Batteries in Parallel

What you want to do is to attack this problem by using nodal equations, as you can see from the drawing this circuit has only one node placed in the junction between the three resistors, you can start by assuming the following:

1. By KCL the current at the node is equal to 0

Since we don't know anything about the direction of the currents, we can safely let the node voltage determine the direction.

2. Let the [tex]V_{x}[/tex] denote the node voltage
Let [tex]I_{1}[/tex] denote the current from 28.3 V source
Let [tex]I_{2}[/tex] denote the current from 14.15 V source
Let [tex]I_{3}[/tex] denote the current in the 16.5Ohm resistance

3. Applying the nodal equations we get the following:

[tex]I_{1}+I_{2}+I_{3}=0[/tex]

Note that the current can take any direction. Now we replace currents with voltage drops divided by resistor values:

[tex]\frac{V_{x}-28.3}{30.6}+\frac{V_{x}-14.15}{4.32}+\frac{V_{x}}{16.5}=0[/tex]

Solving for [tex]V_{x}[/tex] we get 12.9

Now we can solve for the current in the leftmost resistor:

[tex]I_{3}=\frac{V_{x}}{16.5}=0.784A[/tex]

DONE!

 

No problem, I hope you understood the steps towards the solution.

 

Correct and remember if the batteries were switched the other way around, then you would use [tex]V_{x}+V_{battery}[/tex] instead

I won't argue about that, but i thought that making the complete description of the steps including the description of what rules i applied would help him better, so he/she will know what to do next time he/she is faced with the problem.

 

Hehe, shure. First you have to determine the current flowing through the loop, note that the current will be the same at all places. When You've determined the current in the loop, start by writing the equation expressed as voltage drops , you should start at point a and move towards point b, note that the first voltage drop across the battery will be negative, since you're moving from -to +, and it will be positive across the resistors and positive across the battery that is near the b point, since the battery is going from + to -.

 

As all the components are in series you can rearrange the circuit to group all the cells and all the resistors.
This should make it easier to work out the circuit voltage and hence the current.

 

LOL, ok ;)