You wish to power 2 loads from a 12V dc source. Load one; 7V and 3 mA (LED). Load two; 5V and 1 mA (LED). Use the 10 percent rule to construct the voltage divider.
Answer:
1st calculate the bleeder current. It is 10% of the total load current.
total load current = 3mA + 1mA = 4 mA
bleed current is thus = 0.10 x (3 + 1) = 0.4 mA
This is the current through the third resistor; the last bleeder. We can call this I-bleed or I-3 .
The, voltage across R3 is the 5 volts of load 2. So, using Ohm's law; R = V / I = 5 V / .0004 A = 12.5 k ohms.
Now calculate the current through R2. It is the current through load 2 plus the current through R3. We have the current through R3 from the prior calculation. It equals 0.0004 A. The current through load 2 is 1mA. Thus the total current through R2 is 0.0014 A.
The voltage across R2 to the next divider is equal to load 1 voltage drop which is 7 volts, less the load 2 voltage drop which is 5 volts. Answer = 2 volts.
Using ohm's law then, R2 = 2V / 0.0014 A = 1.43 k ohms.
Lastly, we calculate R1. The current through it is the current through load 1 plus the current through R2. That is; 3 mA + 0.0014A = 0.0044A.
Now R1 = V / I = (12 - 7) V / 0.0044 A = 1.13 k ohms.
