# Voltage Drop Across Indicator Lamp

1. Jan 8, 2008

### Lhawx

Hi,
I am new here so please bare with me. I know a bit about electrical and Electronics but I do not recall the formula for this application......it is a long question..please help! Here it is.

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?
answer: 1.5 + 1.5 +1.5 + 1.5 + 1.5 + 1.5 = 9 volts. Since the cells are connected in series, the amps would remain the same at 1 amp.

b. What is the total resistance resulting from the 600ft of AWG 18 wire?
answer: 6.5100 ohms x .6 = 30.906 ohms

c. What is the voltage drop across the total length of wire?
answer: I do not remember hoe to calculate!

d. What is the voltage drop across the indicator lamp?
answer: I do not know how to calculate this!

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

Thanks
Lisa

2. Jan 8, 2008

### stewartcs

Hi Lisa,

a. If the cells are connected in Series Aiding, then 9 volts is correct (your problem doesn't indicate whether they are aiding or opposing so it is probably ok to assume aiding). The 1 amp is the rating of the batteries, not necessarily how much current they are outputting. The load on them will determine how much current is flowing.

b. The typical resistance per foot of 18 AWG is typically around 0.00751 for copper. So 0.00751 x 600 = 4.51 Ohms. Were you given another value for the per foot resistance of copper?

c. First you need to find the total current flowing in the circuit with Ohm's Law and the total resistance in the circuit. I = V/R. Once you have the total current (9 volts / Req Ohms), you can rearrange Ohm's Law to find the voltage drop in the wire (V = IR) with R being the resistance found in part b above.

d. Same as part c except you are given the resistance for the lamp (i.e. 50 ohms).

CS