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Chapter 27: Problem 53

A car battery with a \(12 \mathrm{~V}\) emf and an internal resistance of \(0.030 \Omega\) is being charged with a current of \(40 \mathrm{~A}\). What are (a) the potential difference \(V\) across the terminals, (b) the rate \(P_{r}\) of energy dissipation inside the battery, and (c) the rate \(P_{\mathrm{cmf}}\) of energy conversion to chemical form? When the battery is used to supply 40 A to the starter motor, what are (d) \(V\) and (e) \(P_{r}\) ?

Short Answer

Expert verified
(a) 10.8 V, (b) 48 W, (c) 432 W, (d) 10.8 V, (e) 48 W.

Step by step solution

01

Calculate the Terminal Voltage (V) during Charging

The terminal voltage when charging can be found using the formula: \( V = ext{emf} - I imes r \), where \( I = 40 \) A is the current and \( r = 0.030 \) \(\Omega\) is the internal resistance. Substituting the values, we get: \( V = 12 \, ext{V} - 40 \, ext{A} \times 0.030 \, \Omega = 10.8 \, ext{V}.\)
02

Calculate the Rate of Energy Dissipation (Pr) during Charging

The rate of energy dissipation inside the battery (\(P_r\)) due to internal resistance is given by: \( P_r = I^2 \times r \). Using \( I = 40 \) A and \( r = 0.030 \) \(\Omega\), we have \( P_r = (40)^2 \times 0.030 = 48 \, \text{W}.\)
03

Calculate the Rate of Energy Conversion to Chemical Form (P_cmf)

The power converting to chemical energy, \(P_{\text{cmf}}\), is the difference between the total power supplied and the power lost to internal resistance: \( P_{\text{cmf}} = V_{\text{emf}} \times I - P_r \). Substituting the known values: \( P_{\text{cmf}} = 12 \, \text{V} \times 40 \, \text{A} - 48 \, \text{W} = 432 \, \text{W}.\)
04

Calculate the Terminal Voltage (V) during Discharging

When the battery is discharging to supply 40 A to the motor, the terminal voltage is recalculated as: \( V = ext{emf} - I \times r = 12 \, ext{V} - 40 \, ext{A} \times 0.030 \, \Omega = 10.8 \, ext{V}.\)
05

Calculate the Rate of Energy Dissipation (Pr) during Discharging

The rate of energy dissipation inside the battery during discharging is found using: \( P_r = I^2 \times r = 48 \, \text{W} \), which is the same as during charging because the current and internal resistance have not changed.

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Key Concepts

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

Potential Difference
The potential difference refers to the voltage across components in an electrical circuit. In a car battery scenario, it is essential to understand how the potential difference affects the charging and discharging processes.
When charging the battery, the potential difference is calculated by considering the battery's electromotive force (emf) and subtracting the voltage drop across its internal resistance due to the charging current. This is expressed as:
  • \( V = \text{emf} - I \times r \)
Here, \( V \) denotes the terminal voltage, \( I \) the current, and \( r \) the internal resistance.
During both charging and discharging, the calculation remains consistent if the current and resistance values stay unchanged, resulting, in this example, in a terminal voltage of \(10.8 \text{ V}.\)
Energy Dissipation
Energy dissipation is the process where energy is lost in a circuit, often as heat, due to resistance. In a battery, energy dissipation primarily occurs through its internal resistance.
This is quantitatively described by the formula:
  • \( P_r = I^2 \times r \)
Where \( P_r \) is the rate of energy dissipation, \( I \) represents the current flowing through the resistance, and \( r \) is the internal resistance of the battery.
Calculating this during the charging process, we find that the battery dissipates \(48 \text{ W}\), a result that remains unchanged during discharging, provided no alterations in current or internal resistance occur.
Chemical Energy Conversion
Inside a battery, chemical energy conversion is as crucial as the electrical processes. When charging, electrical energy transforms into chemical energy stored in the battery, increasing its capacity to perform work.
The rate at which this conversion occurs can be calculated by subtracting the power lost to internal resistance from the total supplied power:
  • \( P_{\text{cmf}} = V_{\text{emf}} \times I - P_r \)
Here, \( P_{\text{cmf}} \) denotes the rate of chemical energy conversion, \( V_{\text{emf}} \) the emf of the battery, \( I \) the current, and \( P_r \) the energy dissipation rate.
For the system at hand, this results in a conversion rate of \(432 \text{ W}\), essentially meaning 432 watts are effectively used for charging the battery.
Internal Resistance
Internal resistance is an inherent property of all batteries, representing the opposition to current flow within the battery itself. This can lead to energy loss as heat when a current passes through the battery.
Its effect is evident when we calculate the voltage drop across a battery's terminals while charging or discharging. Although small, it plays a pivotal role in determining a battery's efficiency and is crucial when dealing with high currents.
Understanding internal resistance helps us adjust for the potential difference across external terminals, ensuring the correct calculations are made for both power dissipation and chemical energy conversion.

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Most popular questions from this chapter

Displays two circuits with a charged capacitor that is to be discharged through a resistor when a switch is closed. In Fig. 27-38a, \(R_{1}=20.0 \Omega\) and \(C_{1}=5.00 \mu \mathrm{F}\). In Fig. \(27-38 b, R_{2}=10.0 \Omega\) and \(C_{2}=8.00 \mu \mathrm{F}\). The ratio of the initial charges on the two capacitors is \(q_{02} / q_{01}=1.75\). At time \(t=0\), both switches are closed. At what time \(t\) do the two capacitors have the same charge?

A \(5.7\) A current is set up in a circuit for \(15.0 \mathrm{~min}\) by a rechargeable battery with a \(6.0 \mathrm{~V}\) emf. By how much is the chemical energy of the battery reduced?

Six \(18.0 \Omega\) resistors are connected in parallel across a \(12.0 \mathrm{~V}\) ideal battery. What is the current through the battery?

An isolated charged capacitor may gradually discharge as charge leaks from one plate to the other through the intermediate material, as if it were discharging through an external resistor. (a) What is the resistance of such an external resistor if a \(2.00 \mu \mathrm{F}\) capacitor's potential difference decreases to \(60.0 \%\) of its initial value of \(50.0 \mathrm{~V}\) in \(2.40 \mathrm{~d} ?\) (b) What is the corresponding loss of potential energy in that time interval? (c) At the end of that interval, at what rate is the capacitor losing potential energy?

A \(1.0 \mu \mathrm{F}\) capacitor with an initial stored energy of \(0.60 \mathrm{~J}\) is discharged through a \(1.0 \mathrm{M} \Omega\) resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time \(t\), (c) the potential difference \(V_{c}\) across the capacitor, (d) the potential difference \(V_{R}\) across the resistor, and (e) the rate at which thermal energy is produced in the resistor.
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