Calculate resistivity of the material of the wire

[looi76]

[SOLVED] Calculate resistivity of the material of the wire

Question:
0.75A Current flows through an iron wire when a battery of 1.5V us connected across its ends. The length of the wire is 5m and Area is 2.5 \times 10^{-7} m^2. Calculate resistivity of the material of the wire.

Equation:
R = \frac{Pl}{A}

Attempt:
I = 0.75A \ , \ V = 1.5V \ , \ l = 5m \ , \ A = 2.5 \times 10^{-7} m^2

R = \frac{Pl}{A}

P = IV

P = 0.75 \times 1.5

P = 1.125

R = \frac{1.125 \times 5}{2.5 \times 10^{-7}}

R = 2.3 \times 10^7

Is my answer correct?

 

[Hootenanny]

You have your P's mixed up

The P here represents the resistivity of the material

R = \frac{Pl}{A}

Whereas the P here represents the power dissapated by the wire.

P = IV

 

[looi76]

Thanks Hootenanny!

I = 0.75A \ , \ V = 1.5V \ , \ l = 5m \ , \ A = 2.5 \times 10^{-7} m^2

R = \frac{V}{I} = \frac{1.5}{0.75} = 2\Omega

R = \frac{Pl}{A}

2\Omega = \frac{P \times 5}{2.5 \times 10^7 \Omega{m}}

P = \frac{2 \times 2.5 \times 10^{-7}}{5}

P = 1 \times 10^{-7} \Omega{m}

is this correct?

 

[Hootenanny]

Thanks Hootenanny!

I = 0.75A \ , \ V = 1.5V \ , \ l = 5m \ , \ A = 2.5 \times 10^{-7} m^2

R = \frac{V}{I} = \frac{1.5}{0.75} = 2\Omega

R = \frac{Pl}{A}

2\Omega = \frac{P \times 5}{2.5 \times 10^7 \Omega{m}}

P = \frac{2 \times 2.5 \times 10^{-7}}{5}

P = 1 \times 10^{-7} \Omega{m}

is this correct?

That's what I get