Respuesta :
Answer:
[tex]0.06\Omega/m[/tex]
Explanation:
Firstly, when you measure the voltage across the battery, you get the emf,
E = 13.0 V
In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.
Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ([tex]I=\frac{V}{R}[/tex]). But this is not the case.
Let the resistance of the ammeter be r
Hence, using Ohm's law we get the following 2 equations:
[tex]\frac{13}{20z+r} =7.6[/tex] .......(1)
[tex]\frac{13}{40z+r} =4.5[/tex] ......(2)
Substituting the value of r from (2) in (1), we have,
[tex]13=152z+7.6\times\frac{13-180z}{4.5}[/tex]
which simplifying gives us, [tex]z=0.0589\Omega/m\approx0.06\Omega/m[/tex] (which is our required solution)
putting the value of z in either (1) or (2) gives us, r = 0.5325 [tex]\Omega[/tex]
