Current through an Ammeter in an Electric Circuit

[cnh1995]

Oh!
Yes, you are right.

The current is $$i=\frac {6000}{110}~V \times \frac {110}{12000}{\Omega}^{-1}$$
i.e., $$i=50.0~mA$$

There's again one typo in your equation, but the final result is correct.

 

[phinds]

You're still not getting the obvious simplified circuit. Draw the circuit WITH the voltage source AND the ammeter AND 3 resistors, one in series w/ the voltage source, one in series w/ the ammeter and one in the bottom line.

You want to get to the point where when your eyes see this:

Your brain sees this, pretty much automatically

That IS of course based on the specific question being answered. This reduction might not be as helpful if you were asked to find the voltage across the left-most resistor (although even for that, I would still do this reduction, get the current in that line and then the voltage)