Jan 062020
 

Series and Parallel Resistor Reduction

To make a Series and Parallel Resistor Reduction, it is only necessary to do some simplifications using formulas we already know. The situation is different when you have to simplify a circuit that is composed of combinations of resistors in series and parallel.

Series and Parallel resistor reduction procedure

To simplify a complex circuit and to obtain the equivalent resistance, we follow this procedure:

1 – You need to rearrange the circuit we want to simplify, to see clearly the component parts, and discover the parts that are already connected in series and parallel.
2 – We assign to each of these parts, a new name, such as RA, RB, RC, RD, etc. (look at the diagram).
3 – We obtain the equivalent resistance of each part, using the familiar formulas. (resistors in series and parallel resistances).
4 – We replace the parts in the original circuit with the values of the equivalent resistances (RA, RB, etc.) obtained in the previous step.
5 – We analyze the resulting circuit, and seek other possible combinations of resistance groups (parts) in series and parallel, which have been created.
6 – We repeat the process again from step 2 with different names for equivalent resistances to avoid confusion (e.g. RX, RY, RZ, etc.), until we obtain a single final equivalent resistance of the entire circuit.

Analyze the following diagram.

Series and Parallel Resistor Reduction

R1 = 120, R2 = 250, R3 = 68, R4 = 47, R5 = 68. (all in Ohms).
R6 = 5, R7 = 4, R8 = 2, R9 = 1.2. (all in K ohms).

  • RA = R1//R2 = R1 x R2 / (R1 + R2) = 120 x 250 / (120 + 250) = 81 ohms.
  • RB = R4 + R5 = 47 + 68 = 115 ohms.
  • RC = R6//R7//R8 = 1/( 1/R6 + 1/R7 + 1/R8) = 1/( 1/ 5K + 1/4K + 1/2K) = 1053 ohms.

We replace the equivalent values, obtained in the previous steps, in the original circuit, and we get:

Series and Parallel Resistor Reduction

We simplify the circuit again, and we get new resistances equivalent. then:

  • RD = RA + R3 = 81 + 68 = 149 ohms.
  • RE = RC + R9 = 1053 + 1200 = 2253 ohms.

And replacing these latest data, the following circuit is obtained:

Series and Parallel Resistor Reduction

In the latter circuit it can be seen that RB and RE are in parallel, and we get a new RF equivalent resistance that is in series with RD:

RF = RB//RE = RB x RE / (RB + RE) = 115 x 2253 / (115 + 2253) = 109 Ohms.

RF is in series with RD. We add their values to obtain the final resistor value. That is the total equivalent resistance value of all the circuit.

Series and Parallel Resistor Reduction

Then: Req = RF + RD = 109 + 149 = 258 ohms.

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