There is no "low resistance path from the battery directly to ground" in your circuit.
Current is always "looking to go to ground", but it cannot get there through the inside of the battery. Battery construction techniques are widely varied, but they all consist of some form of internal barrier that prevents the internal mobile charges of one polarity from being able to reach the internal net charges of the opposite polarity. If that were possible, the battery would not contain any potential energy (the ability to do useful work).
Because the positive and negative charges in the battery can't recombine internally, the only way they can get to each other is to travel through the external circuit (path) you provide. Thus the charges are "forced" to do useful work in your external pathway because they are "looking to go to ground".
Current (in EE convention) will flow from the positive terminal. So the only way down is through your resistor chain. Interestingly enough, a circuit attached to "ground" (Earth or otherwise) will not necessarily see no lower voltages.
Current does not "stop" at ground. This is Kirchhoff's Voltage Law. In your schematic it would make no difference if you turned your battery upside down. The voltages (relative to "ground") would just now be negative.
Ok, so lets say we flip the battery...
So lets say we flip the battery and current flows from the bottom of
the battery, approaching the first resistor. Wouldn't current choose
the ground path over the first resistor?
It might be interesting (if not amusing) to think of the electrons as little people. Each person is motivated by two things:
- Getting away from other similar people that it finds annoying (other electrons)
- Hanging out with (going towards) people that it finds attractive (positive charges).
That process is why currents circulate -- (must!) flow in loops. Forget "ground" for a minute. I think the circuit and concepts make sense to you.
You're getting stuck on the concept of ground. The reason is that electrons and "ground" are unintuitive because it's a mixed metaphor. "ground" is supposed to represent the "lowest potential" in the system. It doesn't always. Nothing in physics demands that -- it's just terminology.
What it really represents in practice in a simple circuit like this one (and why we teach the concept of "ground" or better "reference") is that it represents the place the mobile charges would most like to be. In this case it's the negative terminal of your battery.
In electrical engineering, we are stuck with the convention that current is defined in terms of the flow of positive charges. Therefore, the positive charges would most like to be at "ground" (the most negative -- lowest -- point in the system). Actual electron flows are reversed from this convention, which is, I suspect, the source of your confusion.
But even that isn't the whole story since "ground" is just a reference potential and nothing forces it to be the absolute lowest potential in the circuit. Really it is just an arbitrary point where we say the potential is 0 Volts. Voltage is a differential measurement and cannot be defined without a point of comparison.
Looking at your photo, you seem to have +Vs of the LM35 connected to +5V on the Arduino, so if you are reading 3.84V there, that is definitely a problem to address before trying to fix anything else.
The likely problem is this: Te 7 segment displays (or some wiring fault on the breadboard) is drawing too much current from the Arduino, and ultimately from the USB connector. This means that the voltage regulation provided by the USB power supply is no longer working effectively, and so the supply voltage that you measured as 3.84V is likely wandering, and in any case low.
While many if not all the chips in this project will operate off voltages lower than the 5V that V+ is supposed to be at, there are some aspects that will be sensitive to lower V+.
One notable V+ -sensitive feature is the A-to-D convertor in the Atmega328. (As The Photon commented.) The convertor can be set to choose between a couple of different reference voltages. Your code indicates that you have it set to use the V+ (VCC) value (5V) as the reference. If that reference is low (eg: 3.8V) then your input voltage from the LM35 will measure as a number that is scaled up accordingly.
And I'm fairly certain that if your V+ is low, it will change according to how many LEDs are illuminated and drawing current, so your measurements will vary according to what digits are displaying, and vice versa, of course.
Once you've fixed the power supply issue, you might assess whether to set the A/D convertor to use the built-in 1.1V reference instead, as this will give you higher resolution and more stable converted values. See the Atmega328 docs for more details on the A/D convertor and options.
Best Answer
Voltage dividers work great if and only if all of the current that flows through R1 also flows through R2.
simulate this circuit – Schematic created using CircuitLab
However, as soon as another current path forms between the resistors, the divider no longer works. Since some of the current is diverted away from R2, the voltage drop changes. What's worse, if the current to the load is variable, the voltage of the divider will be variable too:
simulate this circuit
It's for this reason that resistor dividers are almost never used as a power source. Instead, you would use a voltage regulator of some type. Voltage regulators have active feedback to maintain the output voltage regardless of current draw (up to a point, of course). And some regulators, such as SMPS, can have very high efficiencies, so you don't have to burn off so much heat to drop the voltage.
For your simple circuit, a linear regulator is probably best suited. They're very simple to use (typically a single IC chip plus two capacitors) and inexpensive. They're great for most low-current digital applications. The 7805 is a very common linear regulator that outputs a fixed 5V, however, it is ancient technology and there are plenty of modern devices available as well.
simulate this circuit