I know someone else asked the same question, but I wasn't able to understand it.
My question is, why do you need 2 resistors (1 to ground), to divide the voltage?
Just 1 resistor would do the same task. Suppose I have 9V and want to get 5V. I could do it with 1 resistor, too. Excuse my ignorance, I'm not experienced.
.
If you have one resistor and no load, then the 9v in will give you 9v at the other end.
If you have any sort of load, then that's behaving like your second resistor. Note that as the load changes, the division ratio and so the load voltage changes. That's one of the reasons why we tend not to use single resistors for dropping the voltage into a load.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Various loads on a series resistor.
So the answer is that sometimes you can omit your R2 but you have to understand the implications. Even when R2 is there you need to be aware that with a low resistance load the load on the divider will increase and the voltage will drop from the unloaded value.
Resistors only drop voltage if current flows through them and the amount dropped depends on the current. Assuming your load draws zero current, which is a good approximation of most voltage inputs (which your Vout feeds into), without R2 no current flows through R1 so no voltage drops across R1 so the voltage on both sides of R1 are the same.
If you divide anything, you always get at least two parts.
Also a voltage divider must consist of two parts, one for the first voltage and one for the second voltage.
That‘s why there are two resistors involved.
Once the role of the second resistor R2 has become clear (to cause a current to flow through R1 and, accordingly, a voltage drop across it to be subtracted from the input voltage), now we can combine the two resistors into one to get the so-called "potentiometer". Here is a 4-step scenario for "inventing" and investigating the famous device. It is extracted from my Wikibooks story about Ohm's experiment (the pictures are drawn by me).
A tapped pipe (there is no flow). Let's begin with considering a well-known hydraulic analogy (plumbing) - that we can see everywhere around us. For example, imagine a large vessel filled of water that supplies a long thin pipe; let's first the pipe to be tapped (Fig. 1). The question is: "What is the pressure inside the pipe?" And more precisely speaking, "What are the local pressures along the pipe?" There are not so many people that will answer rightly these simple questions.
Fig. 1. The local pressures along the tapped pipe are equal to the input pressure.
We can get to know, if we drill small holes at equal intervals along the pipe (if we want to be more precise, we might stick vertically thin glass pipes acting as local manometers). The result is expectable for us: all the water levels (accordingly, all the local pressures along the pipe) are equal. This picture shows the pressure distribution along the pipe; we can name it "pressure diagram".
An opened pipe (there is flow). Now open the pipe (Fig. 2); the water will begin flowing. This is a well-known situation from our routine where someone opens a faucet somewhere in the end of the plumbing. At the left end the water pressure is maximum; at the right end it is minimum. But what are the local pressures along the pipe now? Our intuition suggests that the local pressures will decrease gradually from left to the right.
Fig. 2. The local pressures along the opened pipe decrease gradually.
Really, the levels of the water bars (accordingly, the local pressures along the pipe) decrease gradually from left to the right. The envelope of the pressure diagram is a triangle.
Let's now transfer these notions to our electrical domain, in order to see if the voltages along a resistive wire are distributed in the same way. That means to reproduce the genuine Ohm's experiment under the conditions of today.
Now, fix the two ends of a wire in porcelain insulated terminals (holders) and apply voltage (for instance, 10 V) first to the left end of the wire (Fig. 3).
Fig. 3. Investigating the local voltages along a wire
What can we investigate now in this arrangement? What do we measure with the voltmeter? The usual viewpoint is to think of a resistor as of a point, as of something that has not dimensions, as of a two-terminal element that has only a property of resistance. But here we have the unique chance to peep inside the "resistor"! What will you "see" along the wire? What will the voltmeter show when we slide its active probe from right to left? What are the local voltages along a resistor, if there is no current - zero, 10 V or something else?
An opened circuit (there is no current). Remember what a resistor does - it "resists". What does it resist? It resists, obstructs, disturbs current by dissipating power. But no current flows in an open circuit. So, there is nothing to resist; as though, the resistor is not a resistor but "conductor" that transfers the entire voltage from the left to right end.
Fig. 4. The local voltages along the resistive film of an opened circuit are the same.
We can apply the idea of a pressure diagram to present in a similar way the voltage distribution along the resistive wire. We can think of voltage as a kind of pressure; so, we may present the local voltages by local voltage bars in exactly the same way as we presented the local pressures by local water bars (Fig. 4)! As above, the lengths of the voltage bars are proportional to the magnitudes of the local voltages regarding to ground (we might set the zero voltage level at the height of the resistor and then to draw the positive voltage bars above and the negative voltage bars below the resistor's level). The set of these voltage bars forms the whole voltage diagram. We can use the envelope of the voltage diagram instead the set of voltage bars to simplify the image.
Closing the circuit. Now, ground the right end of the wire. Move the voltmeter probe along the wire and measure the local voltage drops; Ohm did exactly the same. He moved the probe from one position to other, measured the corresponding potentials, made the difference between them and calculated the ratio (V2 - V1)/(L2 - L1) = (V2 - V1)/(r2 - r1) = dV/dR = I. Thus he has established that this ratio (it was the current I) is constant along the wire; so, Ohm has concluded that V/R = I. We can see that, when moving the voltmeter probe, voltage drops decrease gradually from 10 to 0 volts; accordingly, the bars of our voltage diagram decrease their length gradually (Fig. 5).
Fig. 5. The local voltages along the resistive film of a closed circuit decrease gradually.
We can draw such a picture for every real conductor with some resistance that convey big current to a powerful load (Fig. 6).
Fig. 6. Voltage distribution along a line
Because if you do it with 1 resistor, it's not a voltage divider anymore. It's something else. And as a result, all the rules you learned about voltage dividers won't work.
The major feature of a resistor ladder / voltage divider is that it gives a quasi-correct voltage when quiescent i.e. with no load being drawn from Vout. A reasonable amount of load from Vout will also cause a reasonable voltage at the tap. The load is influencing the voltage, because it is in parallel with R2, so its effective conductance (1/resistance) will add to R2's conductance (1/resistance).
However with R1 and R2 selected sensibly relative to all possible ranges of load current, the tapped voltage will remain in reasonable limits.
You want to junk the whole idea of a voltage divider by getting rid of R2. Now, the load is not in series with R2, so the load's effective resistance will add to R1 to decide the voltage at the tap point. The load's current draw (the more it draws the lower its effective resistance), will cause wild shifts in inlet voltage. If the load's current draw nears zero, R1's voltage drop stops being much at all, and the inlet voltage nears the supply voltage.
So "dump the ladder; hot to resistor to load to return" only works with loads of reasonably well-known impedance. However, the entire circuit must be ready to experience full system voltage.
A great example of this is an old streetcar headlight resistor; it uses a large resistor box "R1" (placed in the motorman's compartment to heat the compartment) to drop 600V down to the 32V needed for the headlight. So normally R1 is dropping 568 volts. When the headlight burns out, motormen think "I'll change that! 32 volts from ground isn't dangerous!" Now how much voltage is actually on that socket? If R1 was dropping 568 volts at (let's say) 5.68 amps, how much is R1 dropping at 0 amps? Anyone? Anyone? Bueller? That's right. E=IR, E=0 when I=0 regardless of R. So with R1 dropping 0V, the voltage at the headlight socket is 600V. It better be insulated for that.
You should never feel ashamed for trying to know better about something. no matter your technical level. A simple answer would be: voltage is divided by series. Ideally, when you have only 1 resistor in parallel with the Power supply, its voltage is equal to its power supply's voltage only. Every time you add a resistor in series, the voltage gets divided among these resistors.(ergo the voltage divider rule). So in your case, it would be 9V. You should recheck your circuit for a possible mistake. Also, on real-life circuits, the resistance of the cable is taken into account to be in series with the load's resistance. so you can have 5V using 1 resistor only if you have a very long and thin wire, or a very large current.
why do you need 2 resistors (1 to ground), to divide the voltage?
One resistor can be used to drop voltage (if the load draws current) but to divide voltage you need something to create a division ratio. To be a voltage divider the output voltage needs to be a constant proportion of the input voltage. So a voltage divider that drops 9 V to 5 V should also drop 18 V to 10 V, 90 V to 50 V etc. always maintaining the same division ratio.
The only thing that can do that is another resistor (or a load which acts like a resistor, and so effectively is one). Why? Because to maintain the division ratio you need a second component which has the same property of proportionality between voltage and current as R1 - in other words, another resistor.
Note that this need for two resistors only applies to DC. In an AC circuit capacitors and inductors have reactance, which is like resistance but varies with frequency. In an AC circuit you can have a capacitor or inductor voltage divider, which works just like the resistor divider except that it uses two capacitors or two inductors.
