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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

Grounding a shell ** A conducting spherical shell has charge \(Q\) and radius \(R_{1}\). A larger concentric conducting spherical shell has charge \(-Q\) and radius \(R_{2}\). If the outer shell is grounded, explain why nothing happens to he charge on it. If instead the inner shell is grounded, find its final charge.

Capacitance of raindrops \(N\) charged raindrops with radius \(a\) all have the same potential. Assume that they are far enough apart so that the charge distribution on each isn't affected by the others (that is, it is spherically symmetric). What is the total capacitance of this system? How does this capacitance compare with the capacitance in the case where the drops are combined into one big drop?

Tapered rod Two graphite rods are of equal length. One is a cylinder of radius \(a\). The other is conical, tapering (or widening) linearly from radius \(a\) at one end to radius \(b\) at the other. Show that the end-to-end electrical resistance of the conical rod is \(a / b\) times that of the cylindrical rod. Hint: Consider the rod to be made up of thin, disk-like slices, all in series. (This result is actually only an approximate one, valid in the limit where the taper is slow. See Problem \(4.6\) for a discussion of this.)

Image charges for two planes : A point charge \(q\) is located between two parallel infinite conducting planes, a distance \(d\) from one and \(\ell-d\) from the other. Where should image charges be located so that the electric field is everywhere perpendicular to the planes?

Transatlantic telegraphic cable *? The first telegraphic messages crossed the Atlantic in 1858 , by a cable \(3000 \mathrm{~km}\) long laid between Newfoundland and Ireland. The conductor in this cable consisted of seven copper wires, each of diameter \(0.73 \mathrm{~mm}\), bundled together and surrounded by an insulating sheath. (a) Calculate the resistance of the conductor. Use \(3.10^{-8}\) ohmmeter for the resistivity of the copper, which was of somewhat dubious purity. (b) A return path for the current was provided by the ocean itself. Given that the resistivity of seawater is about \(0.25\) ohm-meter, see if you can show that the resistance of the ocean return would have been much smaller than that of the cable. (Assume that the electrodes immersed in the water were spheres with radius, say, \(10 \mathrm{~cm}\).)
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