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

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V. (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

Short Answer

Expert verified
Internal resistance is 0.36 ohms, and emf is 8.94 volts.

Step by step solution

01

Understand the Problem

We have a battery with two scenarios of potential difference and current. We need to find the internal resistance and the electromotive force (emf) of the battery. Use the formula that links these concepts: \[ V = \mathcal{E} - Ir \] for Scenario 1, and \[ V = \mathcal{E} + Ir \] for Scenario 2.
02

Write Equations for Each Scenario

For Scenario 1, with potential difference 8.40 V and current 1.50 A:\[ 8.40 = \mathcal{E} - 1.50r \]For Scenario 2, with potential difference 10.20 V and current 3.50 A (reverse direction, meaning potential difference increases):\[ 10.20 = \mathcal{E} + 3.50r \]
03

Solve for Internal Resistance (r)

Subtract the first equation from the second to eliminate \(\mathcal{E}\):\[ (10.20 - 8.40) = (\mathcal{E} + 3.50r) - (\mathcal{E} - 1.50r) \]\[ 1.80 = 5r \]Solve for \( r \):\[ r = \frac{1.80}{5} = 0.36 \text{ ohms} \]
04

Solve for Emf (\(\mathcal{E}\))

Use the value of \( r = 0.36 \) ohms in one of the original equations to find \(\mathcal{E}\). We'll use the first equation:\[ 8.40 = \mathcal{E} - 1.50(0.36) \]\[ 8.40 = \mathcal{E} - 0.54 \]Solve for \(\mathcal{E}\):\[ \mathcal{E} = 8.40 + 0.54 = 8.94 \text{ volts} \]
05

Compile the Solution

The internal resistance of the battery is 0.36 ohms, and the emf of the battery is 8.94 volts.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromotive Force (emf)
Electromotive Force, or emf, is a crucial concept in understanding how batteries, or cells, function within an electrical circuit. It represents the energy provided by the battery per coulomb of charge. Unlike the potential difference, which can vary, the emf is considered a constant value representing the maximum potential that the source can provide.
In essence, emf is the ability of a battery or generator to push electrons around a circuit, akin to the force that makes a pump move water through pipes. This is why it's often labeled as a force—even though it technically isn't one in the traditional physics sense.
  • Measured in volts (V), emf is denoted by the symbol \( \mathcal{E} \).
  • It's calculated using the equation: \( \mathcal{E} = V + Ir \), where \( V \) is the terminal voltage, \( I \) is the current, and \( r \) is the internal resistance.
  • In the context of the problem, the emf tells us how much voltage the battery can ideally provide without any losses.
  • Using our calculated internal resistance \( r = 0.36 \text{ ohms} \), the emf can be found by rearranging and solving the formula with given potential differences and currents.
Potential Difference
Potential difference, commonly known as voltage, is the amount of energy needed to move a charge from one point to another in a circuit. It's the key reason why electrons move through a wire, leading to the flow of electricity.
In practical terms, potential difference is what causes an electrical appliance to work when connected to a power source. It can be thought of as the pressure that pushes charges through a circuit, analogous to how water pressure pushes water through a pipe.
  • The potential difference is measured in volts (V).
  • It's often denoted by the symbol \( V \), but don't confuse it with emf, which is the maximum potential difference provided by a battery.
  • In our original problem, potential difference values change when the direction and magnitude of the current change, stating values of 8.40 V and 10.20 V.
  • These changes depend on how the internal resistance of the battery affects the flow of current, emphasizing that potential difference is not always constant.
Ohm's Law
Ohm's Law is an essential principle in electrical circuits, providing a relationship between current, voltage, and resistance. It is fundamental for analyzing circuits and solving problems related to electrical behavior.
The law states that the current flowing through most materials is directly proportional to the voltage across it and inversely proportional to its resistance.
  • Expressed mathematically, Ohm's Law is \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
  • In our problem, internal resistance is a crucial aspect. It affects how potential difference and current behavior changes inside the battery. Therefore, influencing the overall voltage output across the terminals.
  • For batteries, Ohm's Law helps define how well they can deliver current under different conditions by factoring in internal resistance.
  • Understanding Ohm's Law is vital for solving any circuit problem as it connects the three primary electrical properties: voltage, current, and resistance, making it easier to derive unknowns from known quantities.

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 \(3.20 \times 10{^6} A/m{^2}\). The density of free electrons for copper is \(8.5 \times 10^{28}\) electrons per cubic meter. Calculate (a) the current in the wire and (b) the drift velocity of electrons in the wire.

Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500 V and produce currents of 80 mA (or even larger). A typical pulse lasts for 10 ms. What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is 0.104\(\Omega\). (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has \(8.5 \times 10{^2}{^8}\) free electrons per cubic meter, find the average drift speed under the conditions of part (b).

A copper transmission cable 100 km long and 10.0 cm in diameter carries a current of 125 A. (a) What is the potential drop across the cable? (b) How much electrical energy is dissipated as thermal energy every hour?

A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at 25 V for a very short time, usually about 3.0 ms. (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

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