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. E26.11, the battery has emf 35.0 V and negligible internal resistance. R1=5.00Ω. The current through is R1, 1.5 A and the current through R3is 4.50 A. What are the resistances R1and R3?

Short Answer

Expert verified

The value of resistances isR2=2.5Ω and R3=6.11Ω.

Step by step solution

01

Definition of Internal Resistance

Internal resistance refers to the opposition to the flow of charges offered by the cells and batteries themselves resulting in the generation of heat.

02

Determine the resistance R2

Given data:

  • ε=35.0V
  • I3=4.5A
  • I1=1.5A
  • r = 0

AsR1 andR2 are in parallel, this means they have the same voltage that equals to the voltage across their combination R12. So, the voltage acrossR2 is given using Ohm’s law as:

V2=V1=I1R1

Plug the values,

V2=1.5A5.0Ω=7.5V

Now, the combination is in series with , which means the current is the same for both as:

I3=I12=I1+I2

Plug the values,

l2=l3-l1=4.5-1.5=3.0A

Now, use Ohm’s law to calculate resistance as:

R2=V2l2=7.53.0=2.5Ω

03

Determine the resistance R3

The voltage drop of the battery is due to R3and the combination R12. And as the internal resistance is zero, the Ohm's law can be used as:

ε=I(R12+R3)R3=εl=R12 (1)

The combination can be calculated as:

R12=R1R2R1+R2=5.02.55.0+2.5=1.67Ω

Plug the values in equation (1),

R3=35.0V4.5A-1.67Ω=6.11Ω

Thus, the value of resistances is R2=2.5Ωand R3=6.11Ω.

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

Why does an electric light bulb nearly always burn out just as you turn on the light, almost never while the light is shining?

In a cyclotron, the orbital radius of protons with energy 300keVis 16.0cm. You are redesigning the cyclotron to be used instead for alpha particles with energy 300keV. An alpha particle has chargeq=+2e and mass m=6.64×10-27kg. If the magnetic filed isn't changed, what will be the orbital radius of the alpha particles?

A 140-g ball containing excess electrons is dropped into a 110-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.300 T and direction from east to west. If air resistance is negligibly small, find the magnitude ond direction of the force that this magnetic field exerts on the ball just as it enters the field.

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours .(Ah)A50AhA battery can supply a current of50Afor 1.0h,or25Afor2.0hor for and so on. (a) What total energy can be supplied by a 12-v,60-Ahbattery if its internal resistance is negligible? (b) What volume (in litres) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 900kg/m3.) (c) If a generator with an average electrical power output ofrole="math" localid="1655719210000" 0.45kW is connected to the battery, how much time will be required for it to charge the battery fully?

Small aircraft often have 24 V electrical systems rather than the 12 V systems in automobiles, even though the electrical power requirements are roughly the same in both applications. The explanation given by aircraft designers is that a 24 V system weighs less than a 12 V system because thinner wires can be used. Explain why this is so.

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