Page 94 - Teach Yourself Electricity and Electronics

74 Basic dc circuits

Problem 4-14
Suppose 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)?

4-8 Three resistors in series
(Problem 4-14).

Just add the values, getting a total of 112 + 470 + 680 1262 . You might round
this off to 1260 . It depends on the tolerances of the components—how precise their
actual values are to the ones specified by the manufacturer.

Resistances in parallel

When 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 all
the ohmic values to siemens, you can add these figures up and convert the final answer
back to ohms.
The symbol for conductance is G. This figure, in siemens, is related to the resis-
tance R, in ohms, by the formulas:
G 1/R, and
R 1/G

Problem 4-15
Consider five resistors in parallel. Call them R1 through R5, and call the total resistance
R as shown in the diagram Fig. 4-9. Let R1 100 , R2 200 , R3 300 , R4 400
and R5 500 respectively. What is the total resistance, R, of this parallel combi-
nation?
Converting the resistances to conductance values, you get G1 1/100 0.01
siemens, G2 1/200 0.005 siemens, G3 1/300 0.00333 siemens, G4 1/400
0.0025 siemens, and G5 1/500 0.002 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/G
1/0.0228 43.8 .