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. E26.17,the voltage across the2Ωresistor is 12 V. What are the emf of the battery and the current through the6Ωresistor?

Short Answer

Expert verified

the EMF of the battery is 18.00V

The Current in 6 ohms resistor is 3A

Step by step solution

01

Step 1:About Ohms law

Ohm's law states that thecurrentthrough aconductorbetween two points is directlyproportionalto thevoltageacross the two points. Introducing the constant of proportionality, theresistance, one arrives at the usual mathematical equation that describes this relationship:

Given
Let us label the next on the

R1=1.0ΩR2=2.0ΩandR3=6.0Ω

-R2isV2=12,V

The voltage across R2 is Vg = 12.0 V

02

Determine the EMF of the battery and the cureent through 6 ohms resistor

Solution

thecurrentI3andtheemfofthebattery8-ThetworesistorsR1andR2areinseriesandthecurrent

through these two resistors is the same through every resistor and equals the current through the combination, and We could

calculate this current using Ohm's law

R1andR2l12=l1=II2=V2R2122=6.0A

AsshOWnbythe?gure,thecombinationR12isinparallelwithR3andthepotentialdifferenceVacrossresistorsconnected
in the parallel is the same for every resistor and equals the potential difference across the combination as next
V12=V3=V123
The combination voltage V123isthevoltagedropofthebatteryandi^=v123 as the internal resistance of the battery is zero,therefore the emf of the battery equals where We can get the combination voltage Vlz by Ohm's law in the next form
m we parallel IS me same nor every resuswr ano equals me pomenual omerence across me comomauon as The combination voltage V123 is the voltage drop of the

battery and as the internal resistance of the battery is zero,

therefore the emf of the battery equalsV12-V3=V123where We can get the combination voltage V12 by Ohm's law in the

next form

V12-V3=V123i^=v123=I12(R1+R2)=6(1+2)=18V

Therefore the EMF of the battery is 18.00V

Now We can use the value of V to get the currentI3 where we can plug our values for V3 and R3 into Ohm's law to get I3

I3=V3R3=186=3A

The Current in 6 ohms resistor is 3A

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

If a “75-W” bulb (see Problem 25.35) is connected across a 220-V potential difference (as is used in Europe), how much power does it dissipate? Ignore the temperature dependence of the bulb’s resistance.

An idealized voltmeter is connected across the terminals of a15.0-Vbattery, and arole="math" localid="1655719696009" 75.0-Ω appliance is also connected across its terminals. If the voltmeter reads11.9V (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

Questions: A conductor that carries a net charge has a hollow, empty cavity in its interior. Does the potential vary from point to point within the material of the conductor? What about within the cavity? How does the potential inside the cavity compare to the potential within the material of the conductor?

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.)

The battery for a certain cell phone is rated at3.70V.According to the manufacturer it can produce3.15×104Jof electrical energy, enough for 2.25hof operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

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