Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 25: Problem 23

When a \(9-\mathrm{V}\) battery is temporarily short-circuited, a \(200-\mathrm{mA}\) current flows. What's the battery's internal resistance?

Short Answer

Expert verified
The battery's internal resistance is \(45 \, \Omega\).

Step by step solution

01

Understand the Given Quantities

In this problem, the given quantities are: The voltage (\(V\)) of the battery, which equals 9 volts and the current (\(I\)) that flows when the battery is short-circuited, which is 200 milliampere or 0.2 Ampere.
02

Apply Ohm's Law

Ohm's law states that \(V = I \cdot R\), where \(R\) is resistance, \(V\) is voltage, and \(I\) is current. Here, we need to find the resistance \(R\). So, we rearrange the equation as \(R = \frac{V}{I}\)
03

Substitute the Given Values

Substitute the given values into the equation from step 2: \(R = \frac{V}{I} = \frac{9 \, \text{volts}}{0.2 \, \text{ampere}}\)
04

Solve for R

After substituting the given values into the equation, solve for \(R\) to find out the battery's internal resistance: \(R = 45 \, \Omega\)

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.

Start your free trial

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

A parallel-plate capacitor has plates of area \(10 \mathrm{cm}^{2}\) separated by a 0.10 -mm layer of glass insulation with resistivity \(\rho=1.2 \times 10^{13} \Omega \cdot \mathrm{m}\) and dielectric constant \(\kappa=5.6 .\) Because of the finite resistivity, charge leaks through the insulation. (a) How can such a leaky capacitor be represented in a circuit diagram? (b) Find the time constant for this capacitor to discharge through its insulation, and show that it depends only on the properties of the insulating material and not on its dimensions.

Show that only half the total energy drawn from a battery in charging an \(R C\) circuit ends up stored in the capacitor. (Hint: What happens to the rest? You'll need to integrate.)

A resistor draws \(1.00 \mathrm{A}\) from an ideal \(12.0-\mathrm{V}\) battery. (a) If an ammeter with \(0.10-\Omega\) resistance is inserted in the circuit, what will it read? (b) If this current is used to calculate the resistance, by what percent will the result be in error?

Show that a capacitor is charged to approximately \(99 \%\) of the applied voltage in five time constants \((5 R C)\)

What's the emf of a battery that delivers \(27 \mathrm{J}\) of energy as it moves 3.0 C between its terminals?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Atoms and Radioactivity

Read Explanation

Medical Physics

Read Explanation

Astrophysics

Read Explanation

Fields in Physics

Read Explanation

Turning Points in Physics

Read Explanation
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