Help with current flow, voltage, resistors in basic circuit

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1. Jan 15, 2016

marciokoko

Im reading this stack exchange question:

http://electronics.stackexchange.co...o-choose-value-of-resistor-in-voltage-divider

So there is a 5V circuit with 2-100 Ohm resistors that yields a voltage drop across the divide of 2.5V. The current through the circuit is 25mA.

There is a second circuit the same 5V potential with 2-1M Ohm that yields a voltage drop across the divide of 2.5V. The current through the circuit is 2.5uA.

That is due to the fact we added those resistors. The larger the resistor, the smaller the current because more electrical energy is wasted as heat energy.

Then he goes on to explain that he adds a load to the first circuit, a 1K Ohm load-resistor to what used to be a 2.5V line. Now the potential is 2.38V instead of 2.5V and it has a 2.38mA current through it. It explains because: because the two bottom resistors are in parallel. I don't understand why. I know the total resistance in a circuit with parallel resistors is 1/Rt = 1/R1 + 1/R2 but how do I get 2.38mV and 2.38mA?

This is what I came up with:

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2. Jan 15, 2016

Merlin3189

Your calculation for RT is wrong.
You have $\frac{1}{R_T} = \frac{1}{100} +\frac{1}{1000}$
Then $R_T = 100 + 1000$ and then $R_T = 1100$
The first line is correct, but then your arithmetic goes astray. When you put any resistance in parallel with 100Ω, the result must be less than 100Ω
When you get the correct resistance for $R_T$ the voltage and current work out ok.

3. Jan 15, 2016

marciokoko

Oh! 90.9.
I mistakenly tried to elevate everything to the -1 power and flip everything over. Got greedy!

4. Jan 15, 2016

marciokoko

One more thing about resistors in parallel. In this circuit I can understand that the voltage is split between both resistors in series and the 5V goes to 2.5V across each resistor with a 2.5mA current.

but then when I add a 3rd resistor in parallel to the second resistor in the circuit, the new voltage across the divide is 1.67V. So how should I view this new circuit?

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5. Jan 16, 2016

Merlin3189

The two 1k resistors are in parallel, = 500Ω
then you have 1k in series with 500Ω,
so V1k = 5V x 1k/(1k + 500) = 3.33V
and V500= 5V x 500/(1k+500) =1.67V

I'm not sure what your diagram is showing: It looks like the potential difference with respect to ground, the negative of the battery being grounded. So it is showing the potential difference across the two parallel resistors.
Why it shows 0 A is a puzzle. If the circuit is broken there, then ok there will be zero current, but then the voltage should be either 0V or 5V. If the circuit is joined there, the voltage should be 1.67V wrt ground, but the current would be 3.33mA.

6. Jan 16, 2016

marciokoko

Yeah, but what is confusing me is this. The V drop across the 1k resistor should be 0.0033A x 1000 Ohms = 3.3V but the picture shows 1.67V.

Then of course the drop of the 0.0033A x 500 Ohm resistor (2-1k in parallel) would be 1.67V.

Why am I seeing it backwards?

7. Jan 16, 2016

Merlin3189

Well this is the thing. The picture shows voltage at a point rather than across a component or between two points. But in circuits voltage is always between two points.
As I said in #5, it looks as if the negative of the battery is connected to ground, an arbitrary 0V point, then voltages can be given at other points meaning the voltage between that point and ground.
So the 1.67V is between the blue bars point and the negative of the battery (ground.) Clockwise the only components between these two points are the two 1k resistors in parallel. So 1.67V is the voltage across these.
Anticlockwise there are two components, the 1k resistor on the left and the 5V battery. So -1.67V = -5V across battery + 3.33V across the 1k resistor. (Connect the common of the voltmeter to the blue bars point, the other side of 1k will be + 3.33V, then the battery takes you 5V negative going from its + to -, and you will read -1.67V when you contact the battery negative (ground.)

In situations like this it may be helpful to imagine connecting your voltmeter (common or black) to one point, then think what voltage you read as you go round the circuit. The voltage across any component will be the difference between the ends. Here you would get, starting from battery negative= 0V, at battery +ve =+5V (so 5V across battery), at left of 1k = +5V (connected to battery), at right of 1k =1.67V (so 1.67-5=-3.33V across the 1k), at left junction of resistors = 1.67V, at right junction of resistors= 0V (so 0-1.67= -1.67V across the pair of resistors), and back to 0V along the bottom wire. (Going round the other way the voltages would be opposite in sign.)

8. Jan 16, 2016

marciokoko

Here is the current across 2 points. But I still get 0V after the first resistor, which isn't correct.

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9. Jan 16, 2016

sophiecentaur

I can't help you with the arithmetic but Hyperphysics has a page that could be just what you need. An algebraic approach, before putting numbers in the problem is always going to take you further in Science. The best thing to do is to wait until the end before plugging the numbers in.

It's 0V because it's connected to the -terminal of the battery and someone is assuming that is your reference ground. The current calculation is perfectly correct for 5V across 1k5
If someone is not aggreeing with that then they are talking about a different circuit. (I'm discounting incompetence / typos but don't totally rule it out)

10. Jan 16, 2016

Merlin3189

It is correct if the negative battery is called 0V, If you look at this bit of wire, it is connected to negative of battery, so is 0V.
The 3.33mA through the pair of 1k adds 1.67V on the left end of the pair.
The 3.33mA through the single 1k adds 3.33V to that, giving 5V on its left end, which is connected to the battery positive which is +5V from its negative terminal.

I've put some voltages on your diagram - as would be shown by your software and as would be shown by voltmeters.

11. Jan 16, 2016

marciokoko

Merlin

I think it's more a matter of me not knowing how to use icircuit. I just got it yesterday. Yes I see it makes sense that it's giving me 0V.

Those voltmeter markings you added yourself with a graphics program right? That's not a voltmeter option on icircuit right?

I think I need to draw a wire across the two points and then place that blue meter thing over the wire that I draw going across the 2 points I want.

12. Jan 16, 2016

marciokoko

13. Jan 17, 2016

Merlin3189

Yes, I just altered your .png with Paint.

I'm not sure whether putting the blue line between two points either side of a component will work? It may short out the component?
Maybe there will be a voltmeter in icircuit?
Your software is ok and you can see what you need, so long as you bear in mind what it is telling you. Presumably you have to set a ground connection somewhere? Then you could try setting it to different points and see what you get. Enjoy playing with it.

14. Jan 19, 2016

marciokoko

Ok Ive been playing around with the software and Im confused about something else but its related Im sure.

I understand that the leg before the first resistor is 25mA and 5V between IT and Ground.
I understand the resistor accounts for a V drop to 2.5V as measured in the wire between both resistors.
I understand the V drops to 0V after the second resistor because there is no longer a V difference between that point and GND.

But why does the current reverse sign when I monitor the current over the 2nd resistor, from -25mA to 25mA and -2.5V to 2.5V.

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15. Jan 19, 2016

davenn

try again

16. Jan 20, 2016

marciokoko

Ok here is a larger version:

17. Jan 20, 2016

marciokoko

18. Jan 21, 2016

davenn

19. Jan 23, 2016

Tom.G

Regarding posts 14 & 16:

I'm not familiar with icircuit, but here is my speculation on the sign reversal.

Simulators tend to distinguish between the the leads of a component. Resistor (and capacitor) leads are usually numbered, "1" and "2".
Since icircuit represents a meter as an overlay on a component, it's likely sensitive to the rotation of that component.
For instance, the 100 ohm resistor that shows reversed polarity; you may have rotated it before wiring it into the circuit.
This is seen in simulators that use a "probe" representation for a meter. The current polarity that is displayed depends on which resistor terminal is measured.

20. Jan 23, 2016

sophiecentaur

I do not like simulators very much. They can get in the way of understanding because they obscure those patterns that the Maths can show so well. However, they do achieve one important thing for the user. Because they are essentially GIGO, they demand absolute consistency from the user and you can't get away with an intuitive approach to what's up and what's down. By assigning numbers to every node in a circuit, you are forced to think, which can't be a bad thing.