# Resistance D-C circuit [closed]

This question seems simple, but gives two plausable answers:

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads 12.5 V . You cut off a 20.0-m length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A . You then cut off a 40.0-m length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.40 A . Even though the equipment you have available to you is limited, your boss assures you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance.

From this i took the voltage of the battery to be 12.5 V, and using the relation shape V=IR, i found the resistance of the wire per meter to be 0.089 Ohms using the first set of data, and 0.071 ohms using the second set. Which, if any, is right?

## closed as off-topic by Bill N, Sebastian Riese, CuriousOne, ACuriousMind♦, GertMar 3 '16 at 2:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Bill N, Sebastian Riese, CuriousOne, ACuriousMind, Gert
If this question can be reworded to fit the rules in the help center, please edit the question.

• Hi Padraig and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie Mar 2 '16 at 16:47
• If you are encountering a exercise like this, then you have probably seen the terms internal resistance of battery and meter resistance. But, you have chosen to ignore them. If you haven't seen those terms, why are you trying to work an exercise like this? Something is out of kilter. And "Which, if any, is right?" is not a good question for this site. Ask about a specific concept which puzzles you. – Bill N Mar 2 '16 at 17:00
• FYI: This is not a practical way to measure resistance because the voltage at the terminals of any practical battery is going to be a function of current. The correct way to measure, with the equipment provided, is to simultaneously measure the voltage (between the two ends of the wire, not between the battery terminals) and the current. Then compute $R=\frac{V}{I}$. – Solomon Slow Mar 2 '16 at 20:01

This feels like a homework question, so let me give you a hint first:

You actually need both pieces of data to determine the resistance of the ammeter and the internal resistance of the battery. This resistance is in series with the resistance of the wire. Assuming that it is constant, two datapoints are sufficient to solve for the resistance of just the wire (two equations with two unknowns).

Putting the resistance of battery plus ammeter = $R_i$, and the resistance of the external wire $R_1$ and $R_2$ with measured currents $I_1$ and $I_2$, the equations you have to solve are

$$V = I_1\left(R_i + R_1\right)\\ V = I_2\left(R_i+R_2\right)$$

With a little manipulation, you will find that $R_i$ drops out, and you get an expression for $R_1-R_2$. Which gives you the resistance of 20 m of wire.

Note - the ammeter is said to have low resistance, but no mention was made of the voltage across the terminals of the battery with the wire connected across... An ohmmeter does exactly that - it measures both the voltage across the load, and the current through it. If you set up your experiment that way (rather than measuring the unloaded voltage of the battery), then the result would follow more obviously.

Apparently, although "the equipment is of high quality", the process used for measuring could be improved...