Use a series voltage dropping resistor

• 1). Use the Ohms Law to calculate the "load current" in amperes (load amps = watts/volts). Load current = 6/6 = 1 ampere
  
  
• 2). Calculate the resistance of the "series voltage dropping resistor." R = E/I where: R = resistance in ohms, E = voltage, and I = load current in amperes. Therefore R = 6/1 = 6 ohms.
  
  
• 3). Calculate the power dissipation rating of the power resistor and add a 25 percent safety factor. P = 1.25(I)(E) = 1.25(1)(6) = 7.5 watts. Select the next closest standard resistor wattage rating value, which is 10 watts.
  
  
• 4). Connect one end of the voltage dropping resistor to the negative battery terminal using a jumper. Connect the other end of the voltage dropping resistor to one of the terminals on the sealed beam lamp with a jumper. Complete the circuit by connecting the other terminal on the lamp to the positive battery terminal. The light will illuminate.
  
  
• 5). Check the circuit voltages. Place the voltmeter probes across the resistor. The meter will indicate 6 volts. Place the probes across the terminals on the lamp. The meter will read 6 volts.
  
  

Constructing a voltage divider using two fixed value resistors(R1 and R2)

• 1). Calculate the "bleeder current." The bleeder current is the current flowing in the voltage divider network, none of which flows through the load. The rule of thumb for designing a voltage divider is to make the bleeder current 10 percent of the load current. Our load current is 1 amp, therefore our bleeder current equals 0.1 amps. Calculate the total resistance for the voltage divider network. Total resistance equals the source voltage divided by the bleeder current. R total = E source/I bleeder = 12/0.1 = 120 ohms.
  
  
• 2). Calculate the current flowing through R1. The current flowing through this resistor will be equal to the sum of the "bleeder current" plus the "load current." I R1 = I bleeder + I load = 0.1 + 1.0 = 1.1 amps.
  Calculate the resistance value of R1. R1 = 6/1.1 = 5.4545 ohms. In this case we would round down to 5 ohms, which would provide 5.995 volts to our load. That is close enough to the 6 volt rating of the load. Calculate the power rating for R1. P = 1.25(1.1)(6) = 8.25 Watts. Use the next closest value, or 10 watts.
  
  
• 3). Calculate the resistance of R2. R2 = RT - R1 = 120 - 5.45 = 114.55. In this case we'll round up to 115 ohms. Calculate the power dissipation rating for R2. P = 1.25(0.1)(6) = 0.75 = 1 watt.
  
  
• 4). Connect R1 and R2 in series using a jumper lead. Connect this series circuit between the positive and negative terminals of the battery. Using two more jumpers connect the seal beam lamp across R1. The lamp will illuminate.
  
  
• 5). Connect the DMM across both resistors and it will read 12 volts. Connect the volt meter across R1 or R2 and it will read 6.0 volts when the load is attached.