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Chapter 3: Problem 30

Automobile battery If the voltage at the terminals of an automobile battery drops from \(12.3\) to \(9.8\) volts when a \(0.5\) ohm resistor is connected across the battery, what is the internal resistance?

Short Answer

Expert verified
The internal resistance of the automobile battery is approximately 0.127 ohms.

Step by step solution

01

Understanding the Real Battery Model and Given Values

In a real battery model, the battery is considered as a combination of an ideal battery with an internal resistor \( \text{r} \). When a load resistor \( R \) is connected, the battery delivers a current and the terminal voltage drops (due to internal resistance). This can be mathematically expressed as \( V = E - I*r \) where\( V \) is the terminal voltage, \( E \) is the electromotive forces (emf), \( I \) is the current and \( r \) is the internal resistance of the battery. Here, we are given the initial terminal voltage\( V1=12.3 \) volts (when no external resistance is connected, i.e., the emf), the final terminal voltage \( V2=9.8 \) volts (when the load resistance of \(0.5 \) ohm is connected), and the external resistance \( R=0.5 \) ohms.
02

Calculate Current Flowing in the Circuit

When the load resistor \( R \) is connected, the current flowing in the circuit can be calculated using Ohm's law as follows: \( I = \frac{V2}{R} \). Substituting given values: \( I = \frac{9.8}{0.5} = 19.6 \) amperes.
03

Determining the Internal Resistance

The formula for voltage in a real battery can be used to find the internal resistance\( r \) given by the equation: \( r = \frac{E - V2}{I} \). Substituting given values: \( r = \frac{12.3 - 9.8}{19.6} = 0.127 \) ohms.

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Key Concepts

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

Understanding Internal Resistance
When dealing with real batteries, it's crucial to understand that they are not perfect energy sources. Every battery has something called internal resistance.
  • Think of it as a tiny resistance inside the battery itself.
  • This internal resistance is part of what reduces the voltage actually available at the terminals.
Let's imagine a battery like a pump that pushes water through a pipe. If the pipe is narrow (high resistance), less water (or electrical current, in the case of a battery) will come through.
This is similar to how internal resistance works. When a battery is connected to a circuit, its internal resistance uses up some of the energy.
So, the actual voltage you measure across the terminals is lower than the battery鈥檚 emf (or ideal voltage without any load connected).
This concept is crucial when performing calculations that involve real-world circuits and batteries.
Explaining Terminal Voltage
The terminal voltage refers to the voltage output that you measure across the terminals of a battery.
It is effectively the usable voltage after accounting for losses due to the internal resistance of the battery.
  • It is important to note that terminal voltage decreases when the battery is put under load.
  • This decrease happens because the internal resistance "eats" up some voltage, causing what you measure at the terminals to be lower than the natural emf of the battery.
For instance, if you have a battery with a natural emf of 12.3 volts and notice that when a device is connected, the available voltage drops to 9.8 volts, the difference of 2.5 volts is lost inside the battery.
This loss is due to the internal resistance, and this is why terminal voltage is a critical factor in designing and understanding circuits.
Essentials of Current Calculation
Current calculation is a core application of Ohm鈥檚 Law, given by the formula \[ I = \frac{V}{R} \]where:
  • \(I\) is the current flowing through the circuit,
  • \(V\) is the terminal voltage, and
  • \(R\) is the resistance in the circuit.
In practical scenarios like battery connections, this calculation helps us figure out how much current is flowing in the circuit when a specific resistance is applied.
For example, if our terminal voltage is 9.8 volts, and a resistance of 0.5 ohms is connected, we can quickly find the current:\[ I = \frac{9.8}{0.5} = 19.6 \text{ amperes} \]Understanding how to calculate current is crucial as it impacts how devices function or are designed in electronic systems. Knowing the current helps ensure that the components in the circuit can handle the load without overheating.

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Most popular questions from this chapter

Attenuator chain Some important kinds of networks are infinite in extent. Figure \(4.49\) shows a chain of series and parallel resistors stretching off endlessly to the right. The line at the bottom is the resistanceless return wire for all of them. This is sometimes called an attenuator chain, or a ladder network. The problem is to find the "input resistance," that is, the equivalent resistance between terminals \(A\) and \(B\). Our interest in this problem mainly concerns the method of solution, which takes an odd twist and which can be used in other places in physics where we have an iteration of identical devices (even an infinite chain of lenses, in optics). The point is that the input resistance (which we do not yet know - call it \(R\) ) will not be changed by adding a new set of resistors to the front end of the chain to make it one unit longer. But now, adding this section, we see that this new input resistance is just \(R_{1}\) in series with the parallel combination of \(R_{2}\) and \(R\) Use this strategy to determine \(R\). Show that, if voltage \(V_{0}\) is applied at the input to such a chain, the voltage at successive nodes decreases in a geometric series. What should the ratio of the resistors be so that the ladder is an attenuator that halves the voltage at every step? Obviously a truly infinite ladder would not be practical. Can you suggest a way to terminate it after a few sections without introducing any error in its attenuation?

A charge inside a shell * Is the following reasoning correct or incorrect (if incorrect, state the error). A point charge \(q\) lies at an off-center position inside a conducting spherical shell. The surface of the conductor is at constant potential, so, by the uniqueness theorem, the potential is constant inside. The field inside is therefore zero, so the charge experiences no force.

A capacitor consists of two coaxial cylinders of length \(L\), with outer and inner radii \(a\) and \(b .\) Assume \(L \gg a-b\), so that end corrections may be neglected. Show that the capacitance is \(C=\) \(2 \pi \epsilon_{0} L / \ln (a / b)\). Verify that if the gap between the cylinders, \(a-b\), is very small compared with the radius, this result reduces to onethat could have been obtained by using the formula for the parallelplate capacitor.

Equipotentials * A point charge is located in the vicinity of a neutral conducting sphere. Make a rough sketch of a few equipotential surfaces; you need only indicate the qualitative features. How do the surfaces make the transition from very small circles (or spheres, in space) around the point charge to very large circles around the whole system? Explain why there must be points on the surface of the sphere where the electric field is zero.

Synchrotron current * In a 6 gigaelectron-volt \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)\) electron synchrotron, electrons travel around the machine in an approximately circular path 240 meters long. It is normal to have about \(10^{11}\) electrons circling on this path during a cycle of acceleration. The speed of the electrons is practically that of light. What is the current? We give this very simple problem to emphasize that nothing in our definition of current as rate of transport requires the velocities of the charge carriers to be nonrelativistic and that there is no rule against a given charged particle getting counted many times during a second as part of the current.
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