current division in circuit

[scientifico]

Hello, i'm new to electronics and to check if I have correctly learnt the basics I drawed a simple circuit where I writed the values of the current around all the circuit, can you tell me if they are correct? (R3 is 1776 ohm not 1776 kohm)

Thank you!

 

[jackorocko]

Looks good to me.

 

[(*steve*)]

The figures seem reasonable. I didn't actually do the math, but by approximation they are well within the range I would expect.

Perhaps you can show us your working and we can confirm that you're doing the calculations the right way. If you are then you've got both the right answer (less important) and also the right method (vastly more important).

I note that two of your currents in one part of the circuit don't add up to the current in another part, but that seems to be simply an issue of using a different number of significant digits. You probably should try to keep that constant.

 

[Laplace]

Your math is correct, subject to rounding errors. The placement of current meters within the circuit confirms that the current in a loop is the same everywhere in the loop. Where would you place volt meters in the circuit to confirm that the sum of voltage drops around a loop equals zero?

 

[john monks]

You get an "A" for today!!!

 

[scientifico]

I think voltages should be like this...
In a parallel circuit voltage is the same in every component but why 0.0011*1176 isn't equal to 0.0049*390 ?

Thank you all for answers!

 

[(*steve*)]

But you do -- the results are exactly the same given the number of significant digits you're using.

Part of my mental math was that the 390 ohm resistor was somewhat less than 1/4 of the resistance of the 1776 ohm resistor, so the current should be a little more than 4 times as large. Plus, the currents should add up to around 6mA. Both of these hold true.

Likewise the parallel combination of the 390 ohm resistor and the 1776 ohm resistor is going to be around 80% of the value of the 390 ohm resistor so say approx 310 ohms (it's actually closer to 320, but I had to calculate that) and that is just under half the resistance of the 673 ohm resistor, so I'd expect the voltage across the 673 ohm resistor to be a fraction over double the voltage across the pair of parallel resistors. And it is.

You can check your work in your head (and that's good practice for exams -- if you can estimate a value that turns out to be close to the answer you calculate then you can be more confident in your calculations)

 

[scientifico]

But you do -- the results are exactly the same given the number of significant digits you're using.

They aren't the same
0.0011*1176 = 1.2936
0.0049*390 = 1.911

 

[(*steve*)]

Look at your original circuit. It's 1.776 k, in your second it has morphed to 1.176k

Using the values in your original circuit it's right. If you change the values of the resistors then other stuff will change.

I didn't notice the change in the resistor value at first. I worked off the original diagram because I wanted to check that you had copied the currents across properly and I didn't think about the resistors.

 

[scientifico]

Yes I mistaked the resistance value, excuse me

 

[scientifico]

Hello, today I tryed in the same circuit to put the voltmeter in series between the first resistance and the other two, like an amperometer and it visualized the total voltage of source (6 V) with no drops.
I know that the voltmeter must be putted in parallel, but does it shows the total voltage because it absorbs so little current that resistances can't resist to this very little amount of current?

Thank you!

 

[(*steve*)]

Well, the can and they do, but the effect is so small in relation to the impedance of the meter that it would be well below its resolution.

 

[scientifico]

So, if the voltmeters in the picture were extremely exact they would measure both a voltage drop? will the drop be the same in both voltmeters?

 

[Laplace]

If by 'extremely exact' you mean an ideal voltmeter, then the meter would have infinite resistance. Voltmeter 1 would read 6 V while Voltmeter 2 would read zero.

 

[scientifico]

Why would the multimeter 2 read 0 V if the potential difference across the resistance is 6 V?
Thanks!

 

[Laplace]

How much current is flowing through R1 with Voltmeter #1 connected in this unconventional manner?

 

[(*steve*)]

...remembering that a perfect voltmeter is an open circuit that therefore does not allow current to pass through it.

 

[scientifico]

How much current is flowing through R1 with Voltmeter #1 connected in this unconventional manner?

The current in the circuit should be 6/150 = 0.04 A but the voltmeter 2 is connected in the right manner so why you say it shows 0 V ?

 

[jackorocko]

The current in the circuit should be 6/150 = 0.04 A but the voltmeter 2 is connected in the right manner so why you say it shows 0 V ?

answer laplace's question, there is zero current through R1 so there is no voltage drop. The reason why was posted by steve in the post previous to yours.

 

[scientifico]

is it possible say that the current in the resistance of these 2 circuits is the same? I mean that the wire present in the left circuit doesn't matter...
Thank you