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Chapter 4: Problem 22

A \(1.58-\mathrm{V}\) alkaline cell with a \(0.200-\Omega\) internal resistance is supplying \(8.50 \mathrm{~A}\) to a load. (a) What is its terminal voltage? (b) What is the value of the load resistance? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable or inconsistent?

Short Answer

Expert verified
The battery cannot deliver such a high current without a voltage drop exceeding its EMF.

Step by step solution

01

Understand the problem

An alkaline battery is providing a current of 8.50 A to a load. The battery has an internal resistance of 0.200 惟 and an EMF (electromotive force) of 1.58 V. We need to determine the terminal voltage, load resistance, and evaluate the reasonableness of the results.
02

Calculate the Terminal Voltage

The terminal voltage, \( V_t \), of a battery can be found using the equation:\[ V_t = EMF - I \cdot r_i \]where \( I \) is the current, and \( r_i \) is the internal resistance.Substituting the given values: \[ V_t = 1.58 \, ext{V} - 8.50 \, ext{A} \cdot 0.200 \, \Omega \]\[ V_t = 1.58 \, ext{V} - 1.70 \, ext{V} \]\[ V_t = -0.12 \, ext{V} \]
03

Calculate the Load Resistance

Using Ohm's Law, the load resistance \( R_L \) can be calculated as:\[ R_L = \frac{V_L}{I} \]Since the expression for \( V_L \) (load voltage) is the same as the terminal voltage: \[ R_L = \frac{-0.12 \, ext{V}}{8.50 \, ext{A}} \]This yields:\[ R_L = -0.0141 \, \Omega \]
04

Analyze Reasonableness of Results

The calculated terminal voltage is negative, which is physically unreasonable since a battery cannot have a negative voltage value across the load terminals. Additionally, a negative load resistance is not possible. These results suggest something is inconsistent.
05

Identify Unreasonable Assumptions or Inconsistencies

The inconsistency arises from an unreasonable assumption that such a small battery (1.58 V) can supply a current as large as 8.50 A to a load. With the given internal resistance, the battery cannot supply this much current without the internal voltage drop exceeding the EMF.

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Key Concepts

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

terminal voltage
When discussing batteries, the terminal voltage is the voltage output of a battery when it is connected within a circuit. It's critical to understand that terminal voltage is generally less than the electromotive force (EMF) because of the internal resistance within the battery. The formula to calculate terminal voltage is:\[ V_t = EMF - I \cdot r_i \]where:
  • EMF is the electromotive force of the battery, representing the ideal voltage output.
  • I is the current flowing through the circuit.
  • ri is the internal resistance of the battery.
To solve for terminal voltage, subtract the voltage dropped across the internal resistance from the EMF. This results in the actual voltage available to power a load in a circuit.
load resistance
Load resistance is an important concept in circuits that helps determine how much current a component in a circuit draws. It's the resistance that is external to the power source and is connected in series with internal resistance. To find the load resistance, apply Ohm's Law:\[ R_L = \frac{V_L}{I} \]where:
  • RL is the load resistance.
  • VL represents the voltage across the load (equivalent to terminal voltage).
  • I is the current flowing through the load.
In cases where the result for load resistance is negative鈥攁s in the original exercise鈥攊t signals an inconsistency, because resistance cannot be negative in a real-world scenario.
Ohm's Law
Ohm's Law serves as a fundamental principle to relate the voltage, resistance, and current in an electrical circuit. It is given by the formula:\[ V = I \cdot R \]where:
  • V is the voltage across the resistor or load in volts.
  • I is the current flowing through the resistor in amperes (A).
  • R is the resistance in ohms (惟).
Ohm's Law is essential for calculating different parameters in circuits. It is a straightforward relationship that demonstrates how a change in one parameter affects the others, making it a powerful tool for analyzing circuits. In practice, if you know any two quantities, you can calculate the third. This law applies to most electric circuits and helps in determining if calculated results are logical and feasible.
electromotive force
The electromotive force (EMF) is often misunderstood. While commonly referenced in voltages, EMF is a measure of energy provided by a battery or other energy source per unit of electric charge. Unlike terminal voltage, EMF represents the maximum potential difference when no current is flowing.
  • EMF is intrinsic to the source and remains constant in ideal conditions.
  • It describes the capability of the power source to provide energy.
  • EMF is crucial for understanding the potential energy converted into electrical energy by a power source.
It's important to remember that once a circuit is closed and current starts flowing, the internal resistance of the battery will cause the terminal voltage to be less than the EMF. This internal resistance is what causes a drop, thus explaining why the terminal voltage is often observed to be lower than the EMF under operation.

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Most popular questions from this chapter

Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on \(120 \mathrm{~V}\) and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on \(120 \mathrm{~V}\) and has 39 remaining identical bulbs, what is then the operating voltage of each?

(a) What resistance would you put in parallel with a \(40.0-\Omega\) galvanometer having a \(25.0-\mu \mathrm{A}\) sensitivity to allow it to be used as an ammeter that has a full-scale deflection for \(10.0-\mu \mathrm{A} ?\) (b) What is unreasonable about this result? (c) Which assumptions are responsible?

A person with body resistance between his hands of \(10.0 \mathrm{k} \Omega\) accidentally grasps the terminals of a \(20.0-\mathrm{kV}\) power supply. (Do NOT do this!) (a) Draw a circuit diagram to represent the situation. (b) If the internal resistance of the power supply is \(2000 \Omega\), what is the current through his body? (c) What is the power dissipated in his body? (d) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in this situation to be \(1.00 \mathrm{~mA}\) or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning.

Suppose you want to measure resistances in the range from \(\mathbf{1 0 . 0} \Omega\) to \(10.0 \mathrm{k} \Omega\) using a Wheatstone bridge that has \(\frac{R_{2}}{R_{1}}=2.000\). Over what range should \(R_{3}\) be adjustable?

A \(160-\mu \mathrm{F}\) capacitor charged to \(450 \mathrm{~V}\) is discharged through a \(31.2-\mathrm{k} \Omega\) resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is \(2.50 \mathrm{~g}\) and its specific heat is \(1.67 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot{ }^{\circ} \mathrm{C}}\), noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance seem significant?
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