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Chapter 19: Problem 1

(I) Calculate the terminal voltage for a battery with an internal resistance of 0.900 \(\Omega\) and an emf of 6.00 V when the battery is connected in series with (\(a\)) a 71.0-\(\Omega\) resistor, and (\(b\)) a 710-\(\Omega\) resistor.

Short Answer

Expert verified
The terminal voltages are approximately 5.93 V for the 71-\(\Omega\) resistor and 5.99 V for the 710-\(\Omega\) resistor.

Step by step solution

01

Understanding the Problem

We need to calculate the terminal voltage of a battery with a given electromotive force (emf) and internal resistance when connected to resistors of different resistances. Terminal voltage is the voltage output of a battery at its terminals.
02

Formula for Terminal Voltage

The terminal voltage \( V \) of a battery can be calculated using the formula: \( V = \text{emf} - I \times r \), where \( \text{emf} \) is the electromotive force, \( I \) is the current through the circuit, and \( r \) is the internal resistance of the battery.
03

Calculate Current for Each Case

For each resistor, calculate the current \( I \) using Ohm's Law: \( I = \frac{\text{emf}}{R + r} \), where \( R \) is the resistance of the external resistor and \( r \) is the internal resistance.
04

Current Calculation for 71.0 \( \Omega \) Resistor

For \( R = 71.0 \Omega \), calculate the current: \( I = \frac{6.00}{71.0 + 0.900} \approx 0.083 \) A.
05

Terminal Voltage with 71.0 \( \Omega \) Resistor

Substitute \( I \) and \( r \) into the terminal voltage formula: \( V = 6.00 - 0.083 \times 0.900 \approx 5.93 \) V.
06

Current Calculation for 710 \( \Omega \) Resistor

For \( R = 710 \Omega \), calculate the current: \( I = \frac{6.00}{710 + 0.900} \approx 0.00843 \) A.
07

Terminal Voltage with 710 \( \Omega \) Resistor

Substitute \( I \) into the terminal voltage formula: \( V = 6.00 - 0.00843 \times 0.900 \approx 5.99 \) V.

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Key Concepts

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

Internal Resistance
When discussing batteries and circuits, internal resistance is a vital concept to grasp. Every battery has some resistance to the flow of electric current within itself, known as internal resistance, denoted as \( r \). This resistance is caused by the materials and processes inside the battery, which can impede the free flow of charge.Here鈥檚 why understanding internal resistance is important:
  • It affects the actual voltage output of the battery, reducing the terminal voltage compared to its emf value.
  • High internal resistance can cause the battery to heat up and reduce efficiency.
  • The internal resistance can influence how much current flows through the circuit for a given load.
Internal resistance is usually a small value, like in our example of 0.900 \( \Omega \), but as you can see, it plays an important role in calculating terminal voltage.
Ohm's Law
Ohm's Law is a fundamental principle in electronics, stating that the current \( I \) flowing through a conductor between two points is directly proportional to the voltage \( V \) across the two points, and inversely proportional to the resistance \( R \) of the conductor. The law is represented by the equation: \[ I = \frac{V}{R} \]In the context of this exercise, Ohm's Law allows us to determine the current flowing through a circuit when a battery is connected to an external resistor. For instance:
  • When the external resistance is 71.0 \( \Omega \), the current is calculated by \( I = \frac{6.00}{71.0 + 0.900} \), leveraging both internal and external resistances.
  • Similarly, with 710 \( \Omega \), it becomes \( I = \frac{6.00}{710 + 0.900} \), showcasing how variations in resistance influence current.
Ohm鈥檚 Law simplifies such calculations, offering insight into how voltage, current, and resistance interrelate.
Electromotive Force (emf)
Electromotive force, often shortened to emf, and symbolized as \( \epsilon \), is the total energy supplied per charge by a source to drive a current around a complete circuit. Unlike terminal voltage, which accounts for losses due to internal resistance, emf is the ideal voltage value of a battery when no current flows, or when it鈥檚 not connected to any load.Why is emf important?
  • It's the upper limit of potential energy a battery can provide. This sets a starting point for evaluating actual terminal voltage.
  • Emf helps estimate the theoretical maximum efficiency of a battery since no energy is lost inside the battery when no current is drawn.
  • In calculations, it gives a reference to determine how much voltage "drops" due to internal resistance when current is flowing.
In our exercise, the battery's emf is listed as 6.00 V. It serves as the starting point to determine the real terminal voltages once internal resistance and load are considered during the current flow.

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Most popular questions from this chapter

(II) Two 3.8-\(\mu\)F capacitors, two 2.2-k\(\Omega\) resistors, and a 16.0-V source are connected in series. Starting from the uncharged state, how long does it take for the current to drop from its initial value to 1.50 mA?

(II) A circuit contains a single 250-pF capacitor hooked across a battery. It is desired to store four times as much energy in a combination of two capacitors by adding a single capacitor to this one. How would you hook it up, and what would its value be?

(II) An electric circuit was accidentally constructed using a 7.0-\(\mu\)F capacitor instead of the required 16-\(\mu\)F value. Without removing the 7.0-\(\mu\)F capacitor, what can a technician add to correct this circuit?

(I) (\(a\)) Six 4.8-\(\mu\)F capacitors are connected in parallel. What is the equivalent capacitance? (\(b\)) What is their equivalent capacitance if connected in series?

(II) A battery for a proposed electric car is to have three hundred 3-V lithium ion cells connected such that the total voltage across all of the cells is 300 V. Describe a possible connection configuration (using series and parallel connections) that would meet these battery specifications.
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