Measuring obstacle size in a parallel circuit of different lengths of wire

In summary, the conversation discusses a lab activity in which students were instructed to create a circuit using nichrome wires and bulbs to measure the obstacle caused by an object. The number of pieces of wire in parallel is found to be inversely proportional to the obstacle size, and this concept needs to be generalized for wires of different lengths. The conversation then goes on to discuss finding the number of wires needed for specific obstacle sizes and simplifying the final expression.
  • #1
utd223
4
0

Homework Statement



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.

Homework Equations



Leq = L/N


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!
 
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  • #2
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
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.
 

What is a parallel circuit?

A parallel circuit is a type of electrical circuit in which the components are connected in multiple branches, allowing current to flow through different paths simultaneously.

Why do we need to measure obstacle size in a parallel circuit?

Measuring obstacle size in a parallel circuit is important because it helps us understand how the different components in the circuit affect the flow of current and how we can optimize the circuit for better performance.

What is the role of wire length in measuring obstacle size in a parallel circuit?

The length of wire in a parallel circuit affects the resistance and thus the flow of current. By measuring obstacle size in a parallel circuit of different wire lengths, we can understand how resistance changes and how it impacts the overall circuit.

How do we measure obstacle size in a parallel circuit?

To measure obstacle size in a parallel circuit, we can use a multimeter to measure the resistance of the circuit at different points. This will give us an idea of how the size of the obstacle affects the flow of current.

What are the potential applications of measuring obstacle size in a parallel circuit?

Measuring obstacle size in a parallel circuit can have various applications, such as designing more efficient circuits, troubleshooting faulty circuits, and understanding the impact of different components on the overall performance of the circuit.

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