1
\$\begingroup\$

I thought I understand electricity but some thing don't add up so I'm hoping someone can tell me what I am doing wrong. So, to me, voltage represents a potential ability/energy of a charge to do some work/to transfer energy. Resistor uses some of that energy to heat itself or do something else (doesn't matter what really), in order to drop the voltage of the charge. By dropping voltage, it also reduces the current (because the charge is now 'less needy' to get to the other terminal). With this understanding of electricity, I tried to understand the following circuit:

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

The Voltage source is 3V. The current is easily calculated as 1A.

So the story goes: Source is giving 1 amper of current an 'ability' of 3 volts. The first resistor causes a 2V voltage drop. Now that amper has only 1 volt of 'ability' left. The current reaches the branch. It divides equally, 0.5A goes to each branch. But the resistance of the resistors in parallel branches is 2 ohms, and I only have 1V left per ampere! Not only that, but my current that passes through the resistor is not even 1A anymore, now it's only 0.5A, which means it has only 0.5V of 'ability' left! So what's going on here?

A side question that arises: By Kirchhoffs laws, the sum of voltage drops has to equal the voltage of the source. So in the end, the voltage of a charge is 0 (after it passes thorough the last resistor). Why does it even move then, if it's 'willingness' to get to the other terminal is now 0?

\$\endgroup\$
  • \$\begingroup\$ Hint: use the integrated circuit editor to make a clear, well-labeled circuit next time. It's hard to refer to each of your three resistors separately if you don't label them, \$\endgroup\$ – Marcus Müller Nov 18 '16 at 21:14
  • \$\begingroup\$ "the resistance of the resistors in parallel branches is 2 ohms" remember that n equal resistors in parallel are the same as a single series resistor of ohms/n. What you have there is the same as a single one ohm resistor. Sum it to r3 and you get a total resistance of 3 ohms which in turns, as per ohm's law, given your 3 volts results in exactly 1A. Reverse it and you get, from the 1A, your initial 3 volts. \$\endgroup\$ – zakkos Nov 19 '16 at 0:40
  • 1
    \$\begingroup\$ Voltage (electric potential) is very much like altitude above ground. But altitude alone cannot push an object, and voltage alone cannot push an electron. The 'ability' you're talking about is not the voltage of each charged particle in the wire. Instead it's the voltage-drop; the change in voltage measured lengthwise along an entire resistor. (A sliding boulder is driven by the slope of a hill, and charges within wires are driven by the "voltage-slope." Altitude of a single location is irrelevant, "slope" is what matters.) \$\endgroup\$ – wbeaty Nov 19 '16 at 2:37
  • \$\begingroup\$ The key thing to see is that If you look at R3 and R1/R2 as a unit, the same current must pass through each, but each can have a different voltage across them. However, if you look at R1 and R2, each must have the same voltage across them, but they can each have different amount of current going through them. The voltage is split between R3 and R1/R2, but R1 and R2 have the very same voltage across them. The current is split between R1 and R2, but R3 has the very same current passing through it as R1 and R2 have combined. \$\endgroup\$ – David Schwartz Feb 23 '17 at 10:29
2
\$\begingroup\$

I'm going to zero in on your mistake:

Source is giving 1 ampere of current an 'ability' of 3 volts.

Funny use of terms, but okay. In my terms, 3 volts per ampere is the same thing as saying 3 Ohms.

The first resistor causes a 2V voltage drop.

Yup. It takes 2 V of motive force to cause 1 A of current to flow through 2 Ohms of resistance.

Now that ampere has only 1 volt of 'ability' left.

Now you are 'okay' but starting to tread on thin ice. But yes, there is 1 volt of potential difference remaining and all of the 1 A of current that has arrived at this point, must then also leave this point. And to do so, it must only face one remaining Ohm of resistance.

The current reaches the branch. It divides equally, 0.5A goes to each branch.

Yes. The remaining potential difference (1 V) can only cause 0.5 A in a 2 Ohm resistor. Luckily, there are two of those resistors, so luckily all of the incoming current can find an outgoing way to move given the remaining potential difference.

But the resistance of the resistors in parallel branches is 2 ohms, and I only have 1V left per ampere!

Here's where you suddenly jump tracks. One volt per ampere means one Ohm. That's what it means. And it turns out that there is only one Ohm there, too. Good thing.

Let me re-write your statement: "But the resistance of the resistors in parallel branches is 2 ohms, and I only have 1 Ohm!"

I didn't change one whit of meaning there. It says exactly what you just wrote. But now you can see that there is a conflict. You are simultaneously saying there is 2 Ohms and there is 1 Ohm. And that doesn't make sense.

The fact is that there is 1 Ohm between the two points under consideration, not 2 Ohms. Yes, there are two galvanic paths and each of these paths represents 2 Ohms, apiece. And the current will only be 0.5 A for each of these two paths, like it should be. But the combined result is still 1 Ohm and the combined current will still be 1 A.

You just need to fix your mental models.

\$\endgroup\$
2
\$\begingroup\$

Regarding:

Not only that, but my current that passes through the resistor is not even 1A anymore, now it's only 0.5A, which means it has only 0.5V of 'ability' left! So what's going on here?

You are correct in that the current splits, and each resistor (in this particular case) has 0.5 A going through it, but the voltage there does not split, by \$ V = IR\$, half an amp of current through a two Ohm resistor means there is still 1 Volt across those resistors.

\$\endgroup\$
  • \$\begingroup\$ :) don't - you fixed it! \$\endgroup\$ – Marcus Müller Nov 18 '16 at 21:18
0
\$\begingroup\$

By dropping voltage, it also reduces the current

nope.

The current stays the same – the overall amount of charges must stay constant. But because each of these charges now "carry" less energy, there's a voltage difference between the two ends of a resistor.

You're citing Kirchhoff's Law – but that can only work if you acknowledge that current must stay the same.

With that in mind:

You're right, symmetry tells us that through each of the parallel 2 Ohm resistors R1 and R2, half the current in the circuit must flow – 0.5 A.

0.5A * 2 Ohm = 1V

And together with the 2V drop over the serial 2 Ohm resistor R3 that gives 1V + 2V = 3V voltage drop, which nicely aligns with your 3V source.

Everything's fine!

\$\endgroup\$
  • \$\begingroup\$ Ok I made a mistake there, but I don't think it's that relevant to the problem I presented. \$\endgroup\$ – M. Wother Nov 18 '16 at 21:12
  • \$\begingroup\$ It very much is. You need to add up the currents, but the voltage at a node must be the same from every perspective. \$\endgroup\$ – Marcus Müller Nov 18 '16 at 21:13
  • \$\begingroup\$ Oh you are correct, thank you. Now if you know the answer to the side question that would also be great \$\endgroup\$ – M. Wother Nov 18 '16 at 21:19
  • \$\begingroup\$ Which would be? \$\endgroup\$ – Marcus Müller Nov 18 '16 at 21:19
  • \$\begingroup\$ ah right, that. You should really ask this in a separate question, with an even simpler circuit (just voltage source - resistor). \$\endgroup\$ – Marcus Müller Nov 18 '16 at 21:20
0
\$\begingroup\$

When we talk about voltages, we're really talking about energy. And the thing about energy is that it's not always good at telling us what happens to a single particle at every point in the journey. Your introductory physics class might have used a roller coaster to describe this:

Roller coaster

You don't have to know the forces on the roller coaster along the entire track. All you have to know is the difference in height between the start and the end. Voltage is kind of like that.

Physically, here's what happens. The battery creates an electric field between the terminals. When the wires are connected, the charges in the wires and resistors arrange themselves to channel the field through the circuit. The field pushes on electrons and moves them through the wires and resistors. The force from the field does work on the electrons, giving them energy:

$$W = \vec F \cdot \vec d = q\vec E \cdot \vec d$$

As the electrons move through the resistors, they bump into things, which slows them down. In other words, they lose energy!

Working with electric fields and charges directly would be very, very hard. Fortunately, just as with the roller coaster, we can use energy (voltage) by itself. And instead of looking at individual electrons, we can look at the total flow of charge (current). With that in mind, here are some things that might help:

  • Voltage isn't something the electrons "have". A voltage difference represents the strength of the electric field through part of the circuit.
  • Two resistors in parallel have the same voltage. This is a property of the electric field -- you gain or lose the same amount of energy when you travel between two points, no matter what path you take. (This is also true with gravity, by the way.)

If you'd like a more detailed look at what's going on with the fields and charges in a circuit, check out this document, which uses those ideas to introduce circuit theory. If you don't want to read the whole thing, just look at the diagrams.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.