Basic DC Electrical Theory 03

Harold Hall


I should add that whilst the explanations are in terms of resistors the circuits may be built from other types of components, for example, a heater, light bulb, or the input to a transistor. In these cases we are talking about the numerical value of the device's resistance in ohms. A DC motor can also have dropper resistors but this is a slightly different case.


Choice of values

Considering Sk 6, the viewer may question why is resistor R2 required, as surely the load (R3), together with resistor R1, will provide the required voltage divider? This is acceptable providing the load is fixed. A typical example being a 6 volt bulb powered from a 15 volt supply. In this case the resistor would require to drop 9 volts. If the bulb had a resistance of 12 ohms then the resistor would need to be 18 ohms as it requires to drop 1 1/2 times the bulb voltage. Incidentally, it is not possible to accurately measure the working resistance of a cold bulb so it will be necessary to work out the resistance from its quoted wattage. Similarly, one cannot measure the apparent resistance of an electric motor whilst stationary, more about that later.


Two speed motor

At this stage one could easily fall into the trap of considering that as the ratio is the important factor the actual resistor values are unimportant, this is not totally the case. Above we have explained how for a fixed load the resistor R2 is superfluous but in many cases the load is variable and R2 performs a valuable function.


Consider a small DC motor to power a milling machine table feed. Ideally, this should run at one speed whilst a component is being machined but faster when traversed away so as to save time.


Workshop Data

Let us consider that the motor requires 12 volts (fast) and 3 volts (slow) and draws a current of 0.5 amps at the lower speed. Sk 8 shows the basic circuit.

One of Kirchoff's two laws states that "the algebraic sum of all the currents meeting at a point in a circuit equals zero" or to put it in a way that I prefer the sum of the currents arriving equals the sum of the currents leaving. Relating that principle to Sk 8 you will see that the current in R1 is equal to the current through the motor (Im) plus that through R2 (Ir).


If now the value of R2 is chosen to have a value that results in the same current through it as the motor (0.5 amps), the current through R1 will be 1 amp. The value of R1 then being chosen to drop 9 volts at 1 amp. If however the load on the motor increases to 0.6 amps (say due to changes in the shape of the component increasing the depth of cut) then the current through R1 would appear (see later clarification) to increase to 1.1 amps. This would increase the volt dropped by R1 proportionately, that is from 9 to 9.9 volts and the voltage at the motor therefor drop to 2.1 volts. The speed dropping as a result.