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A car battery with a \(12 \mathrm{~V} \mathrm{emf}\) and an internal resistance of \(0.040 \Omega\) is being charged with a current of \(50 \mathrm{~A}\). What are (a) the potential difference \(V\) across the terminals, (b) the rate \(P_{\underline{r}}\) of energy dissipation inside the battery, and (c) the rate \(P_{\mathrm{emf}}\) of energy conversion to chemical form? When the battery is used to supply 50 A to the starter motor, what are (d) \(V\) and (e) \(P_{r} ?\)

Short Answer

Expert verified
(a) 14 V; (b) 100 W; (c) 600 W; (d) 10 V; (e) 100 W.

Step by step solution

01

Calculate Potential Difference When Charging

To calculate the potential difference across the terminals when charging, use the formula \( V = ext{emf} + Ir \), where \( I \) is the current and \( r \) is the internal resistance. Substitute the given values: \( V = 12 + 50 \times 0.040 \). Simplifying this gives \( V = 12 + 2 = 14 ext{ V} \).
02

Determine Energy Dissipation Rate Inside Battery When Charging

The rate of energy dissipation inside the battery, when charging, is calculated using the formula \( P_r = I^2 r \). Substitute the given values: \( P_r = 50^2 \times 0.040 \). Calculate this to get \( P_r = 100 ext{ W} \).
03

Calculate Rate of Energy Conversion to Chemical Form

The rate of energy conversion to chemical form, \( P_{emf} \), is calculated using \( P_{emf} = ext{emf} \times I \). Substitute the given values: \( P_{emf} = 12 \times 50 \). This results in \( P_{emf} = 600 ext{ W} \).
04

Calculate Potential Difference When Discharging

When using the battery to supply the starter motor, the potential difference across the terminals is given by \( V = ext{emf} - Ir \) since the battery is discharging. Substitute the values: \( V = 12 - 50 \times 0.040 = 12 - 2 = 10 ext{ V} \).
05

Determine Energy Dissipation Rate Inside Battery When Discharging

The rate of energy dissipation inside the battery when discharging is calculated in the same way as when charging: \( P_r = I^2 r \). Using the same current and resistance, \( P_r = 100 ext{ W} \).

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle used in circuit analysis. It describes the relationship between voltage, current, and resistance. The law is given by the equation:
  • \( V = IR \)
where:
  • \( V \) is the voltage across the electrical component in volts,
  • \( I \) is the current flowing through the component in amperes,
  • \( R \) is the resistance of the component in ohms.
Ohm's Law applies to many practical scenarios like this exercise, where it is used to find the potential difference across the terminals of a battery. In both charging and discharging states, being able to apply Ohm's Law helps in determining how the voltage changes based on internal resistance and current flow. Remember, when a device is being charged, the potential difference increases by the internal voltage drop (\( Ir \)), whereas when it is discharging, it decreases.
Electric Power
Electric power quantifies the rate of energy transfer within an electrical circuit. It is calculated using several formulas, but a key equation when considering internal resistance and current is:
  • \( P = I^2 r \)
In this exercise, the power dissipated as heat inside the battery is calculated this way, both when charging and discharging. The formula \( P = I^2 r \) highlights how power dissipation depends directly on the square of the current and internal resistance.Another important formula for electric power involves the electromotive force (emf):
  • \( P_{emf} = E \times I \)
where \( E \) is the emf of the battery. This allows us to determine how much energy per second is being transformed into chemical energy inside the battery during charging. Understanding these concepts of power helps in analyzing the efficiency and functioning of electrical devices.
Internal Resistance
Internal resistance is an inherent part of every electrical power source, like batteries. It is a measure of how much the battery resists the flow of current internally:
  • Affects both charging and discharging processes.
  • In the formula \( V = \text{emf} ± Ir \), internal resistance causes a decrease or increase in terminal voltage based on the direction of current flow.
With a specified internal resistance value, such as \( 0.040 \Omega \) in this case, we assess how it impacts power loss (or energy dissipation) and adjust voltage calculations accordingly.In practical terms, higher internal resistance implies greater losses, which can lead to reduced efficiency. Understanding internal resistance is critical for designing efficient power delivery systems and for predicting how a power source performs under different loads.
Battery Charging and Discharging
Battery charging and discharging involve the flow of current in and out of the battery, affecting both voltage and energy conversion inside the battery.
  • During charging, current flows into the battery, increasing its stored energy.
  • The formula \( V = \text{emf} + Ir \) calculates terminal voltage while charging.
  • Energy is converted into chemical energy inside the battery at a rate of \( P_{emf} = E \times I \).
  • When discharging, current flows out, reducing stored energy and terminal voltage is computed as \( V = \text{emf} - Ir \).
Understanding both processes is key to managing battery performance. Rates of charging and discharging affect the efficiency and lifespan of batteries. Proper analysis ensures reliability in applications like vehicles where batteries need to be both charged and rapidly discharged to start engines.

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Most popular questions from this chapter

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

A \(120 \mathrm{~V}\) power line is protected by a 15 A fuse. What is the maximum number of 500 W lamps that can be simultaneously operated in parallel on this line without "blowing" the fuse because of an excess of current?

A capacitor with initial charge \(q_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first one-third of its charge and (b) two-thirds of its charge?

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a \(0.220 \mu \mathrm{F}\) capacitor. The capacitor is charged to \(5.00 \mathrm{~V},\) then discharged through the resistor. The time for the potential difference across the plates to decrease to \(0.800 \mathrm{~V}\) is measured by a clock inside the game. If the range of discharge times that can be handled effectively is from \(10.0 \mu \mathrm{s}\) to \(6.00 \mathrm{~ms}\) what should be the (a) lower value and (b) higher value of the resistance range of the resistor?

When resistors 1 and 2 are connected in series, the equivalent resistance is \(16.0 \Omega .\) When they are connected in parallel, the equivalent resistance is \(3.0 \Omega .\) What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

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