I ONLY CAN NOT DO PART D. I GOT A,B AND C. Consider the problem of using a low-v

Question

I ONLY CAN NOT DO PART D. I GOT A,B AND C.

Consider the problem of using a low-voltage system to power a small cabin. Suppose a 12-V system powers a pair of 100-W lightbulbs (wired in parallel). <?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /?>

a. What would be the (filament) resistance of a bulb designed to use 100 W

when it receives 12 V?

b. What would be the current drawn by two such bulbs if each receives a full 12 V?

c. What gage wire should be used if it is the minimum size that will carry the current.

d. Suppose a 12-V battery located 80-ft away supplies current to the pair of bulbs through the wire you picked in (c). Find:

1. The equivalent resistance of the two bulbs plus the wire resistance to and from the battery.

2. Current delivered by the battery

3. The actual voltage across the bulbs

4. The power lost in the wires

5. The power delivered to the bulbs

6. The fraction of the power delivered by the battery that is lost in the wires.

THANKS!!!!

Explanation / Answer

current required is 200/12 = 16.6667A

thus gauge is 16 with resistance4.016 ohms/ 1000 feet

resistance of bulbs = 12*12/2*100 = 0.72 ohms

resistance of wire = 4.016*80*2/1000 = 0.64256 ohms

thus total resistance=1.36256 ohms

current delivered = 12/1.36256= 8.80695 A

actual voltage = 8.80695*0.72=6.341004 V

power of bulbs= 6.341004*8.80695=55.8449051778 W

thus power lost = 12* 8.80695 - 55.8449051778= 49.8384948222 W

thus fraction delivered = 55.8449051778/(12* 8.80695) * 100 = 52.8417%

change gauge ans resistance if it is diffenent than the AWG system

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