# Problem with equivalent resistance

I have a problem determining equivalent resistance of this circuit. The previous assignment had the battery between C and 3R1 resistor, and i knew how to calculate the equivalent resistance. I calculated the series resistance between R1 and 3R1 and then calculated this result in parallel to R1. To this result i added R1 and 3R1. I thought this assignment, which has the battery between C and D would be solved the same way. But no, solutions say that the result is just parallel resistance between 2R1 and 6R1 as it's seen on a picture below. So now i wonder where has R1 between A and B points gone. It's completely omitted in the solutions. Just like if it wasn't there. How is it possible?

• You need to tell between which two points the resistance should be seen. It is not obvious which it is, so it seems fair to give it a downvote. Jun 25, 2022 at 16:44 simulate this circuit – Schematic created using CircuitLab

Here's the circuit redrawn. R1 and R2 form a voltage divider, as do R3 and R4. Since the ratios are the same for each voltage divider, the voltage at the midpoint of each divider (NODE1 and NODE2) is the same since they are both being supplied by the same voltage V1. If the voltage between two points is the same, no current will flow between them, no matter what resistance connects them. If there is a resistor connected between them, as in your example, it doesn't participate in the circuit and can (with the usual caveats: assume ideal components, assume zero resistance/inductance/capacitance interconnects, assume RF doesn't exist, etc) be ignored the same as a resistor connected on one only end or a resistor that's shorted out.

• Thanks. It's actually logical. But what if the ratios weren't the same? How would i calculate equivalent resistance then? Jun 7, 2022 at 19:38
• @PavleHribar One way is to do a delta-to-wye conversion on, say, the resistors between A, B, and C and combine the resulting resistances. Or do you could do a mesh analysis of the circuit to find the current I Jun 7, 2022 at 19:55
• "...assume RF doesn't exist, etc." Every time I do that my WiFi stops working and I can no longer access CircuitLab.
– cjs
Jun 8, 2022 at 8:58

If we redraw the schematic as: simulate this circuit – Schematic created using CircuitLab

I can work you through the process of finding the above equation using KCL and the freely available SymPy:

var('r1 r2 r3 r4 r5 I V va vb')
eq1 = Eq( V/r1 + V/r2, I + va/r1 + vb/r2 )
eq2 = Eq( va/r1 + va/r3 + va/r5, V/r1 + vb/r5 )
eq3 = Eq( vb/r2 + vb/r4 + vb/r5, V/r2 + va/r5 )
simplify( solve( [ eq1, eq2, eq3 ], [ V, va, vb ] )[V]/I )
(r1*r2*r3 + r1*r2*r4 + r1*r2*r5 + r1*r3*r4 + r1*r4*r5 + r2*r3*r4 + r2*r3*r5 + r3*r4*r5)/(r1*r3 + r1*r4 + r1*r5 + r2*r3 + r2*r4 + r2*r5 + r3*r5 + r4*r5


Then the impedance it presents is:

$$R=\frac{R_1\,R_2\,R_3 + R_1\,R_2\,R_4 + R_1\,R_3\,R_4 + R_2\,R_3\,R_4 + R_5 \left(R_1\,R_2 + R_1\,R_4 + R_2\,R_3 + R_3\,R_4\right)}{R_1\,R_3 + R_1\,R_4 + R_2\,R_3 + R_2\,R_4 + R_5\left(R_1 + R_2 + R_3 + R_4\right)}$$

If $$\R_5=0\:\Omega\$$ then it reduces to:

$$R=\frac{R_1\,R_2\,R_3 + R_1\,R_2\,R_4 + R_1\,R_3\,R_4 + R_2\,R_3\,R_4}{R_1\,R_3 + R_1\,R_4 + R_2\,R_3 + R_2\,R_4}$$

Which is the exact same thing as $$\\left(R_1\mid\mid R_2\right)+\left(R_3\mid\mid R_4\right)\$$.

If $$\R_5=\infty\:\Omega\$$ then it reduces to:

$$R=\frac{R_1\,R_2 + R_1\,R_4 + R_2\,R_3 + R_3\,R_4}{R_1 + R_2 + R_3 + R_4}$$

Which is the exact same thing as $$\\left(R_1+ R_3\right)\mid\mid\left(R_3+ R_4\right)\$$.

In your case, this really does simplify down to $$\R=\frac32 R_1\$$.