# Homework Help: Measuring obstacle size in a parallel circuit of different lengths of wire

1. Oct 13, 2012

### utd223

1. The problem statement, all variables and given/known data

In lab, we were instructed to create a circuit that uses nichrome wires and identical bulbs to measure the obstacle caused by an object. To do this, we slid the alligator clip along the nichrome wire until the 2 bulbs in parallel are the same brightness. The object that we measured the obstacle of is actually another nichrome wire, of different length than the first nichrome wire. These wires are in parallel with each other. I will call the first nichrome wire L1 and the second L2.

If the number of pieces of wire (let's label this N) present an obstacle L then we found that putting the N pieces in parallel gave an obstacle of size Leq = L/N. This expression needs to be generalized to the case of pieces of wire of different length in parallel.

1) Write an expression for the number of wires of length L cm that (when put in parallel) would present the same obstacle as a wire of length L1. Call this number of wires, N1 and give your answer in terms of L and L1. So, N1 = ?

2) Similarly do the same to find N2, using terms of L and L2. N2 = ?

3) Now suppose N1 + N2 wires of Length L cm are put in parallel. What size of obstacle (called Leq) is presented by them? (give your answer in terms of L, N1, and N2.)

4) Use your previous answers to write your Leq in terms of L, L1, and L2. Then simplify so that your answer is in terms of L1 and L2.

2. Relevant equations

Leq = L/N

3. The attempt at a solution

My physics lab frustrates me, because we do not use the actual terms to describe what happens in circuits. I believe "obstacle" is another way of saying resistance, and so I assume the answer to 4 should go along the lines of (1/Rtot) = (1/R1) + (1/R2). However, I can't simply write this down. I tried to work backwards from this equation to get the answers for 1-3, but I don't know how to go about it. Here are my attempts:

1) since Leq = L/N, I thought that L1 = L/N1, making N1 = L/L1
2) Similarly, N2 = L/L2
3) Here, I got stuck, because if I follow the same route, I end up with Leq = L / (N1 + N2), and manipulating this equation to get to number 4 seemed impossible.

I have an inkling I'm going about this the wrong way. If anyone understands the necessary way of thinking to answer these questions, please share with me, because I am absolutely lost. I really need to learn this, as the next few labs need me to be adept in understanding parallel circuits. Thank you!

2. Oct 14, 2012

### CWatters

Your approach looks ok to me.

For 4).. What happens if you take your equation Leq = L / (N1 + N2) and substitute N1 = L/L1 and N2 = L/L2 ?

PS I agree with you. The use of the word obstacle is a bit naff.

3. Oct 14, 2012

### utd223

Ah, ok. I figured it out. My final answer ended up being 1/Leq = (L1+L2) / L1L2, which will simplify to 1/Leq = 1/L1 + 1/L2. Thanks! I guess I just needed someone to confirm I was on the right track.