Given: An ammeter with a range of 0-10mA

Need to turn it into an ohmmeter to measure resistances of 0-1000 ohms. It must read 0A when R=0 ohms, and 10mA when R=1000 ohms. It needs to have a commonly-available voltage source (I chose 9V, but I can change it).

Now V=IR or I=V/R, so as R goes to infinity, current goes to zero and vice versa (inverse relationship). Since I need the opposite of that, it gets a lot harder. I'm guessing I need some inductors or capacitors in there somewhere, but I can't figure out how to make a circuit with a direct relationship between R and current. Am I missing something obvious?

Given: An ammeter with a range of 0-10mA

Need to turn it into an ohmmeter to measure resistances of 0-1000 ohms. It must read 0A when R=0 ohms, and 10mA when R=1000 ohms. It needs to have a commonly-available voltage source (I chose 9V, but I can change it).

Now V=IR or I=V/R, so as R goes to infinity, current goes to zero and vice versa (inverse relationship). Since I need the opposite of that, it gets a lot harder. I'm guessing I need some inductors or capacitors in there somewhere, but I can't figure out how to make a circuit with a direct relationship between R and current. Am I missing something obvious?

Here (http://www.allaboutcircuits.com/vol_1/chpt_8/6.html) you go. The answer to all your questions.

The potential difference across a resistance is equal to the current times the resistance so if you know the voltage drop and the current you can tell the ohm value of the resistor. Other than that go look at the link Dngrmse supplied. Perhaps put two resistors in the circuit (a known one and X), then go look at the voltage drops across both.

Thanks for the link, but you didn't read the question.

I need to alter the ammeter so it reads 10mA when the resistance is 1000 ohms, and it reads 0A when the resistance is 0 ohms.

V=IR (I=V/R) doesn't apply here, since current goes to infinity (not zero) as resistance goes to zero. Adding resistances in series or parallel does no good.

Thanks for the link, but you didn't read the question.

I need to alter the ammeter so it reads 10mA when the resistance is 1000 ohms, and it reads 0A when the resistance is 0 ohms.

Why? Is the 0 ma at 0 ohms a requirement in a class or something? Ohmeters, (which only measure current anyway), always have their scale reversed, in that zero ohms is on the right, and infinity is on the left. If you apply ten milliamps to your meter, the needle will deflect all the way to the right, correct? This will be 'zero' measured ohms. If you add additional resistance, the meter will deflect less. The logarithmic scale on the link posted shows that it will be fairly accurate in the range you specified, (zero to 1000 ohms). To clean things up some, you should also add a small trim resistor, in series with whatever current limiting resistor is used to zero the meter when the test leads are shorted.

Why? Is the 0 ma at 0 ohms a requirement in a class or something?

Yes.

Yes.


This is an edit. I think you're looking for a DC bridge circuit, (explained in some detail here (http://physics.ucsd.edu/~drs/Classwork/120A_Lab2_Week3_Spr_04.pdf) ).

I'm pretty sure this will answer.

It looks like current would still be inversely proportional to resistance with that setup. Or am I wrong?

It looks like current would still be inversely proportional to resistance with that setup. Or am I wrong?

Consider a bridge, with four balanced resistors. Connect an ammeter between two opposite 'corners' of the bridge. Connect a power source to the remaining two corners. Since the bridge, (at this point), is perfectly balanced, there is no voltage differential between the two nodes our meter is connected to, and no current flow. Now we'll break one of those legs, and add our test leads. Short the leads out, (a short, or zero resistance), and we still have our balanced bridge, so no current flow results. Zero resistance, zero current. When you measure a resistor with this, you unbalance the bridge, and as resistance increases, current flow increases as a result.

After much searching, I found an almost suitable java simulation....it will work to demonstrate the ideas. It's right here (http://www.oersted.dtu.dk/personal/ldn/javalab/Circuit08.html) .
When you're looking at this, imagine that you've inserted test leads, in series, with the lower left resistor in the bridge.


Now, first thing to do, is set the supply voltage to 7.6 volts. Set the meter resistance all the way to zero. Set each resistor in the bridge to 200 ohms. The simulation should inform you that no current is flowing through the ammeter. Now set the lower left resistor to 1.2 k, (which is your 200 ohm resistor, plus your 1k resistor in series. The 1k resistor is your test resistor). The sim should now inform you that 10 milliamps of current is flowing through your ammeter.

As you can see, with no test resistor in the circuit, you have a balanced wheatsone bridge, and no current will flow through a connected ammeter. When you test a 1k resistor, you get 10 milliamps. The circuit is not linear, so you'll have to scale your meter appropriately.

This should give you a rough idea how to proceed. In real life, your 10 ma movement has some internal resistance that will have to be accounted for, and you might also want to add an additional resistor to your power source to limit maximum current.