Now by looking at the schematic, 33 kΩ and 15 kΩ resistors have no current though them. Thus voltage across is zero.

But when voltmeter is connected to terminal A and B, what happens to the 33 kΩ resistor? Since the voltmeter is essentially a resistor?

 

A new resistance appears connected from a to b. The value is the resistance of the voltmeter.

 

Now by looking at the schematic, 33 kΩ and 15 kΩ resistors have no current though them. Thus voltage across is zero.

A battery with nothing connected to it also has no current. But it has voltage across it. So the assumption is incorrect.

(a) will be at the same potential as the middle of voltage divider on the left, and (b) will have the same potential as (c). The voltage drop from (a) to (b) is the difference between these two potentials.

 

What kind of volt meter? An old Simpson 260 series was rated at 20K ohms per volt (2.5 volt range would put a 50K ohm load into the circuit) on DC and if I recall properly, it was 5K ohm per volt for AC. In order to properly figure out what's happening, you need to know your test equipment.

 

This is a classic example of knowing what happens when you connect your meter to the circuit.

Like Bill said, the old Simpson 260 voltmeter series basic spec was 20,000 ohms per volt. Search for the schematic diagram and you will see why it was 20,000 ohms per volt.

So, if in the diagram above, you have quite a few different ways to connect a voltmeter ... A-C, B-C, A-B. Assuming a 20k/volt meter, and using the 10 V scale, you can substitute the appropriate resistance at each different node connectins and compute what the meter should read.

You can do the same with your digital multimeter and see the effects of different meters on your circuit measurements.