I thought I understood the basics of DC circuits and Ohm's law but now I'm confused [closed]

Question

So I was taught in school and also read a lot of articles of the basics of DC circuits. I know the formula R=V/I, and therefore that resistance is directly proportional to voltage and inversely proportional to current. So if I increase the resistance the voltage increases too right?

Now I'm working on an old RC car I used to play with. The battery pack consists of 6 AA batteries which provide 9V, but I noticed that there's another output that outputs 4.5V. I guess it's being used to power the driver board. I tried reading the voltage but I also put a 10 megaohm resistor in the circuit. I was surprised when I saw that the multimeter was reading around 2.2V. Isn't the voltage supposed to be higher because of the resistor?

I am completely confused now and I'm pretty sure that I have a lack of understanding or just confusing some things. Please explain why did this happen.

Answer 1

I know the formula R=V/I, and therefore that resistance is directly proportional to voltage and inversely proportional to current. So if I increase the resistance the voltage increases too right?

We use our mathematical equations to describe, not dictate what happens in the physical world.

Allow me to tackle this in a more abstract way as I've had similar confusions in the past.

Suppose I gave you the following equation: A = B/C. I then give you some numbers to plug into the equation: A = 2, B = 4, C = 2. So we now have 2 = 4/2, which makes sense so far.

Then I tell you that A has increased from 2 to 4 and now you have to determine what changes occurred in B and/or C that account for A becoming 4. So we are now faced with this problem: A = 4 = B/C. Solving this is impossible since we now have 2 unknowns - we don't know what B or C are anymore. B could be 8 and C could still be 2, or B could still be 4 and C is now 1, or B and C are any (infinite) number of fractions whose result is 4.

What a mess...but hold on, what even are A, B and C? And that's exactly the point - all the equation A = B/C does for you is describe the relationship between these mysterious variables I just made up, but it tells you nothing about the underlying physical phenomenon associated with them.

Now let's draw a parallel to R = V/I. Suppose we start with R = 2, V = 4, I = 2. This gives us R = 2 = 4/2. Now lets increase our resistance to 4. Mathematically we are faced with the same impossible situation above since we don't know what changes in I and/or V resulted in R becoming 4.

Except this time we understand something about the underlying physical phenomenon. We know we have a DC circuit and all we did was replace our 2Ω resistor with a 4Ω one. So let's entertain two possible outcomes of doing this:

Possibility 1 claims that by putting in that 4Ω resistor, our voltage source has become stronger and the circuit is now delivering double the energy (Power = 16W). At this point we are trillionaires since we have magical resistors that create free energy for us.

Unfortunately for us, it's more likely that Possibility 2 has occurred: The 4Ω resistor has resulted in the current being halved and we are now only able to deliver half the energy (Power = 4W).

The main point is that we have two levels of understanding to solve such problems. First is our understanding of DC circuits and the physical phenomena of voltage, resistance, current etc.. And on top of that is our mathematical equations that describe the relationships between those physical phenomena.

We use our mathematical equations to describe, not dictate what happens in the physical world.

Answer 2

I know the formula R=V/I, and therefore that resistance is directly proportional to voltage and inversely proportional to current.

I would say that, "The resistance defines the relationship of voltage across a resistor to the current through the resistor".

So if I higher the resistance the voltage increases too right?

Only if the current remains constant. In very many cases the voltage is fixed so increasing the resistance decreases the current as you would expect from \$ I = \frac V R \$.

Now I'm working on an old RC car I used to play with. The battery pack consists of 6 AA batteries which provide 9 V, but I noticed that there's another output that outputs 4.5 V.

It sounds as though they are tapping off at the battery pack mid-point or they have a voltage regulator stepping down from 9 V to 4.5 V.

I guess it's being used to power the driver board.

It could be. Many types of logic will run will on about 5 V.

I tried reading the voltage but I also put a 10 megaohm resistor in the circuit. I was surprised when I saw that the multimeter was reading around 2.2V. Isn't the voltage supposed to be higher because of the resistor?

Bring this to its logical conclusion: with an infinite resistance (no resistor present) you should get infinite voltage?

No, you have connected 1 MΩ in series with a multimeter with what appears to be a 1 MΩ input impedance. This has divided the available voltage in two.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) Direct measurement. (b) Voltage measurement through a series resistor.

You can use Figure 1b as a way of measuring the internal resistance of the multimeter.

Answer 3

So if I higher the resistance the voltage increases too right?

You're not thinking about it quite right. Increasing the resistance doesn't increase the voltage UNLESS the current is kept constant (i.e. a current source) is being used).

If a voltage source is being used (i.e. the voltage is being kept constant) then increasing the resistance will reduce the current.

Do you understand F = ma? It is the same thing. Increasing the force being applied doesn't increase the mass of whatever you are pushing. If the mass of whatever you are pushing is held constant though (as is usually the case) and you increase the force then its acceleration will increase.

But if the acceleration was somehow kept constant instead of the mass then increasing the force must also be accompanied by an increase in mass, otherwise the acceleration cannot remain the same. This is a lot more difficult to do since we apply forces which result in accelerations rather than the other way around. There is one exception to this: gravity. Gravity automatically applies more force to objects with more mass but they have more mass to accelerate so the acceleration due to gravity stays constant regardless of the object's mass. In a sense, gravity applies the same acceleration to everything and a force results from this which is the reverse of most cases we see.

Answer 4

what you made was a voltage_divider.

the math on that is

Vout = Vin * R1 / (R1 + R2)

and your DVM is measuring the Vout.

With both your resistors (the external 10Megohm and the DVMs internal 10MegOhm, the math becomes

Vout = 4.5v * x / (x + x) = 4.5 * 0.5 = 2.25

Try other resistors instead of your 10Meg OHm.

Imeg ohm, 100Kohm, 330K ohm, 10Kohm

If you go below 10Kohm, your DVM will just read the battery voltage (within 0.1%).

Have fun.

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If you remove the DVM, then there is no current thru that 10Meg OHm resistor you had connect to the 4.5v node.

If there is no current, then there is no voltage drop across the resistor.

Both ends of the resistor are at 4.5 volts. Not a very useful thing, because you cannot connect anything to it ...... ahhh a gold-leaf-electrometer is acceptable.