# Mesh Analysis for Current Source in between “blocks”

I can only find one equation with two unknowns.

$$-1(I_1) - 3(I_1) + 2(I_2) = 0$$

$$6\text{A} = (I_2) + (-I_1)$$

I need to find the voltage across the $1\Omega$ resistor. I am currently stuck on finding the current that flows through each loop. You can see it as a current divider consisting of the 2Ω resistor and the other two resistors that add up to 1Ω + 3Ω = 4Ω.

So the proportion of the conductances and thus the proportion of currents of both branches are
1/4 : 1/2 or 1:2,
i.e. 1/3 of the total 6A are going one way and 2/3 the other way, i.e. 2A are goind through the left branch and 4A through the right branch.

If 2A are going through a 1 Ω resistor the voltage across it is 2V.

1. You have a sign error in your second equation. To match the direction convention used in the first equation, it should be

$$6 \mathrm{A} = I_2 + (-I_1)$$

1. You have two equations in two unknowns and you can solve them using the normal procedures of linear algebra.

2. You don't even need to do mesh analysis on this simple circuit; you can just use the current divider rule.

3. To get the voltage across the 1$\Omega$ resistor, you just need to use Ohm's law, keeping track of the sign convention for the mesh currents and the passive sign convention:

$$V_1 = (-I_1)(1\Omega)$$