potential difference of 0.15Vwhen a 1000resistor is substituted.

(a) What is the internal resistance?

(b) What is the emf of the solar cell?

(c) The area of the cell is5.0cm2 , and the rate per unit area at which it receives energy from light is2.0mW/cm2 .What is the efficiency of the cell for converting light energy to thermal energy in the1000 external resistor?

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.

Sign-up for free