You just landed a job as an assistant to an electrician who is working on a building site. He is making use of long single-conductor insulated wire for some of the connections in the complicated "smart building." A new shipment of spools of this wire has been delivered, and the electrician is not happy that it came from a manufacturer that he suspects is producing inferior wire. The electrician asks you to determine the resistance per unit length of the wire (in Ω/m) and provides you with a battery and a high-quality multimeter. Despite the quality of the meter, he suggests that you not use the ohmmeter to measure resistance, because the value of the resistance is so low that the meter is not likely to measure it accurately. You are puzzled as to how to go about this task. As the electrician runs off to attend to another task, he says, "Cut off a couple of different lengths of wire and use the battery and the multimeter." You put the multimeter in voltage mode across the terminals of the battery without a wire connected and measure 7.70 V. You then cut off two lengths of wire, one 5.00 m in length, and one 10.0 m in length. Connecting them one at a time to the battery through the multimeter in current mode, you find that there is a current of 5.47 A in the 5.00 m length of wire, and 3.25 A in the 10.0 m length of wire. Ah-ha, now you have enough information!

Respuesta :

Answer:

R/l = 0.25925 Ω / m

Explanation:

Ohm's law says that the potential difference is proportional to the product of the resistance by the current

         V = I R

         R = V / I

In this case, since we have two lengths, we can have two lengths, we can find the resistance for each

         L = 5 m

         R = 7.70 / 5.47

         R = 1,408 Ω

         L = 10 m

         R = 7.70 / 3.25

         R = 2,369 Ω

We can make a direct proportions rule (rule of three) to find the resistance per unit length

For L = 5 m

         R/l = 1,408 / 5

         R/l = 0.2816 Ω / m

For L = 10 m

        R/l = 2,369/10

        R/l = 0.2369 Ω / m

We can see that the value is similar that differs from the second decimal place, in this case the value for the longer re wire is more accurate because it has a lower joule effect.

One way also to find the average value

       R/l = (0.2816 + 0.2369) / 2

       R/l = 0.25925 Ω / m

Answer:

Resistance per unit length = (0.2593 ± 0.022) ohms/m

Explanation:

V = potential difference across the wire of each = 7.7c

Current reading in 5m wire = 5.47A

Current reading across 10m wire = 3.25

Resistance R= V/A

Where v = volt, A = ampere

So

Resistance across 5m wire = 7.7/5.47 =1.408ohms

Resistance across 10m wire = 7.7/3.25 = 2.369ohms

Resistance per unit length of 5m wire =1.408ohms/5 = 0.2816ohms/m

Resistance per unit length of 10m wire = 2.369ohms/10 = 0.2369ohms/m

Range of error between wires = 0.24ohms per m which is acceptable

Average of two tests = 0.2593ohms/m

Lower range = average of test - 5m wire resistance per unit length

=0.0223= upper range

So resistance per unit length of wire =(0.2593 ± 0.022) ohms/m