- #1

Ryaners

- 50

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1. Homework Statement

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).