Problem 4-14Suppose the following resistances are hooked up in series with each other: 112 , 470, and 680 . What is the total resistance of the series combination (Fig. 4-8)?74 Basic dc circuits4-8Three resistors in series(Problem 4-14).Just add the values, getting a total of 112 + 470 + 680 1262. You might roundthis off to 1260 . It depends on the tolerances of the components—how precise theiractual values are to the ones specified by the manufacturer.Resistances in parallelWhen resistances are placed in parallel, they behave differently than they do in series.In general, if you have a resistor of a certain value and you place other resistors in par-allel with it, the overall resistance will decrease.One way to look at resistances in parallel is to consider them as conductances in-stead. In parallel, conductances add, just as resistances add in series. If you change allthe ohmic values to siemens, you can add these figures up and convert the final answerback to ohms.The symbol for conductance is G. This figure, in siemens, is related to the resis-tance R, in ohms, by the formulas:G1/R, andR1/GProblem 4-15Consider five resistors in parallel. Call them R1 through R5, and call the total resistanceR as shown in the diagram Fig. 4-9. Let R1 100, R2 200, R3 300, R4 400and R5 500respectively. What is the total resistance, R, of this parallel combi-nation?Converting the resistances to conductance values, you get G1 1/1000.01siemens, G2 1/2000.005 siemens, G3 1/3000.00333 siemens, G4 1/4000.0025 siemens, and G5 1/5000.002 siemens. Adding these gives G 0. 01 + 0. 005+ 0.00333 + 0.0025 + 0.002 0.0228 siemens. The total resistance is therefore R 1/G1/0.022843.8.