Finding the power supplied by a battery

[snoweangel27]

I have a multiple loop circuit, which I was able to apply Kirchhoff's rules to in order to find the current of each resister. However, got stuck trying to determine the power supplied by the batteries (one in each loop). I tried using the equation p= I \epsilon, where \epsilon is the volts of the battery and I is volts of the battery divided by equivalent resistance of the loop that I am solving. Something seems to be wrong in my calculations, and I am not quite sure what I am doing wrong.

 

[John Creighto]

I have a multiple loop circuit, which I was able to apply Kirchhoff's rules to in order to find the current of each resister. However, got stuck trying to determine the power supplied by the batteries (one in each loop). I tried using the equation p= I \epsilon, where \epsilon is the volts of the battery and I is volts of the battery divided by equivalent resistance of the loop that I am solving. Something seems to be wrong in my calculations, and I am not quite sure what I am doing wrong.

You probably did your nodal analysis wrong. If you calculated the currents correctly then you can use P=IV=I(IR)=I^2R .

Note that V is the power across the resister.

 

[rcgldr]

Note that battery voltage and total capacity will drop as the load increases.

 

[uart]

Note that battery voltage and total capacity will drop as the load increases

I suspect that he's using the term "battery" where he really means ideal DC voltage source.

I is volts of the battery divided by equivalent resistance of the loop that I am solving

Well that sounds like the problem right there, why the heck would you do that? Why did you even bother calculating all the currents (presumably) correctly if you were then going to just ignore them and do something silly like what you stated above? The correct value of I to use here is the actual value of I flowing through the voltage source (or battery) in question. You can use Kirchovs current law to find the current in each voltage source from the other currents you have calculated.