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 the circuit shown in Fig. E26.32 both batteries have insignificant internal resistance and the idealized ammeter readin the direction shown. Find the emf E of the battery. Is the polarity shown correct?

Short Answer

Expert verified

the EMF of the Battery is 52.35V

Step by step solution

01

About polarity in electric charges

Polarity is defined as the property in a molecule, or compound through which they are either attracted or repelled by an electric chargebecause of an asymmetrical arrangement of electropositive or electronegative atoms around the center of the species

02

Determine the Current in  the battery

Solution
Find the unknown emf E- The current measured by the ammeter is I = 1.5 A and this is the current flow in
resistance 12 ohms and we will use Kirchhoff's rules in our solution to get E The figure below shows the directions of currentand the traveling paths of loops-
apply the loop rule to get the variables where the loop rule is a statement that the electrostatic force is conservative
Suppose around a loop, measuring potential differences across circuit elements as we go and the algebraic sum oftheese differences is zero when We return to the starting poin use loop 1 and apply equation 26.6 as showin the figure below where the direction of our travel is counterclockwiseto get the current flows in resistance 48 Q which will help us to get the current shows in 15 ohms

V=0-75V+l48Ω+1.5A12Ω

,Visnegativebecausethedirectionoftravelingisfrompositivetonegativeterminalinthebattery(Seefigure26.8a).Theterms(I43ohms)(48ohms)and(1.5A)(12ohms)arepositivebecausethetravelingdirectionisoppositetothedirectionofthe
current Now we can solve the summation for and we will get

l48==1.188A
Now l use the junction rule where the junction rule is based on conservation of electric charge and the current enter:ajunction point is equal to the current flOWs out from this point, so in our circuit, We could get the current flow in , whereenters point a, as shown in the figure below while

l=l48+lili=l-l48Ωli=1.50A-1.188A=0.312A

03

Determine the EMF of the battery

Again apply the loop rule

Let the outer loop 2 and apply equation 26.6 as shown in the figure below
where the direction of our travel is counterclockwise
:V=0-75V+li15Ω+i+1.5A12Ω=0

,V is negative because the direction of traveling is from positive to negative terminal in the battery is positive because the direction of traveling is from negative to positive terminal in the batteryThe terms (IE)(15 ohms) and (1.5 A) (12 ohms) are positive because the traveling direction is opposite to the direction of thecurrent in each resistor (See ?gure 26.8b )- Now we can solve the summation for and we will get
The emf is positive, so the polarity in the figure is right

E=52.35V

Therefore the EMF of the Battery is 52.35V

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

Question: A conducting sphere is placed between two charged parallel plates such as those shown in Figure. Does the electric field inside the sphere depend on precisely where between the plates the sphere is placed? What about the electric potential inside the sphere? Do the answers to these questions depend on whether or not there is a net charge on the sphere? Explain your reasoning.

A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction pependicurlar to its original direction (Fig. E27.24). The beam travels a distance of 1.10 cm while in the field. What is the magnitude of the magnetic field?

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer’s heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

A typical small flashlight contains two batteries, each having an emf of1.5V, connected in series with a bulb having resistance17Ω. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for1.5hwhat is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

A resistor with resistance Ris connected to a battery that has emf 12.0 V and internal resistance r=0.40Ω. For what two values of R will the power dissipated in the resistor be 80.0 W ?

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