I have the following question: Calculate the current for each battery in the circuit below and state wether it represents a charge or a discharge. Now I have a basic understanding of mesh currents, I know I have to label each mesh. Now in this circuit I originally thought that there were three meshes. And as i started to try and work it out i'm getting more confused. Mesh 1 Starts at the 8V battery goes through the 12ohm resistor and goes down towards the 6V battery. If this is right I dont know how to write the first equation since every other example i've done has had another resistor there to use as a value. Mesh 2 is the 6V battery going through the 220ohm resistor and coming back to itself. Mesh 3 is the 12V batery going through the 220ohm resistor and then through the 47 ohm resisitor and back to itself. Can someone please put me on the right path. The more i look at it th more confused i'm getting.
Hi alley: This is a fun question! I have gone over this question-it teaches some important circuit analysis concepts. The question states for you to solve for the charge or discharge state of the batteries. Then you mention mesh analysis; were you actually instructed to use this analysis or do you just want to? Remember that a dc source is ideally looked at as 0 ohms to the system when applying other analysis tools like for example superposition. If you still do not get the problem, you could use an electronic software analysis tool-set the circuit up and view the currents-then go back to the question and try by hand again. Remember no pain no gain (emotionally speaking). And I dont know maybe you are already at that point, but your post does not convey this. Also you should post the work you have attempted-easier to guide you!
ally, No need to get tangled up in the mesh net. Let's take it step by step and see if you agree with my reasoning. OK, we can easily see that there is 2 volts across the 12 ohm resistor for a current of 0.16666.... amps which the 8 volt battery is supplying. Also we can easily see that there is 6 volts across the 470 ohm resistor for a current of .012765957 amps which the 12 volt battery is supplying. Node B is locked to 6 volts so there is 6/220 or 0.0272727 amps leaving the node through the 220 ohm resistor. So we have 0.1666 + 0.012765 = 0.179432 amps entering the node from the 8 volt and 12 volt batteries, and only 0.0272727 amps leaving the node through the 220 ohm resister. The difference is 0.152159 amps which is forced through the 6 volt battery. Ratch