Need help with question 2 1. Let’s say that you’d like to augment the dash of yo

Question

Need help with question 2

1. Let’s say that you’d like to augment the dash of your 1974 Dodge Monaco with a nifty pilot light to indicate that the automatic seat warmer is working. The problem is that your car’s electrical system operates nominally at 13.5 V, but the only lamp assembly you have on hand is a #47 incandescent bulb rated at 6.3 V. It looks as though you will have to use a dropping resistor to get the right voltage. Let’s pretend that the filament of an incandescent light bulb behaves as a resistor, even though this isn’t precisely true. According to the data sheet, the #47 lamp draws 150 mA at 6.3 V. Draw the circuit, determine the proper dropping resistor, select the nearest 10% resistor value, and determine its required power rating.

Hints:
• Since the lifespan of an incandescent light bulb is inversely proportional to the power dissipated in its filament, take this into consideration when selecting your 10% resistor value.
• When calculating power dissipation of the resistor, be sure to recalculate operating parameters with your actual chosen resistor, not the theoretical one, and choose a commercially available power rating.

2. With the information in number 1. Well, let’s suppose you would like this lamp to have a soft turn-on characteristic, as opposed to slamming full-blast when you throw the switch. One way to accomplish this is to connect a capacitor across the light bulb. Determine the correct capacitance value and voltage rating if you’d like the lamp to reach 50% of its final voltage in one second. Select a component that is commercially available, and do a little research to determine a typical retail price for the part. Draw the circuit.

Hint:
• When determining the voltage rating of the capacitor, consider what
might happen if the light bulb burns out, which is inevitable.

Explanation / Answer

1 question per post please.

1) OK, so you have 13.5V., but the bulb requires 6.3V. So the first thing is see how much voltage has to be dropped in the series resistor.
13.5 - 6.3 = 7.2V.
You have the current the bulb requires@ 150mA. Calculate the resistor value.
R = E/I = 7.2/ 0.15 = 48 ohms. The nearest 10% range resistor value to that is 47 ohms. Remember, it could be 10% either way in actual value.
The power rating is W = E x I, which is 7.2 x 0.15 = 1.08 watts. A 2W resistor would be used, as a 1W is too small and would overheat.

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