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.
No. If you imagine a 3-by-3 grid of small loops and consider the loop current in the center loop there is no element or node that has only that current flowing through it.
I'm not sure what is meant by "passive configuration", but the current supplied by the voltage source is opposite the direction of \$i_s\$. This is because
You are using superposition, so for the analysis of \$i_1\$ you have a single source circuit. Since the current from the voltage source is in the opposite direction of \$i_s\$, \$i_s\$ is negative.
As for the case where the voltage source is turned off, the current through the (upper) \$5\Omega\$ resistor is zero because it is short-circuited by the voltage source (\$v_1\$ and \$v_2\$ are the same node). You can view these two paths as a current divider with the short-circuited path as a \$0\Omega\$ resistor -- if the current entering the two paths is \$i\$, then the current through the short-circuited path is
$$\frac{5\Omega}{5\Omega + 0\Omega}i = i$$
Alternatively, the current through the \$5\Omega\$ path is
$$\frac{0\Omega}{5\Omega + 0\Omega}i = 0$$
In other words, all the current flows through the short circuited path and none of it through the \$5\Omega\$ resistor. This means you can ignore the upper \$5\Omega\$ resistor as it has no effect on the circuit (no current passes through it, nor is there any voltage across it). Without this \$5\Omega\$ resistor, you should see that there are only two resistors and the desired current is easily solved with a current divider as you have stated.
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.