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A \(5.0 \mathrm{~A}\) current is set up in a circuit for \(6.0\) min by a rechargeable battery with a \(6.0 \mathrm{~V}\) emf. By how much is the chemical energy of the battery reduced?

Short Answer

Expert verified
The chemical energy of the battery is reduced by 10800 J.

Step by step solution

01

Understanding the Problem

We need to find out the amount of reduction in chemical energy stored in the battery. The key variables given are the current \(I = 5.0 \mathrm{~A}\), the emf of the battery \(V = 6.0 \mathrm{~V}\), and the duration \(t = 6.0 \text{ minutes}\).
02

Convert Time to Seconds

Since power calculations require time in seconds, convert 6.0 minutes into seconds. \[ t = 6.0 \text{ min} \times 60\text{ s/min} = 360 \text{ s} \]
03

Calculate Energy Consumed

Use the formula \(E = V \, I \, t\) to calculate the energy consumed from the battery. \[ E = 6.0 \mathrm{~V} \times 5.0 \mathrm{~A} \times 360 \mathrm{~s} \] Substitute the values and calculate. \[ E = 10800 \mathrm{~J} \]
04

Conclusion

The chemical energy of the battery is reduced by \(10800 \mathrm{~J}\) during this period.

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Key Concepts

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

Electrical Energy
Electrical energy is the energy that results from the movement of electric charges. We commonly refer to this as the energy used by or generated in the operation of electrical devices. One way to quantify electrical energy is by considering how much work is done or energy is transferred when charges move through a potential difference.

In the context of the exercise, the battery in the circuit provides electrical energy. The amount of energy transferred in a given time can be calculated using the formula:
  • \[ E = V \, I \, t \]
Where:
  • \(E\) is the electrical energy,
  • \(V\) represents the potential difference or voltage,
  • \(I\) is the current, and
  • \(t\) is the time.
These elements help in finding out how much energy the battery loses, quantified in Joules (\(J\)), due to chemical reactions that push electrons around the circuit.
Electromotive Force (EMF)
The electromotive force (EMF) of a battery is a crucial concept in understanding how it operates. It denotes the potential difference between the terminals of a battery when no current is flowing. Essentially, EMF represents the battery's ability to supply energy per unit charge and can be considered as the source voltage when the battery is powering a circuit.

In practical terms, EMF isn't exactly the same as the actual voltage across the battery's terminals due to internal resistance. Often, we find that as the battery operates, the internal resistance causes a voltage drop, making the terminal voltage (what we measure with a voltmeter) slightly less than the EMF. However, for an ideal battery or low currents, EMF can be approximated as the terminal voltage.

In the problem, an EMF of \(6.0 \, \mathrm{V}\) signifies that the battery can theoretically provide this much energy to each coulomb of charge flowing through the circuit, emphasizing its role as a driving force for the current.
Circuit Analysis
Circuit analysis involves examining the behavior and interaction of components within an electrical circuit. It allows us to determine the various electrical properties, such as current, voltage, and power, at different points.

To analyze the circuit in the problem, we need to understand how the concepts of electrical energy and EMF interrelate:
  • The given current is \(5.0 \, \mathrm{A}\).
  • The battery’s EMF is \(6.0 \, \mathrm{V}\).
  • Time is converted to seconds, resulting in \(360 \, \mathrm{s}\).
Applying these values to the formula \(E = V \, I \, t\) allows us to determine the total energy consumed by the circuit. For each component in the circuit, such as the battery, understanding these calculations is critical to predict performance, efficiency, and overall functionality.

Ultimately, circuit analysis provides a detailed picture of how energy flows and transforms within the electrical network, closely tying into numerous practical applications like power distribution and device operation.

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Most popular questions from this chapter

The starting motor of a car is turning too slowly, and the mechanic has to decide whether to replace the motor, the cable, or the battery. The car's manual says that the \(12 \mathrm{~V}\) battery should have no more than \(0.020 \Omega\) internal resistance, the motor no more than \(0.200 \Omega\) resistance, and the cable no more than \(0.040 \Omega\) resistance. The mechanic turns on the motor and measures \(11.4 \mathrm{~V}\) across the battery, \(3.0 \mathrm{~V}\) across the cable, and a current of \(50 \mathrm{~A}\). Which part is defective?

Four \(18.0 \Omega\) resistors are connected in parallel across a \(25.0 \mathrm{~V}\) ideal battery. What is the current through the battery?

In an \(R C\) series circuit, emf \(\&=12.0 \mathrm{~V}\), resistance \(R=\) \(1.40 \mathrm{M} \Omega\), and capacitance \(C=1.80 \mu \mathrm{F}\). (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to \(16.0 \mu \mathrm{C}\) ?

The following table gives the electric potential difference \(V_{T}\) across the terminals of a battery as a function of current \(i\) being drawn from the battery. (a) Write an equation that represents the relationship between \(V_{T}\) and \(i\). Enter the data into your graphing calculator and perform a linear regression fit of \(V_{T}\) versus \(i\). From the parameters of the fit, find (b) the battery's emf and (c) its internal resistance. $$ \begin{array}{llllllll} \hline i(\mathrm{~A}): & 50.0 & 75.0 & 100 & 125 & 150 & 175 & 200 \\ V_{T}(\mathrm{~V}): & 10.7 & 9.00 & 7.70 & 6.00 & 4.80 & 3.00 & 1.70 \\ \hline \end{array} $$

Nine copper wires of length \(l\) and diameter \(d\) are connected in parallel to form a single composite conductor of resistance \(R .\) What must be the diameter \(D\) of a single copper wire of length \(l\) if it is to have the same resistance?

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