1. The problem statement, all variables and given/known data Your task is to design a "dummy" indicator light for your car. The cheapest bulb available is a #47 incandescent lamp rated at 6.3V for 150 mA. The problem is, your car produces 13.5 V when it's on. Choose the nearest 10% resistor that will reduce the voltage across the bulb to 6.3V. 2. Relevant equations V = IR 3. The attempt at a solution I modelled the bulb as a resistor: 6.3 V = .150 A * R Which gives R = 42 [itex]\Omega[/itex]. From there, I plugged this value into a new circuit with a 13.5 V source instead of a 6.3 V source, and added the unknown resistor to the equation: 13.5 V = .150 A * R + 6.3 V 7.2 V = .150 A * R R = 48 [itex]\Omega[/itex]. The nearest 10% resistor value is 47 [itex]\Omega[/itex], but this wouldn't reduce the voltage across the bulb to 6.3 V. The next closest 10% resistor is 56 [itex]\Omega[/itex]. It just seems that this value is too high ... Did I follow this process correctly? What would you choose?
48 ohms is the theoretical value, but I have to choose a 10% resistor to use. When I use the 47 ohm resistor, I end up with 6.37 V across the bulb.
Close enough! [edit: To the letter of the question the solution does not exist. To the intent of the question, 47-ohms is a good answer, if you stipulate your voltage deviation, you should be fine.]
With a 47Ω resistor, the bulb will have approximately 6.37 volts across it. Light bulbs are not very picky, so this will work fine. With the 56Ω resistor, the bulb will only have about 5.79 volts across it. Still, a workable voltage, but it's the current that we're really concerned with... The 47Ω resistor will limit the current through the bulb to about 151.69 mA which is just over the design rating. The bulb will burn slightly brighter (although probably not noticeably). The 56Ω resistor will limit the current down to about 137.76 mA and the bulb will burn slightly dimmer. You're only looking at about an 8% loss of current though, so either will work. I'd use the 47Ω resistor.
With ±10% resistors there could be a resistor that is 42.3Ω and car battery voltage can spike into the 14.0 volt range. That would cause 166 ma to flow through the bulb or about 10.7% more. Since power is I²R, there can be an extra 23% power dissipated through the bulb which will shorten its life. IMO the design would have had some breathing room built in if you went with the 56Ω bulb.