# Calculate voltage, current and resistance in a complex mixed circuit

How can I calculate voltage and current on each resistor (especially R5) in this circuit? What is its overall resistance? I have no idea how to do this.

I'd redraw it like this to make it easier to visualize: -

simulate this circuit – Schematic created using CircuitLab

And if you are really lucky and all the resistors are the same value then R3 will pass no current and can be omitted. If not, convert R1 and R4 into a voltage source of 5$\frac{R4}{R1 + R4}$ with an output resistance of R1||R4. Do the same for R2 and R5 then you just have a series circuit with no complexity.

• Okay, maybe I'm just too dumb. While using the following values R1=400Ω; R2=100Ω; R3=200Ω; R4=300Ω; R5=500Ω the R1-R4 voltage source has 2.14V, the R2-R5 voltage source has 4.17V and R3 must drop 4.04V. But somehow, this seems to be complete nuts... :D May 19 '14 at 16:46
• R1/R4 source is 2.143. R2/R5 is 4.167 (correct). R1/R4 impedance is 171.43 ohms. R2/R5 impedance is 83.33 ohms. These impedances are directly in series with R3 making total impedance 454.76 ohms and there is (4.167 - 2.143) volts across implying the current thru R3 is 4.451mA. Does that help a bit? May 19 '14 at 18:39
• Makes sense... But when I simulate your circuit with CircuitLab, I get different values (R1-R4: 2.906V; R2-R5: 3.796V) May 19 '14 at 19:59
• Remember the values for voltages were unloaded - the loading effect of R3 means the voltages will be altered due to their effective output resistances. For instance R1/R4 produces an open circuit voltage of 2.143 BUT its source resistance is R1||R4 which is 171.43 ohms and this is having current taken from it via R3 of 4.451mA which, in turn causes a volt drop of 0.763 volts - add this to 2.143 and you get 2.906 v ta-da!! May 19 '14 at 20:28
• Ahh! Now I got the point of the output impedance. But, to be satisfied, can you explain me why this is R1||R4? Why parallel? Anyway, thanks a lot! May 20 '14 at 10:35

You can use the Y-Delta transform. The wikipedia article contains this very same example under the subsection (simplification of networks).

http://en.wikipedia.org/wiki/Y-%CE%94_transform#Simplification_of_networks

This is easy to solve if you break the solution into steps and think Thevenin for each step.

First solve the voltages without R1 and R5. Now model the point between R3 and R4 as a Thevenin equivalent source, and consider the bottom end of R1 as its own Thevenin source. Then connect the two Thevenin sources to find the voltage a the R1, R3, R4 junction.

Given the voltage at that junction, model the R2-R3 junction as a Thevenin source, connect R5 to it, and compute the voltage at the R2, R3, R5 junction.

Now that you have a final value for the R2, R3, R5 junction, go back and compute the final value for the R1, R3, R4 junction. With all the voltages known, you can easily compute the currents.