Question

A certain copper wire has a resistance of 19.0 Ω. At some point along its length, the wire was cut so that the resistance of one piece is 6.0 times the resistance of the other. What is the resistance of the piece with the higher resistance?

1) 3.17 Ω
2) 4.75 Ω
3) 11.40 Ω
4) 114.00 Ω

Answer 1

The resistance of the piece with the higher resistance is 114.0 Ω.

Step-by-step explanation:

Let's call the resistance of the piece with lower resistance R1 and the resistance of the piece with higher resistance R2. We are given that R1 = 6.0 times R2. Let's substitute this into Ohm's law:

R = V/I

Where R is the resistance, V is the voltage, and I is the current. Since the wire is the same material, we can assume the resistivity is constant. Therefore, we can write:

R1 = (V1/I1) = ρL1/A and R2 = (V2/I2) = ρL2/A

Where ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area. Since R1 = 6.0 times R2, we can write:

ρL1/A = 6.0 * ρL2/A

L1 = 6.0 * L2

Let's call the length of the piece with lower resistance L1 and the length of the piece with higher resistance L2. Since the total length of the wire is the sum of the lengths of the two pieces, we have:

L1 + L2 = L

6.0 * L2 + L2 = L

7.0 * L2 = L

L2 = L/7.0 and L1 = 6.0 * L/7.0

Now that we have the lengths of the two pieces, we can calculate their resistances:

R1 = ρL1/A = (ρ * 6.0 * L/7.0) / A = 6.0 * (ρL/A) = 6.0 * R2

R2 = ρL2/A

Since R2 is given to be 19.0 Ω, we can solve for R1:

R1 = 6.0 * 19.0 Ω = 114.0 Ω

Therefore, the resistance of the piece with the higher resistance is 114.0 Ω (option 4).