Electronic – Internal resistance of non ideal current and voltage sources

An ideal voltage source is is "perfectly stiff". Its voltage does not vary with load, and its impedance is zero. A non-ideal voltage source has a nonzero impedance. This will cause its voltage to vary with the current. As more and more current is drawn from the voltage source with a bigger load (smaller resistance), its voltage will drop. So, conversely, if you go the other way and push current back, the voltage likewise rises.

Why the voltage drops when you draw current from the voltage source is that your load is bringing electrons toward ground. Electrons are losing potential as they travel through the non-ideal voltage source's resistance. The voltage loss is V = IR.

When you force current backwards through the voltage source, you are driving that with a potential which is higher than that of the voltage sources. So now electrons are flowing the other way. Your voltage source is closer to the ground than the one driving it and so the current can be regarded as negative. The voltage difference is still V = IR, but the I has changed sign.

The above describes an "ideal non-ideal" voltage source: i.e. a voltage source which is non-ideal in the simplest, first order way (the inclusion of a constant ("ohmic") parasitic resistance).

A regulated power supply will not exhibit this behavior. For example, consider a simple a supply which uses a pair of transistors to create an arrangement where a feedback voltage regulates the current flowing through a pass transistor. Such an arrangement will stiffly (though not ideally) regulate the voltage from falling against increasing current demands. But it will not regulate against a reverse overvoltage. The overvoltage will give rise to a large feedback signal which will simply cut off the pass transistor. But beyond that, the current through the pass transistor will not reverse. The transistor looks like an open circuit at that point, and the supply is then just a passive Thevenin resistance made up of the internal voltage divider that provides the feedback signal.

Some basic voltage regulator circuits given here, with discussion: http://www.circuitstoday.com/controlled-transistor-series-regulator-with-overload-and-short-circuit-protection

The answer should be easy to see if you look at the complete circuit you created:

The battery is modeled by the 9 V voltage source in series with R1. R1 represents the internal resistance of the batter, and V1 is its open circuit voltage. V1 and R1 together form a Thevenin source, which is a better approximation of a battery than just being a perfect voltage source.

R2 is the resistor you put in series with the battery, and R3 is the resistance of your voltmeter. The meter will show whatever voltage is accross R3.

Now put some plausible values on R1 and R3 (you already said what R2 is). R1 for a 9 V battery is probably in the 10-20 Ω range. Let's say 20 Ω, because as you will see shortly, R1 doesn't really matter in this case. A typical modern voltmeter will have about 10 MΩ resistance, so set R3 to that.

Now you have a voltage divider with R1+R2 forming the top resistor and R3 the bottom. What is the attenutation ratio of a divider with a top resistor of 240 Ω and a bottom resistor of 10 MΩ? From the fact that the bottom resistor is much much larger than the top, you should be able to see this is very nearly 1. In reality it is so close to 1 that you don't notice the difference on the voltmeter between the top resistor being 20 Ω (just the battery) and 240 Ω (the battery with your extra resistor in series).