I am having difficulties finding the equivalent resistance between points A and B. While a solution is appreciated, I'm rather looking for an explanation/redrawing and understanding of the circuit. Thank you.
Electrical – Equivalent Resistance between A and B
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Related Solutions
Look at the path from b to e: if you go up along the \$5\Omega\$, \$3\Omega\$ and \$4\Omega\$ resistors, up to there you have a total resistance of \$(5+4+3)\Omega=12\Omega\$. This resistance is in parallel to the resistance in the lower path up to that point, which is \$4\Omega\$. Evaluating the parallel connection of these two resistances gives
$$R_1=\frac{12\Omega\cdot 4\Omega}{(12+4)\Omega}=3\Omega$$
To this resistance you have to add the \$12\Omega\$ resistor leading to point e. So you get a total resistance between b and e
$$R=R_1+12\Omega=15\Omega$$
I guess this should be a way to redraw the above schematic:
$${{R_4 \cdot R_5} \over {R_4 + R_5} } + { {(R_2 + R_3)\cdot R_1} \over R_1 + R_2 + R_3} = {R^2 \over 2R} + {2R^2 \over 3R} = {7 \over 6} R$$ which goes in parallel with R6, so:
$$({7 \over 6}R^2) / ({13 \over 6}R) = {7 \over 13}R$$ should be the equivalent resistance.
Best Answer
simulate this circuit – Schematic created using CircuitLab
From this rearrangement, you can see that there are 4 nodes. One way to solve this is nodal analysis. You put a unit current source between a and b, and set B to 0, giving you three node equations:
$$A:\frac{(V_A-V_3)}{2}+\frac{(V_A-V_4)}{1}+\frac{(V_A-0)}{5}-1=0$$
$$3: \frac{(V_3-V_A)}{2}+\frac{(V_3-V_4)}{2}+\frac{(V_3-0)}{3}=0$$
$$4: \frac{(V_4-V_A)}{1}+\frac{(V_4-V_3)}{2}+\frac{(V_4-0)}{4}=0$$
You then solve for A to get the voltage. Since you know the current going into the system (1 Amp), you can use:
$$\frac{V}{I}=R$$
Since I=1, the resistance will actually just be the voltage at A.