- #1
Ryaners
- 50
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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).
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).
