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A battery delivering a current of 55.0 \(\mathrm{A}\) to a circuit has a terminal voltage of 23.4 \(\mathrm{V}\) . The electric power being dissipated by the internal resistance of the battery is 34.0 \(\mathrm{W}\) . Find the emf of the battery.

Short Answer

Expert verified
The emf of the battery is 24.016 V.

Step by step solution

01

Understanding Given Data

We are given the following information: current \( I = 55.0 \, \mathrm{A} \), terminal voltage \( V = 23.4 \, \mathrm{V} \), and power dissipated by internal resistance \( P = 34.0 \, \mathrm{W} \). We are asked to find the electromotive force (emf) of the battery \( \varepsilon \).
02

Relate Power and Internal Resistance

We know that the power dissipated by the internal resistance is given by the formula \( P = I^2 r \), where \( r \) is the internal resistance. Rearranging this equation gives us \( r = \frac{P}{I^2} \). Substituting \( P = 34.0 \, \mathrm{W} \) and \( I = 55.0 \, \mathrm{A} \), we find:\[r = \frac{34.0}{(55.0)^2} \, \mathrm{\Omega} = 0.0112 \, \mathrm{\Omega}\]
03

Use Terminal Voltage and Current to Relate Emf

The terminal voltage \( V \) is related to the emf \( \varepsilon \) by the formula \( V = \varepsilon - Ir \). We found \( r = 0.0112 \, \mathrm{\Omega} \) from the previous step. Substituting \( V = 23.4 \, \mathrm{V} \), \( I = 55.0 \, \mathrm{A} \), and \( r = 0.0112 \, \mathrm{\Omega} \) into the equation:\[23.4 = \varepsilon - 55.0 \times 0.0112\]
04

Solve for the Emf of the Battery

Rearrange the equation from Step 3 to solve for \( \varepsilon \):\[\varepsilon = 23.4 + 55.0 \times 0.0112 = 24.016 \, \mathrm{V}\]
05

Conclude the Calculation

Therefore, the emf of the battery is approximately 24.016 V.

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Key Concepts

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

Terminal Voltage
The terminal voltage of a battery is a crucial concept when discussing how a battery powers a circuit. It's the voltage output that you can measure across the terminals of the battery, which translates to the energy difference per unit charge between the terminals. This is often less than the battery's electromotive force (EMF) due to energy losses within the battery.

In the case of a battery delivering current to a circuit, the actual voltage can be lower than the EMF because of the voltage drop across the internal resistance. By understanding the terminal voltage, you are essentially seeing how effective the battery is at transferring energy to the external components. This effectiveness can be affected by various factors, including the battery's internal resistance as well as the current flowing through it. Proper management of these factors ensures optimal performance of the battery and the entire electrical circuit.
Internal Resistance
Internal resistance is a very important aspect of a battery's design and functionality. It refers to the inherent resistance to the flow of charge within the battery. This resistance impacts the performance and efficiency of the battery significantly.

The internal resistance of a battery causes heat generation when a current flows through it, resulting in power dissipation as heat rather than useful electrical power. In our example, we calculate the internal resistance by using the formula for power dissipated by the internal resistance: \(P = I^2r\), where \(P\) is power, \(I\) is current, and \(r\) is internal resistance. Solving the given data: \(r = \frac{34.0}{55.0^2}\Omega = 0.0112\Omega\).

By managing the internal resistance, a battery can maintain its efficiency and prolong its operational life, ensuring it can deliver its stored energy effectively.
Electric Power
Electric power is the rate at which electrical energy is transferred by an electric circuit. Understanding electric power within a battery and its circuit is key to comprehending how devices are powered and how much energy they consume.

In this context, electric power describes both how power is supplied by the battery to the circuit and how power is dissipated as heat due to internal resistance. For power dissipated by internal resistance, the formula \(P = I^2r\) quantifies how much energy per second is lost within the battery rather than being delivered to the circuit.

Knowing how much power a device uses can help us understand the battery's efficiency and what percentage of its power is actually being used to perform useful work versus the amount lost as waste heat.
Current
Current is a fundamental concept in electronics and electricity. It's the flow of electric charge around a circuit and is measured in amperes (A). The current within a circuit defines how much charge passes through a point per second.

In our battery example, the current is given as 55.0 A. It describes how much electricity flows from the battery through the circuit. The amount of current flowing impacts other factors like voltage and resistance and helps in calculating power and energy efficiency of a system.

Managing and optimizing current flow in a circuit can lead to greater efficiency and energy savings. Moreover, ensuring the current is at the proper level can prevent overheating and reduce wear and tear on components, extending their lifespan.

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Most popular questions from this chapter

The rear window defogger of a car consists of thirteen thin wires (resistivity \(=88.0 \times 10^{-8} \Omega \cdot \mathrm{m}\) ) embedded in the glass. The wires are connected in parallel to the \(12.0-\mathrm{V}\) battery, and each has a length of 1.30 \(\mathrm{m} .\) The defogger can melt \(2.10 \times 10^{-2} \mathrm{kg}\) of ice at \(0^{\circ} \mathrm{C}\) into water at \(0^{\circ} \mathrm{C}\) in two minutes. Assume that all the power delivered to the wires is used immediately to melt the ice. Find the cross-sectional area of each wire.

A fax machine uses 0.110 A of current in its normal mode of operation, but only 0.067 \(\mathrm{A}\) in the standby mode. The machine uses a potential difference of 120 \(\mathrm{V}\) . In one minute (a) how much more charge passes through the machine in the normal mode than in the standby mode, and (b) how much more energy is used?

A battery has an internal resistance of 0.50\(\Omega .\) A number of identical light bulbs, each with a resistance of \(15 \Omega,\) are connected in parallel across the battery terminals. The terminal voltage of the battery is observed to be one-half the emf of the battery. How many bulbs are connected?

The rms current in a \(47-\Omega\) resistor is 0.50 A. What is the peak value of the voltage across this resistor?

To save on heating costs, the owner of a greenhouse keeps 660 kg of water around in barrels. During a winter day, the water is heated by the sun to \(10.0^{\circ} \mathrm{C}\) . During the night the water freezes into ice at \(0.0^{\circ} \mathrm{C}\) in nine hours. What is the minimum ampere rating of an electric heating system \((240 \mathrm{V})\) that would provide the same heating effect as the water does?

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