Your task is to design a “dummy” indicator light for your car. The cheapest bulb
ID: 1924504 • Letter: Y
Question
Your task is to design a “dummy” indicator light for your car. The cheapest bulbavailable is a #47 incandescent lamp, rated for 6.3 V @ 150 mA. The problem is that
your car battery generates 13.5 V when your car is. Select the appropriate resistor to add
to the circuit in order to reduce the voltage across the bulb to the appropriate level.
Choose the nearest 10-% resistor value and determine what its power rating should be.
The lifespan of an incandescent lamp is inversely proportional to the voltage applied to it,
so round accordingly. Hint: be sure to recalculate the actual operating current with your
selected resistor before determining its required power rating. Draw the circuit.
Explanation / Answer
With a 47O resistor, the bulb will have approximately 6.37 volts across it. Light bulbs are not very picky, so this will work fine. With the 56O 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 47O 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 56O 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 47O resistor. 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 O. 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 O. The nearest 10% resistor value is 47 O, but this wouldn't reduce the voltage across the bulb to 6.3 V.
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