/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q17DQ The energy that can be extracted... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The energy that can be extracted from a storage battery is always less than the energy that goes into it while it is being charged. Why?

Short Answer

Expert verified

The battery has its own internal resistance, there is always energy dissipation during charging due to its internal resistance.

Step by step solution

01

Power and energy of an electrical circuit

Electrical energy is the product of power multiplied by the length of time it was consumed.

Energy(E)=power(P)xtime(t)Power(P)=voltage(V)xcurrent(I)

02

Determine the power dissipation and energy

The batteryhas internal resistance.

Let r be the internal resistance of the battery and I be the current flowing during charging.

Therefore, the power dissipation to Its internal resistance is

P=I2r

And the total energy dissipation during time role="math" localid="1655782126418" tis

E=I2rt

Therefore, there is always energy dissipation during charging due to its internal resistance.

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 91Ó°ÊÓ!

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

An 18-gauge copper wire (diameter 1.02 mm) carries a current

with a current density of 3.2×106Am2. The density of free electrons for

copper is8.5×1028electrons per cubic meter. Calculate (a) the current in

the wire and (b) the drift velocity of electrons in the wire.

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of , the induced emf in the second coil has magnitude 1.65×10-3V. (a) What is the mutual inductance of the pair of coils? (b) If the second coil has 25 turns, what is the flux through each turn when the current in the first coil equals 1.20A? (c) If the current in the second coil increases at a rate of 0.360A/s, what is the magnitude of the induced emf in the first coil?

Question: In the circuit shown in Fig. E26.47 each capacitor initially has a charge of magnitude 3.50 nC on its plates. After the switch S is closed, what will be the current in the circuit at the instant that the capacitors have lost 80.0% of their initial stored energy?

An emf source with E = 120 V, a resistor with R = 80.0 Ω, and a capacitor with C = 4.00 µF are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

Q) A straight, nonconducting plastic wire 8.50 cm long carries a charge density of +175 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 cm directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 cm directly above its center.

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.