What formulas do I need to calculate the resistor equivalent for not just the normal parallel and serial connection?

e.g resistance from A to B og these five resistors: Resistor diagram

Is Kirchhoff's laws and hard work my only option?


Yes, when you're faced with a bridge (which is the common name for this topology), Kirchoff's laws, and pain, are your only choice.

There is one shortcut you can try, which MAY simplify your life. Treat R1,R2 and R4,R3 as independent voltage dividers, and R5 as an open circuit, and see if the two ends of R5 would be at the same voltage. If they are, then the bridge is said to be "balanced", and no current flows in R5. If they are not at the same voltage, then the bridge is "unbalanced", and you get to start calculating loop currents and node voltages, while cursing creatively.

When you see a bridge, ALWAYS calculate the conditions necessary for balance. You may get a pleasant surprise.

  • \$\begingroup\$ It's not true that Kirchhoff is the only way out. Delta-star transformations simplify here. \$\endgroup\$ Jan 6 '12 at 15:52

You can try to use triangle-star transformation in a bridge circuit to get it down to a combination of series and parallel resistor connections.

I've taken the R1, R4 and R5 triangle and turned it into Ra, Rb and Rc star.

Here's how it would look like: enter image description here

The formulas are:

\$\Large R_a=\frac{R_1 R_4}{R_1 + R_4 + R_5}\$

\$\Large R_b=\frac{R_5 R_4}{R_1 + R_4 + R_5}\$

\$\Large R_c=\frac{R_1 R_5}{R_1 + R_4 + R_5}\$

Now you can use series calculations with Rc and R2 and Rb and R3 to turn them into a single resistors and then use parallel formula to turn those two resistors into one.

Here's a Wikipedia article on that.

I could have used same transformation on R2, R3 and R5 and in the end I would have gotten the same result.

There is also the reverse star-triangle transformation. Note that is something is connected to the central node of the star, you can't use the transformation.

To get the original R1, R4 and R5, use following formulas:

\$\Large R_1=\frac{R_a R_b + R_b R_c + R_c R_a} { R_b}\$

\$\Large R_4=\frac{R_a R_b + R_b R_c + R_c R_a} { R_c}\$

\$\Large R_5=\frac{R_a R_b + R_b R_c + R_c R_a} { R_a}\$

The transformations also work for complex impedances and you can sometimes get impedances with negative magnitude when you use it. That is also normal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.