# Calculate resistivity of the material of the wire

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

$$R = \frac{Pl}{A}$$

$$P = IV$$

$$P = 0.75 \times 1.5$$

$$P = 1.125$$

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

$$R = 2.3 \times 10^7$$

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Hootenanny
Staff Emeritus
Gold Member
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.
$$P = IV$$

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?:uhh:

Hootenanny
Staff Emeritus
Gold Member
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?:uhh:
That's what I get