4 Ways To Calculate Series And Parallel Resistance

The formula for calculating a total of n number of resistors wired in series is: Req = R1 + R2 + . . . . Rn That is, all the series resistor values are simply added. For example, consider finding the equivalent resistance in the image below[2] X Research source In this example, R1 = 100 Ω and R2 = 300Ω are wired in series. Req = 100 Ω + 300 Ω = 400 Ω

The equation for combining n resistors in parallel is:Req = 1/{(1/R1)+(1/R2)+(1/R3). . +(1/Rn)}[4] X Research source Here is an example, given R1 = 20 Ω, R2 = 30 Ω, and R3 = 30 Ω. The total equivalent resistance for all 3 resistors in parallel is:Req = 1/{(1/20)+(1/30)+(1/30)} = 1/{(3/60)+(2/60)+(2/60)} = 1/(7/60)=60/7 Ω = approximately 8. 57 Ω.

We see the resistors R1 and R2 are connected in series. So their equivalent resistance (let us denote it by Rs) is: Rs = R1 + R2 = 100 Ω + 300 Ω = 400 Ω. Next, we see the resistors R3 and R4 are connected in parallel. So their equivalent resistance (let us denote it by Rp1) is: Rp1 = 1/{(1/20)+(1/20)} = 1/(2/20)= 20/2 = 10 Ω Then we see the resistors R5 and R6 are also connected in parallel. So their equivalent resistance (let us denote it by Rp2) is: Rp2 = 1/{(1/40)+(1/10)} = 1/(5/40) = 40/5 = 8 Ω So now we have a circuit with the resistors Rs, Rp1, Rp2 and R7 connected in series. These can now simply be added to get the equivalent resistance R7 of the network given to us originally. Req = 400 Ω + 20Ω + 8 Ω = 428 Ω.