How to Calculate Series and Parallel Resistance

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26-10-2016, 04:05
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Need to know how to calculate series resistance, parallel resistance, and a combined series and parallel network? If you don't want to fry your circuit board, you do! This article will show you how in just a few easy steps. Before reading this, please understand that resistors do not actually have an "inside" and an "outside" to them. The use of "in" and "out" is merely a figure of speech to help novices understand the wiring concepts.

Help Calculating Resistance

Series Resistance

  1. What it is. Series resistance is simply connecting the "out" side of one resistor to the "in" side of another in a circuit. Each additional resistor placed in a circuit adds to the total resistance of that circuit.
    • 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
    • In this example, R1 = 100 Ω and R2 = 300Ω are wired in series. Req = 100 Ω + 300 Ω = 400 Ω

Parallel Resistance

  1. What it is. Parallel resistance is when the "in" side of 2 or more resistors are connected, and the "out" side of those resistors are connected.
    • The equation for combining n resistors in parallel is: Req = 1/{(1/R1)+(1/R2)+(1/R3)..+(1/Rn)}
    • 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.

Combined Series and Parallel Circuits

  1. What it is. A combined network is any combination of series and parallel circuits wired together. Consider finding the equivalent resistance of the network shown below.
    • 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.

Tips

  • Remember, when resistors are in parallel, there are many different means to an end, so the total resistance will be smaller than each pathway. When resistors are in series, the current will have to travel through each resistor, so the individual resistors will add to give the total resistance for the series.
  • The equivalent resistance (Req) is always smaller than the smallest contributor for a parallel circuit; it is always greater than the greatest contributor for a series circuit.
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