# Connecting both ends of battery with a resistor is fine?

I recently just started learning about electronics and electricity. After a lot of reading this is my current understanding about a battery. If you buy a battery it is labeled + or - on each end. The + side is just a bunch of atoms with extra electrons(electrons are - but I guess they labeled this side + ?). The - side is a bunch of atoms with missing electrons(protons are + but I guess they labeled this side - ?). So + side of the battery flows to the - side side. Is this correct?

Resistors limits the current/amps but keeps the voltage the same. Correct?

You don't want to connect them together with a wire since there would be too much current flowing through, such that the battery heats up till it explodes or something. So connecting a high ohm resistor between a battery would be fine. Correct?

Can someone also explain to me WHY a voltage divider works? All websites just explain the formula and the circuit.

The main reason I asked these questions was since I was learning about arduino pullup pins. I noticed a voltage divider in the circuit. The digital input pin measures the voltage right? If the voltage is almost 0/ground it returns LOW else HIGH.

• Electron flow is from the - side to the + side of the battery, conventional current is the flow of positive charge, so the opposite direction of the flow of negative charge. And there's no extra electrons or missing electrons, it's a chemical reaction that produces a potential difference. Feb 20 at 5:36
• As for the voltage divider, do you understand Ohm's law? Feb 20 at 5:36
• @Hearth What do you mean by positive charge? The protons don't move from one atom to another right? And I do know what Ohm's law is. Feb 20 at 5:41
• @jujumumu I mean exactly what I say, conventional current is the flow of positive charge. No actual positively charged particles are moving (at least in a metal!), but negative charges moving one direction is equivalent to positive charges moving the other direction. Feb 20 at 5:47
• Maybe I was a little too subtle with my last comment. You need to tell us EXACTLY what battery you are talking about, EXACTLY what mosfet and what is going to be driving it. Or I could guess and suggest a 1/4W 5% 10k resistor. How does one calculate it? The assumption is that the resistor is a 'pulldown' resistor in order to ensure the mosfet is turned off when the MCU is on reset. The actual value is not particularly critical. At 3V3 the current would be I= V/R = 3.3/10000 = 330uA. However, if you're designing a very low power device, then that might be considered excessive. Feb 20 at 10:20

Yes, you can put a carefully selected resistor across the terminals of a battery with no ill effects. Key phrase is "carefully selected."

Current may be described in two seemingly opposing ways. You may want to look at is as if the positive charge moves, or if the negative charge moves. If you are talking about current flow in metals, then you are likely thinking that the protons cannot move and therefore any motion of charged particles must be the electrons. In this case, electrons are actually being pushed out of the negative side of the battery and being sucked into the positive.

Many electrical engineers like to think about "conventional current" where the positive charge moves via the absence of electrons thereby creating a traveling hole. In this case, the current moves out of the positive terminal of a battery and enters into the negative.

Resistors are devices that take more voltage to move either the electrons or the holes than a typical wire, thereby allowing an electrical "pressure" (voltage) to be seen across the device when one is applied.

Current is the resulting flow of electrons or holes when you apply an electrical pressure to a conductor (wire) or other device (resistor, semi-conductor, etc...). If you apply a high pressure to a low resistance, you'll get a high flow. If you apply lower pressure you'll get a lower flow. If you use a higher resistance, you'll also get a lower flow.

All of this follows a set of rules known as Ohms Law which states, Voltage is equal to current times resistance, in it's most basic form.

An additional detail you should know is that when you apply a voltage to a series circuit, the total of all voltage drops in the circuit must equal the applied voltage and that the voltage across any element of the circuit will be equal to the resistance of that element times the current flowing through that element. Think of current as an incompressible fluid. It must be the same in all series components of the circuit. If you know the voltage on the circuit, and you know the value of each resistance, you can add all the resistances to get the total resistance. Now you can calculate the current in the circuit using Ohms law (V/R). Then you can calculate the voltage across each resistive element by multiplying that current by each resistance.

I highly recommend investing in "Getting Started in Electronics" by Forrest M. Mims III. I had the first edition RadioShack version when I was 12, and it was one of my favorite books. As the other folks are trying to suggest, you need to understand how to apply Ohms law in order to calculate the proper values for current limiting resistors or even pull-ups. And there are a couple other details that you might want to know as well when you are driving the gate of a MOSFET.

One of the things that would also be useful, is to draw a schematic of what you are thinking about doing with your components and let the folks make suggestions about how to improve the circuit. If you are limited to "components on hand," Let everyone know which components you have to work with. This site has a schematic drawing feature that should allow you to post basic circuit ideas. Why don't you give that a try?

If you buy a battery it is labeled + or - on each end. The + side [end] is ...

Don't worry about electrons. Current flows from positive to negative when you make a circuit.

Resistors limits the current/amps but keeps the voltage the same. Correct?

Yes, if you don't overload your battery. The battery has some internal resistance and when you connect a load the battery terminal voltage will drop as you have now created a potential divider. More later.

You don't want to connect them together with a wire since there would be too much current flowing through, such that the battery heats up till it explodes or something. So connecting a high ohm resistor between a battery would be fine. Correct?

Yes.

Can someone also explain to me WHY a voltage divider works? All websites just explain the formula and the circuit. simulate this circuit – Schematic created using CircuitLab

Figure 1. A voltage divider.

Here we have a 9 V battery and a total load resistance of 9 kΩ. From Ohm's law we can calculate that a current of 1 mA will flow through the circuit. $$\ I = \frac V R = \frac 9 {9k} = 1 \,\mathrm {mA} \$$.

Again, using Ohm's law we can calculate the voltage across each of the resistors. This will show that there is 7 V across R1 and 2 V across R2. It should be fairly intuitive that the voltage at V2 is 2/9 that of V1. And if you work out how you knew that you'd find that you were thinking along the lines of "what fraction is 2k out of the total?". The answer, of course, is $$\ \frac 2 {2 + 7} = \frac {R_2}{R_1 + R2} \$$.

The main reason I asked these questions was since I was learning about Arduino pullup pins. I noticed a voltage divider in the circuit. The digital input pin measures the voltage right? If the voltage is almost 0/ground it returns LOW else HIGH.

Not quite. Figure 2. Threshold voltages for various logic families. Image source Next.gr.

To guarantee a correct reading of a digital input signal you need to exceed VIH (input high) voltage level for logic '1' and you need to get below VIL (input low) voltage level for logic '0'. See the linked article for more. simulate this circuit

Figure 3. A microcontroller with a 1 MΩ input impedance and an external 1 MΩ pull-up resistor.

I'll leave this as a mental exercise for you.

• The microcontroller's input has a resistance of about 1 MΩ to ground.
• The external circuit consists of a GND connected push-button and a 1 MΩ pull-up.
• When the SW1 is pressed we get 0 V on the input. Since this is < VIL the chip will read a logic '0'.
• Q1: When SW1 is released what is the voltage on the input?
• Q2. Is it enough to guarantee it to be read as logic '1'?
• Q3. If not, what would you change to make it work reliably?

In electronics, you usually don't have to worry about batteries having atoms and electrons. It can be simply modeled as voltage source that can provide current through some amount of internal resistance. Modeling how battery actually works is usually not necessary, unless you are doing something where it matters. Current as we draw it flows from positive battery terminal into negative battery terminal.

Your model of resistors is not correct. Ohm's law says U=R*I so if there is a current through resistance it will drop voltage.

Voltage divider works because full battery voltage must be over the two resistors and equal current flows via both resistors. If the resistances are equal, both resistors have half of the battery voltage over them, if they are unequal then other has more and other has less voltage over it, and Ohm's law can be used to calculate these things.

Yes, digital input pins can be though as they measure voltage.