PRACTICAL APPLICATION OF VOLTAGE DIVIDERS In actual practice the computed value of the bleeder resistor does not always come out to an even value. Since the ruleofthumb for bleeder current is only an estimated value, the bleeder resistor can be of a value close to the computed value. (If the computed value of the resistance were 510 ohms, a 500ohm resistor could be used.) Once the actual value of the bleeder resistor is selected, the bleeder current must be recomputed. The voltage developed by the bleeder resistor must be equal to the voltage requirement of the load in parallel with the bleeder resistor.
The value of the remaining resistors in the voltage divider is computed from the current through the remaining resistors and the voltage across them. These values must be used to provide the required voltage and current to the loads.
If the computed values for the divider resistors are not even values; series, parallel, or seriesparallel networks can be used to provide the required resistance.
Example: A voltage divider is required to supply two loads from a 190.5 volts source. Load 1 requires +45 volts and 210 milliamps; load 2 requires +165 volts and 100 milliamps.
Calculate the bleeder current using the ruleofthumb.
Given:
Solution:
Calculate the ohmic value of the bleeder resistor.
Given:
Solution:
Since it would be difficult to find a resistor of 1451.6 ohms, a practical choice for R_{1} is 1500 ohms.
Calculate the actual bleeder current using the selected value for R_{1}.
Given:
Solution:
Using this value for I_{R1}, calculate the resistance needed for the next divider resistor. The current (I_{R2}) is equal to the bleeder current plus the current used by load 1.
Given:
Solution:
The voltage across R_{2} (E_{R2}) is equal to the difference between the voltage requirements of loads 2 and1, or 120 volts.
Calculate the value of R_{2}.
Given:
Solution:
The value of the final divider resistor is calculated with I_{R3} (I_{R2} + I load 2) equal to 340 mA and E_{ R3} (E_{s}  E load 2) equal to 25.5V.
Given:
Solution:
A 75ohm resistor may not be easily obtainable, so a network of resistors equal to 75 ohms can be used in place of R_{3}.
Any combination of resistor values adding up to 75 ohms could be placed in series to develop the required network. For example, if you had two 37.5ohm resistors, you could connect them in series to get a network of 75 ohms. One 50ohm and one 25ohm resistor or seven 10ohm and one 5ohm resistor could also be used.
A parallel network could be constructed from two 150ohm resistors or three 225ohm resistors. Either of these parallel networks would also be a network of 75 ohms.
The network used in this example will be a seriesparallel network using three 50ohm resistors.
With the information given, you should be able to draw this voltage divider network.
Once the values for the various divider resistors have been selected, you can compute the power used by each resistor using the methods previously explained. When the power used by each resistor is known, the wattage rating required of each resistor determines the physical size and type needed for the circuit. This circuit is shown in figure 367.
Figure 367.  Practical example of a voltage divider.
In figure 367, why is the value of R_{1} calculated first?
In figure 367, how is (a) the current through R_{2} and (b) the voltage drop across R_{2} computed?
In figure 367, what is the power dissipated in R1?
In figure 367, what is the purpose of the seriesparallel network R_{3}, R_{ 4}, and R_{5}?
In figure 367, what should be the minimum wattage ratings of R_{3}and R_{5 }?
If the load requirement consists of both positive and negative voltages, what technique is used in the voltage divider to supply the loads from a single voltage source?

