# Resistance between any two points

I saw a series of simular questions, where was calculated the equivalent resistance of a circuit, but I don't understand a algorithm of calculation for a pair of specific points. I found a solution for calculating a resistance $$\R_{CD}\$$.

$$R_{3,45}=R_3+(R_4 \cdot R_5)/(R_4+R_5 )$$ $$R_{CD}=(R_2 \cdot R_{3,45})/(R_2+R_{3,45} )$$

Here I don't understand why $$\R_{3,45}\$$ and $$\R_2\$$ are considered as parallel-connected resistors. As I understand, these resistors are connected in series. Why $$\R_5\$$ and $$\R_2\$$ are included in this calculation, but $$\R_1\$$ no?

How can I calculate $$\R_{BF}\$$?

• Why does the question ask how to calculate Rbf, since as far as I can tell nodes B and F are directly connected together. Commented Aug 16 at 22:41
• @ChesterGillon the question was formulated exactly this way. There was an answer of 0 Ohm. Commented Aug 17 at 7:54
• @ArturKrush In directed graph theory, which is important here, you don't get to create two differently named vertices that are the same vertex. B and F are the same vertex. So you have to pick one name or the other. But you don't get to keep both. You could keep them named separately if-and-only-if you introduce an edge between them with a finite conductance. But an ideal wire doesn't have a finite conductance. Commented Aug 17 at 22:24

$$\R_{BF}\$$ is shorted directly, so the calculation is pretty easy.

Redraw the circuit and you'll see that the answer for $$\R_{CD}\$$ has to be

$$R_{CD} = R2 \parallel (R3 + R4 \parallel R5)$$

You can ignore any nodes or resistors that don't get connected anywhere as they don't contribute to the circuit.

It might help to redraw the circuit to better represent the series and parallel combinations of resistances found between C and D:

simulate this circuit – Schematic created using CircuitLab

Disregard R1, since it doesn't contribute to resistance between C and D. First replace R4 and R5 with their equivalent combined resistance $$\R_6 = R_4 \parallel R_5\$$:

simulate this circuit

R3 and R6 are in series, so replace them with their combined value $$\R_7 = R_3 + R_6 = R_3 + (R_4 \parallel R_5)\$$:

simulate this circuit

I believe $$\R_7\$$ here is what you called $$\R_{3,45}\$$.

Clearly R2 and R7 are in parallel, so replace them with their equivalent single resistance $$\R_{CD} = R_2 \parallel R_7\$$. The complete expression for $$\R_{CD}\$$ is:

$$R_{CD} = R_2 \parallel \left(\vphantom{\frac{}{}} R_3 + (R_4 \parallel R_5) \right)$$

$$\_{edited}\$$

Computing Rcd means applying a sequential reduction to parallel $$\R_4||R_5\$$ and series $$\+R_3\$$ circuits then applied $$\|| R2\$$.

$$\(R_4||R_5+R_3)||R_2=R_{CD} = (10||10+10) ||10=15||10\$$

B,E,F,G are common while R1 is external.

• No, that's not quite right. You have to combine R4 and R5 before adding that to R3. Commented Aug 16 at 23:55
• Thank you everyone for help! Commented Aug 17 at 7:55
• @td127 TY for the feedback. Commented Aug 17 at 17:35

The circuit is very simple so the various equivalent resistances between the terminals can be calculated by inspection. In general, however, that is for more complex electrical networks, we can apply to each input (or pair of terminals) that interests us, an independent test current generator and perform the analysis of the network on a topological basis and determine voltages and currents in each branch. Once this is done, we leave only one generator and calculate the voltage-current ratio of the relative branch that corresponds precisely to the equivalent resistance between the two terminals to which the current generator was applied.

• thank you for the answer. There is something to think about for me. Commented Aug 19 at 8:17