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Six cells are connected in series to form a battery. Each cell has a rating of 1.5 volts and 1 amp. AWG 18-gauge wire joins the positive terminal of the battery formed by the series of cells to an indicator lamp 300ft away. Another 300ft length of AWG 18 wire runs from the lamp back to the battery's negative pole. The lamp acts as a 50 ohm resistor.

a. What are the total voltage and amperage supplied by the battery?

2. Relevant equations

1.5 + 1.5 +1.5 + 1.5 + 1.5 + 1.5 = 9 volts.

3. The attempt at a solution

Since the cells are connected in series, the amps would remain the same at 1 amp and 9 volts

2. Relevant equations

b. What is the total resistance resulting from the 600ft of AWG 18 wire?

6.3. The attempt at a solution

5100 ohms x .6 = 30.906 ohms

2. Relevant equations

c. What is the voltage drop across the total length of wire?

6.3. The attempt at a solution

I do not remember how to calculate!

2. Relevant equations

d. What is the voltage drop across the indicator lamp?

6.3. The attempt at a solution

I do not know how to calculate this!

I know this is long but could someone please check this for me?

Thanks

Lisa

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# Voltage Drop

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