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. E26.34, the6Ωresistor is consuming energy at a rate of =24when the current through it flows as shown. (a) Find the current through the ammeter A. (b) What are the polarity and emf E of the unknown battery,assuming it has negligible internal resistance?

Short Answer

Expert verified

the current through ammeter A is 2A and the volometer reading is 10 ohms

the EMF of the battery is -46V and the polarity is in oppostite direction of EMF

Step by step solution

01

:About polarity

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

02

Determine the  Current through ammeter A

Given
We are given the poWer consumed by 6.0 ohms resistance
Solution
(a) Find the current I through the ammeter A. As shown by figure the ammeter is connected between1.0 ohms and 19.ohms, so it reads the current flow in both these resistors. Hence, the target here is to find the current throughthese resistors.
First, let use the given value of the power to find the current flow is 6.0 ohms-The power consumed by the resistance isrelated to the current flows through it in the next form
P=I2RI=PR

NowletusplugourvaluesforP69andR69intoequation(1)togetthecurrentthrough6.0ohms

I=PRI6Ω=246=2A

Also, let us reduce the two resistors in parallel 20.0 9 and 20.0 9 to make the circuit more simple, where for tworesistorparallel the equivalent resistance of their combination is given by equation 26.3 in the next form

Req=R1R2R1+R2=20×2020+20=10Ω

Therefore the current through ammeter A is 2A and the volometer reading is 10 ohms

03

Determine the polarity and the EMF of the battery 

find the current flows through 19.ohms, use the loop rule, where the loop rule is a statement that the electrostatforce is conservativeSuppose around a loop, measuring potential differences across circuit elements as we go anthe algebraic sum of these differences is zero when we return to the starting poin use the right loop in the ?gure and apply equation 26.6 as shown in the figure below where the direction of our tra~is counterclockwise

V=0-25V+l6Ω+IReq+l19+l1Ω+l63=0

25,Visnegativebecausethedirectionoftravelingisfrompositivetonegativeterminalinthebattery(Seefigure
The terms (I69) (6.0 ohms), I (19.0ohms), I (1.0 ohms) and (I5ohms) (3.0 ohms) are positive because the traveling direction isopposite to the direction of the current (See ?gure 26.8b )-
Note that, the current direction is from the positive terminal of 25 V to up, where a part goes to the right and flow throughReg and a part goes left to 17.0 ohms. Now we can solve the summation for I and we will get
l = 0.233A

Therefore the ammeter A reads 0.233 A

(b) As We discussed in part (a), the current direction is from the positive terminal of 25 V to up, where apart goes to the right and flows through Reg and a part goes left to 17.0 (2, so at this junction point we will apply thejunction rule, where the junction rule is based on conservation of electric charge and the current enters ajunction point isequal to the current shows out from this point, so in our circuit, we could get the current flows in 17.0 ohms, where Current entersthe junction point, while I Re; and I179 flow out from this point
Let us plug our values for I69 and I Reg into equation (2) to get the current flows

l6=l17+lReql17=l6-IRl17=2A-0.233A=1.767A

Now it is the time to use the loop rule again but for the left loop to get 8 and apply equation 26.6 where the direction of ourtravel is counterclockwise

25 ,V is positive because the direction of traveling is from negative to positive terminal in the battery, while 8 is negative
because the direction of traveling is from positive to negative terminal in the battery(See ?gure 26.8a )-

V=0-2A(3)+25V-2A(6)-1.767A(17)-I^-7.767A13)=0

25 ,V is positive because the direction of traveling is from negative to positive terminal in the battery, while is negativebecause the direction of traveling is from positive to negative terminal in the battery(See ?gure 26.8a )-
The terms (2 A) (3.0 ohms), (2 A) (6.0ohms), (1.767 A)(17.0 ohms) and (1.767 A) (13.0 ohsm) are negative because the travelingdirection is the same direction of the current (See ?gure 26.8b ) Now We can solve the summation and we will get
l^=46V
The final result is = 46 V but as E: is negative, then the polarity in the figure should be in the opposite direction where thenegative terminal of the battery is next to 17.ohms

therefore the EMF of the battery is -46V and the polarity is in oppostite direction of EMF

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

The circuit shown in Fig. E25.33 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction) and (b) the terminal voltage Vabof the 16.0-V battery.

Fig. E25.33

You want to produce three 1.00-mm-diameter cylindrical wires,

each with a resistance of 1.00 Ω at room temperature. One wire is gold, one

is copper, and one is aluminum. Refer to Table 25.1 for the resistivity

values. (a) What will be the length of each wire? (b) Gold has a density of1.93×10-4kgm3.

What will be the mass of the gold wire? If you consider the current price of gold, is

this wire very expensive?

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.

Two coils have mutual inductance M=3.25×10-4H. The current in the first coil increases at a uniform rate of 830 A/S. (a) what is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is 0.104Ω. (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has 8.5×1028free electrons per cubic meter, find the average drift speed under the conditions of part (b).

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