Mesh Analysis for Current Source in between “blocks”

circuit analysis

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.

Best Answer

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.

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