# Unraveling the Mystery of EMF Internal Resistance

• Ryaners
In summary, the problem involves determining the resistance per meter of a long piece of wire using a battery, a voltmeter, and an ammeter. The solution involves using the equations ##V_{ab} = \varepsilon - Ir## and ##R = \frac {V}{I}##, assuming that the battery is ideal and has no internal resistance. The confusion arises from the fact that when the wire is connected to the battery, there is a current flowing and the voltage across the terminals is not equal to the EMF of the battery. However, since the problem states that the voltmeter and ammeter are ideal, the internal resistance of the battery is disregarded and the voltage is assumed to be equal to the
Ryaners
I'm having a difficult time understanding why the internal resistance of the EMF source is disregarded in this problem.

1. Homework Statement

You are asked to determine the resistance per meter of a long piece of wire. You have a battery, a voltmeter, and an ammeter. You put the leads from the voltmeter across the terminals of the battery and the meter reads 12.9V. You cut of a 20m length of the wire and connect it to the battery in series with the ammeter, which reads 6.9A. You then cut off a 40m length of wire and connect it to the battery in series with the ammeter, which reads 4.2A.

Assume the voltmeter and ammeter are ideal.

##R## = resistance of one meter of wire
##I_1## = 6.9 A
##I_2## = 4.2 A
##\varepsilon## = 12.9 V
##r## = internal resistance of battery

Solution from the book:

$$40R - 20R = \frac {V}{I_2} - \frac {V}{I_1} \\ \Rightarrow R = \frac {V}{20} \left( \frac {1}{I_2} - \frac {1}{I_1} \right) \\= \frac {12.9 V}{20} \left( \frac {1}{4.2 A} - \frac {1}{6.9 A} \right) \\= 0.060 \Omega$$

Why I'm confused:

I understand that, when attaching the (ideal) voltmeter across the battery terminals, the resistance is infinite & no current flows, therefore ##V_{ab} = \varepsilon - Ir = \varepsilon## (no current ∴ no dissipation due to internal resistance ∴ the terminal voltage = the EMF ##\varepsilon##).
BUT, when a piece of wire is connected to the terminals, a current will flow, and surely ##Ir \neq 0## and ##V_{ab} \neq \varepsilon##? If I'm right, that means the total voltage should be different in each case (##I## changes, and therefore so would ##Ir##).
But I'm a newbie & trust the book more :) Why is the voltage the same in all 3 setups, even though the total resistance in the circuit changes when the length of wire is changed? Does it have something to do with short circuits? I don't see how connecting a wire with its own resistance to the battery terminals is any different to building a small circuit with a resistor, where the internal resistance of the battery would be relevant (going by the other examples in the book).

Did they give you the internal resistance in the problem? If they didn't, you are supposed to assume that the battery is ideal.

If you include the "r" in the calculations, maybe it just cancels out? It would be strange for them to say the voltmeter and ammeter are ideal, and not mention the voltage source...

EDIT/ADD -- Yeah, that looks like what happened. Just include the two "r" terms on the LHS of your first equation -- what happens to them?

Ryaners

## 1. What is EMF internal resistance?

EMF internal resistance is the resistance that a source of EMF (Electromotive Force) has against the flow of current within itself. It is caused by the internal components of the source, such as the battery or generator.

## 2. How does EMF internal resistance affect circuits?

EMF internal resistance can affect circuits by reducing the overall voltage and current available for the circuit to work with. This can lead to a decrease in the performance and efficiency of the circuit.

## 3. What factors can affect EMF internal resistance?

Some factors that can affect EMF internal resistance include the type of source (battery, generator, etc.), the materials used in the source, the temperature, and the age of the source. These factors can cause variations in the internal resistance of the source.

## 4. How can we measure EMF internal resistance?

EMF internal resistance can be measured by using a multimeter and measuring the voltage drop across the source when a known current is passed through it. The internal resistance can then be calculated using Ohm's Law (R = V/I).

## 5. Why is it important to understand EMF internal resistance?

Understanding EMF internal resistance is important because it can impact the performance and efficiency of circuits and electrical devices. It can also help in troubleshooting issues with circuits and determining the lifespan of a source. Additionally, it is a fundamental concept in the study of electromagnetism and can lead to advancements in technology and applications of electricity.

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