# Circuit analysis

1. Mar 1, 2009

1. The problem statement, all variables and given/known data

Find the three currents in the image.

2. Relevant equations
Khirkoffs laws
Sum of currents equal zero.
The potentials around a closed loop add up to zero.
The equations given in the solution are
$$-2I_1 + 12I_2+6=0$$
$$-12I_2+8 = 0$$
$$I_1+I_2+I_3=0$$

3. The attempt at a solution

When I tried to solve this I had the first equation to be
$$-2I_1-12I_2+6=0$$, by doing a potential walk round the left subcircuit (?).
Why should the second term be positive?

File size:
3.9 KB
Views:
31
2. Mar 1, 2009

### Staff: Mentor

By convention, when you have the current going down through a component, the top of the component will be + and the bottom will be -. Since you are going around the loop on the left against the direction labelled for I1, it is a negative voltage term. Since you are continuing around that clockwise loop on the left and go through the middle resistor in the same direction as the current arrow shown, it is a + voltage drop.

3. Mar 1, 2009