Solving Series-Connected Cells & AWG 18 Wire Circuit

[Lhawx]

1. Homework Statement [

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?

Homework Equations

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

The Attempt at a Solution

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

Homework Equations

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

6.

The Attempt at a Solution

5100 ohms x .6 = 30.906 ohms

Homework Equations

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

6.

The Attempt at a Solution

I do not remember how to calculate!

Homework Equations

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

6.

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

 

[mgb_phys]

Your voltage and current reasoning is correct.
I don't know the wire standard off-hand but I assume you got the number correct.

For the voltage drop it's just Ohm's law, V = I R
Draw the wire, bulb and return wire as three resistors.
The total voltage drop across them must equal the voltage of the battery.
You can then work out the current through the total resistance and so the voltage drop across each.

Note that the 1Amp label on the cells is the maximum it can supply - it does not mean 1A is necessarily flowing in the circuit,

 

[The Electrician]

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

You got the voltage right. But the amperage actually supplied will most likely be different that the 1 amp rating, which is the amperage the battery is CAPABLE of supplying. How much it ACTUALLY supplies will depend on the load, which in this case is a lamp plus some resistance in the wire. You will have to work out the next two parts of the problem to find the amperage supplied to the load.

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

You must look up the resistance per foot of 18 gauge wire from a handbook or the web.
The relevant number is .006384 ohms/foot (you may find slightly different values; don't worry about this. Just use the number you find, or the number I've given). Since you have a total of 600 feet of wire, the resistance of the wire is .006384 * 600 = 3.83 ohms.

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

The lamp has a resistance of 50 ohms, so the total resistance of the load on the battery is 50 + 3.83 = 53.83 ohms. This means the current drawn from the battery is 9 volts / 53.83 ohms = .1672 amps. The current of .1672 amps is present in the full length of wire, so the voltage drop across the total length of wire will be the current in the wire times the resistance of the total length of wire: Vdrop = .1672 * 3.83 = .64 volts.

You need to be more careful with your arithmetic. Above you gave as a sample of your work:

"6.3. The Attempt at a Solution
5100 ohms x .6 = 30.906 ohms"

5100 x .6 does not equal 30.906; it equals 3060. Be careful. There's no sense in losing points for such a mistake in this the age of electronic calculators!

 

[Lhawx]

Thanks a lot for the help.

This was a mistype "5100 ohms x .6 = 30.906 ohms"

It should have been " 6.5100 ohms * .6 = 3.906 ohms