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

Question: You connect a battery, resistor, and capacitor as in Fig. 26.20a, where R = 12.0 Ω and C = 5.00 x 10-6 F. The switch S is closed at t = 0. When the current in the circuit has a magnitude of 3.00 A, the charge on the capacitor is 40.0 x 10-6 C. (a) What is the emf of the battery? (b) At what time t after the switch is closed is the charge on the capacitor equal to 40.0 x 10-6 C? (c) When the current has magnitude 3.00 A, at what rate is energy being (i) stored in the capacitor, (ii) supplied by the battery

Short Answer

Expert verified

Answer:

(a) The emf on the battery is 44.0 V

(b) After 1.20×10-5sthe switch is closed the charge on the capacitor is equal to 40.0×10-6C

(c) When the current has magnitude then the energy being stored in the capacitor is 24W and supplied by the battery is 132 W

Step by step solution

01

EMF of the battery

Electromotive force (Emf) is the total amount of electrical potential energy a battery can supply before it is connected to an external circuit

02

Current through the circuit

Given,

C=5.00×10-6F, R=12.0Ωand time t=0 .

Firstly, we need to find the emf of the battery when the current has a magnitude of I= 3.00 A and the charge on the capacitor isQ=40.0×10-6C

Therefore, the charge over the capacitor is given by the equation:

Q=Cε(1-e-tτ)

Now, the current in the circuit is given by:

I=dQdt=εRe-tτ

03

Calculating the emf

Dividing the above two equations we get:

QA=CR1-e-tτe-tτ

Multiply byetτetτwe get:

QI=CRetτ-1

Solving for t we get:

t=τlnQICR+1

Substituting the values, we get,

t=(6.00×10-5s)ln40.0×10-6C(3.00A)(12.0Ω)(5.00×10-6F)+1=1.20×10-5s

Now, substitute the values in,

ε=IRetτ

We get,

ε=(3.00A)×(12.0Ω)×e1.20×10-5s6.00×10-5s=44.0V

Therefore, the emf is 44.0 V

04

Energy stored in both capacitors

From the above parts, we see that the current in the circuit has a magnitude of I =3.00 A and the charge on the capacitor is Q=40.0×10-6C, now the time at which the circuit has the above magnitude and charge is given by:

t=1.20×10-5s

Now the energy stored is given by:

role="math" localid="1655816376742" UC=Q22C

Therefore, the rate of energy stored in the capacitor is:

PC=QICPC=(40.0×10-6C)(3.00A)5.00×10-6F=24.0W

Therefore, the rate of energy stored in the capacitor is

Again, the rate of energy supplied by the battery is:

Pε=Iε=(3.00A)(44.0V)=132W

Therefore, the energy supplied by the battery is 132 W

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

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?

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?

Electrons in an electric circuit pass through a resistor. The wire on either side of the resistor has the same diameter.(a) How does the drift speed of the electrons before entering the resistor compare to the speed after leaving the resistor? (b) How does the potential energy for an electron before entering the resistor compare to the potential energy after leaving the resistor? Explain your reasoning.

We have seen that a coulomb is an enormous amount of charge; it is virtually impossible to place a charge of 1 C on an object. Yet, a current of 10A,10C/sis quite reasonable. Explain this apparent discrepancy.

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?

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