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

In the circuit in Fig. E25.47, find (a) the rate of conversion of internal (chemical) energy to electrical energy within the battery; (b) the rate of dissipation of electrical energy in the battery; (c) the rate of dissipation of electrical energy in the external resistor.

Questions: When a thunderstorm is approaching, sailors at sea sometimes observe a phenomenon called “St. Elmo’s fire,” a bluish flickering light at the tips of masts. What causes this? Why does it occur at the tips of masts? Why is the effect most pronounced when the masts are wet? (Hint: Seawater is a good conductor of electricity.)

An alpha particle (a He nucleus containing two protons and two neutrons and having a mass of 6.64×10-7kg) travelling horizontally at 35.6km/senter a uniform, vertical,1.80-T magnetic field.(a) What is the diameter of the path followed by this alpha particle? (b) what effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) explain why the speed of the particle does not change even though an unbalanced external force acts on it.

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 ?

A 10.0cm long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

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