Calculate an electric circuit power

[ex3mist]

Hello,

I hope I can have some help for calculating the power of an electric circuit. Here is my query:

I know that if I have for example 7 sources of energy and I connect them parallel then the end result would be that U=U1+U2+U3+U4+U5+U6+U7. And if I connect them consecutively then I=I1+I2+I3+I4+I5+I6+I7. What happens if the electricity sources are connected as it is shown in the picture? (I hope everything is clear with the image.) How much equals U and I?

Thanks!

 

[dlgoff]

I think you may have things backwards.

Battery packs get their desired operating voltage by connecting several cells in series. If higher capacity and current handling is required, the cells are connected in parallel.

[PLAIN]http://buchmann.ca/article29_files/figure1.jpg

[PLAIN]http://buchmann.ca/article29_files/figure3.jpg

http://batteryuniversity.com/learn/article/serial_and_parallel_battery_configurations"

Regards

 

[ex3mist]

Thanks dlgoff,

Really a picture speaks a thousand words. Obviously I've switched the definitions. The mistake is all mine.
Anyway, what would be the result in the end if the batteries are connected as it is shown in the image I attached? All of them have the same parameters: U1=U2=U3=U4=U5=U6=U7 and I1=I2=I3=I4=I5=I6=I7.

 

[mdjensen22]

Since you now know how the series & parallel sources work, you should be able to start combining sources until you are left with a single source.

For example, U4 & U5 combined together give V4 with 2I4 since parallel voltages are the same, but current sources add together.

With that, rewrite the circuit replacing both U4 & U5 with a single source U4 & 2I4. Repeat this process for the rest of the drawing and let us know what you get.

 

[ex3mist]

What I get at the end is 4.I and 3.U, but as long as I know the power of the circiut equals P=U.I

So, in parallel and consecutive circuit P=7.U.I, but in this case P=12.U.I, am I right? What is wrong?

 

[vk6kro]

You haven't said what the currents are. Are they the maximum current the batteries can supply?

If this is right and all the batteries are the same, then the current through the left battery cannot be more than I1 so the total current can't be more than this either.

However, you can't say this is the current that would flow unless you know what the load is.

With no load (ie with the output open circuited) then the voltage will be 3 times the voltage of one battery ie U1 plus U2 plus U4.

So, if all the batteries were 2 volts, then the output will be 6 volts.

If the maximum current allowed was 5 amps, and the batteries still gave 2 volts at this current, then a current of 5 amps will flow if the load was ( 6 volts / 5 amps) or 1.2 ohms.

If the load was greater than 1.2 ohms, then the current would be less than 5 amps.

If it was less than 1.2 ohms, then the current would be greater than the maximum allowed for U1 even though the other batteries would still be delivering less than their maximum currents.

 

[ex3mist]

Thanks vk6kro,

If I understand you right that means the maximus current of the circuit equals the I1? So, the maximal parameters, if the circuit is loaded to its maximum, would be 3U and 1I, wouldn't it? Which is far less then the parameters if the curcuit is connected parallel or consecutive, am I right?

 

[vk6kro]

Yes, I think you have that right.

You understand that the actual current depends on the load resistor and, in this circuit, the maximum current of the battery on the left is what limits the total current.

The other batteries are not working at their full capacity.

 

[ex3mist]

Thanks again!

 

[sophiecentaur]

ex3mist
The problem is with your original diagram.
You can't really specify a battery in your simple terms of "U and I".
The best model for a battery is to consider it as a 'perfect' Voltage source, or emf (which stays the same always) in series with a small resistance (its internal resistance). This internal resistance will cause a drop in available volts across the terminals when current passes through the battery. As the cell gets flat, the emf may drop and the internal resistance may rise - which is an added complication.
Your quantities 'I' are very difficult to specify - except as a rough guide to the maximum current a battery will supply without dropping its output voltage significantly.

If, on of your diagram, you replaced each cell (with its U and I) with a voltage an emf (U)and a series resistor (R), it would be relatively easy to find out what overall voltage you would expect once a particular load is connected.

I don't think many manufacturers commit themselves to saying what sort of value or internal resistance their cells have, though, which makes things more difficult.

On a practical note, the best thing to do would to use your 7 cells differently - with two parallel pairs and one parallel triplet, all connected in series. This would give three times the individual cell voltage and would have twice the current capacity of an individual cell. The odd, seventh, cell wouldn't make a lot of difference.