Expert: Richard J. Raridon - 2/27/2005

Question
Ok I have got a circuit  with three cells  cell 1 has an emf of 6v and a  internal ressistance of  2 ohms cell 2 has  an emf of 2v and an internal resistance of  4 ohms  cell 3 has emf  of  3v  and an  internal resistance of 3 ohms . they is a  6 ohm resistor  a  3 ohm resistor and a 7 ohm resistor also in the circuit . cell 3 is connected the wrong way so  it has  a –emf. So using kirchoffs second law I got

I am suppose to  find the current flowing in the circuit
So using kirchoffs second law I got
6+3+2 = 2I+4I+3I+6I+3I+7I
I = O.2 A

The  ?? then ask to calculate the potential at different  points around the circuit
In confident  I did so correctly
But this part is what has  me stumped.
It says  find the potential difference between the points
Am I just to subtract the potential at the two points ?  are they any special directions or rules I should apply when doing so ?


My second  ?? is when u connect a load  in parallel with a resistor in a  pontential divider  circuit the ressitance of the load should be high compared with the value of resistance of the resistor so that it doesn't decrease the share of the voltage ?  

Answer
I presume all of these are simply connected together in a single series circuit.  If so, you almost got it right, but you need to subtract the 3 volts since it's receiving energy from the circuit rather than contributing it.  So you have 6-3+2 = 25I and I = 0.2 amps which is the answer you listed.  To get the potential difference between two points, you simply add up the potential changes between those two points.  All of the potential changes across the resistors will be negative if you go in the direction of the current flow.  For example, the potential difference across the 6 volt battery would be 6 -0.2(2) = 5.6 volts.
The answer to your second question is yes.